Vector Calculus - Mathematicscrhumber/ma213/vec_book.pdf · Vector Calculus Math 213 Course Notes Cary Ross Humber November 28, 2016

Vector CalculusMath Course Notes

Cary Ross Humber

November ,

Humber ma course notes

ii

Preface

iii


iv

Contents

Preface iii

Linear Algebra Primer

Vectors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

§.. Vector Operations . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

§.. Some Geometric Concepts . . . . . . . . . . . . . . . . . . . . . . . .

§.. Linear Independence, Bases, other definitions . . . . . . . . . . . . .

§.. Projection . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

A little about matrices . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

§.. Determinant Formulas . . . . . . . . . . . . . . . . . . . . . . . . . .

§.. Determinant Geometry . . . . . . . . . . . . . . . . . . . . . . . . . .

§.. Cross product, triple product . . . . . . . . . . . . . . . . . . . . . .

§.. Lines and Planes . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Multivariable functions

§.. Level Sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

§.. Sections . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Derivatives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

§.. What’s wrong with partial derivatives? . . . . . . . . . . . . . . . . .

§.. Directional derivatives . . . . . . . . . . . . . . . . . . . . . . . . . .

v


Tangent vectors and planes . . . . . . . . . . . . . . . . . . . . . . . . . . . .

§.. Surfaces in R3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Coordinates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Parametric Curves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Parametric surfaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Practice problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Exterior Forms

Constant Forms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

§.. 1-Forms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

2-Forms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

§.. Wedge (Exterior) product . . . . . . . . . . . . . . . . . . . . . . . .

k-forms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

§.. Wedge product again . . . . . . . . . . . . . . . . . . . . . . . . . . .

Vector fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Differential forms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

§.. Exterior derivative . . . . . . . . . . . . . . . . . . . . . . . . . . . .

§.. Pullbacks/Change of coordinates . . . . . . . . . . . . . . . . . . . .


Integration and the fundamental correspondence

The correspondence between vector fields and differential forms . . . . . .

Flux integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

§.. Line integrals and work . . . . . . . . . . . . . . . . . . . . . . . . .

§.. Orientations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

§.. Integration of 3-forms . . . . . . . . . . . . . . . . . . . . . . . . . .


vi


Stokes’ Theorem

Surfaces with boundary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

The generalized Stokes’ Theorem . . . . . . . . . . . . . . . . . . . . . . . .

§.. Stokes’ Theorem for 1-surfaces . . . . . . . . . . . . . . . . . . . . .



A Coordinate representations

B Some applications of differential forms and vector calculus

Extreme values . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

§.. Constrained Extrema . . . . . . . . . . . . . . . . . . . . . . . . . . .

Maxwell’s equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

vii


viii

Chapter

Linear Algebra Primer

§ Vectors

The majority of our calculus will take place in -dimensional and -dimensional space.Occasionally, we may work in higher dimensions. For our purposes, a vector is like apoint in space, along with a direction. Other information, such as magnitude or lengthof a vector, can be determined from this point and direction. We visualize a vector as anarrow emanating from the origin, which we often denote as O, and ending at this point.The space (so called vector space)

R2 = {(x1,x2) |x1,x2 ∈R}

consists of pairs of real numbers. Such a pair, which we often denote by a single letter(bold, hatted, arrow on top), is a vector in R

2. The convention taken for these notes is todenote vectors by bold letters. It is typical to express a vector x in column form

x =(x1x2

)

on a chalkboard/whiteboard, or whenever space is not a concern. Whenever space is at apremium, it is just as typical to denote the same vector x in row form x = (x1,x2).

The space R3 consists of 3-tuples of real numbers, or real 3-component vectors. Just as

with R2, we can express R3 as the set

R3 = {(x1,x2,x3) |x1,x2,x3 ∈R}.

Each vector x ∈R3 consists of three components, each of which is a real number.


In general, the space Rn consists of n-tuples of real numbers, or real n-component

vectors,Rn = {(x1, . . . ,xn) |xj ∈R, j = 1, . . . ,n}.

The higher the dimension, the more space is preserved by using row form x = (x1, . . . ,xn).

§.. Vector Operations

There are two basic vector operations, that of vector addition and scalar multiplication.Both operations are defined component-wise. Given two vectors a,b ∈Rn with componentforms a = (a1, a2, . . . , an) and b = (b1,b2, . . . , bn), the vector sum a + b is the vector obtainedby adding the components of a to those of b,

a + b = (a1 + b1, a2 + b2, . . . , an + bn).

Similarly, if α ∈R is a scalar, the scalar multiple αa is obtained by multiplying eachcomponent of a by α,

αa = (αa1,αa2, . . . ,αan).

In what follows, whether we are discussing R2,R3 or Rn, in general, we denote the zero

vector by 0, which is simply the vector with 0 in every component. With respect to vectoraddition and scalar multiplication, the following conditions are satisfied for all α,β ∈Rand a,b,c ∈Rn

V1) a + b = b + aV2) (a + b) + c = a + (b + c)V3) a + 0 = aV4) a + (−a) = 0V5) 1a = aV6) α(βa) = (αβ)aV7) (α + β)a = αa + βaV8) α(a + b) = αa +αb.

Example . Let a = (2,1) and b = (−3,4). Then,

a + b = (2− 3,1 + 4) = (−1,5).

The vector a + b is the diagonal of the parallelogram with sides a and b as depicted inFigure ..

®


x

y

a

ba+b

Figure .: The vector a + b is the diagonal of the parallelogram with sides a,b.

§.. Some Geometric Concepts

Given two vectors x = (x1, . . . ,xn) and y = (y1, . . . , yn) in Rn, we define their inner product

by

〈x,y〉 =n∑

j=1

xjyj .

The term inner product is synonymous with scalar product. If we input two vectors, theoutput is a scalar (real number). This particular inner product is often called the dotproduct in vector calculus texts. So, the same formula may be denoted

x · y =k∑

j=1

xjyj .

Soon, we will see what the inner product tells us about the geometric relationshipbetween two (or more) vectors.

Another important scalar quantity is the length or magnitude of a vector. This is a scalarassociated with a single vector, whereas the inner product is a scalar associated with twovectors. However, these quantities are related. The norm (in particular, Euclidean norm)


of the vector x ∈Rn is

‖x‖ = 〈x,x〉1/2 =

n∑

j=1

x2j

1/2

.

In other words, the quantity ‖x‖2 is the inner product of x with itself. Geometrically, thenorm ‖x‖ represents the length of x.

The Cauchy-Schwarz inequality gives another relationship between the norm and innerproduct, namely

|〈a,b〉| ≤ ‖a‖‖b‖ (.)

for any a,b ∈Rn. Though simple, the Cauchy-Schwarz inequality is very powerful.Another powerful inequality is the triangle inequality

∣∣∣∣∣‖a‖ − ‖b‖∣∣∣∣∣ ≤ ‖a±b‖ ≤ ‖a‖+ ‖b‖. (.)

Theorem . (Properties of the inner product). Let α,β be real numbers and let a,b,c ∈Rn.

. 〈αa,b〉 = α〈a,b〉. 〈a,βb〉 = β〈a,b〉. 〈a + b,c〉 = 〈a,c〉+ 〈b,c〉. 〈a,b + c〉 = 〈a,b〉+ 〈a,c〉

These properties highlight why it is often preferrable to work with inner products,rather than norms. With inner products, scalars factor out of both arguments. Incontrast, the analogous property for norms is

‖αa‖ = |α| ‖a‖,only the absolute value factors out, in general.

Since inner products have nicer properties, whenever it makes sense to do so, we willoften square norms so that they become inner products.

Definition .. A vector a ∈Rn is called a unit vector if ‖a‖ = 1. From any vector b ∈Rnwe can obtain a unit vector by normalizing it. The vector u = b/‖b‖ has norm 1.

Let a = (a1, a2),b = (b1,b2) be two vectors in R2. We want to determine an expression for

the angle, ϕ, between the vectors a and b. Let ϕa,ϕb be the angles between the positivex-axis (e1-axis) and a,b, respectively. To each vector there corresonds a right triangle,


x

y

a

a1

a2

b

b1

b2

ϕa ϕb

ϕ

Figure .: The angle ϕ between a and b

whose side lengths correspond to the components of the vector and hypotenuse is thenorm, as depicted in Figure .. This gives the following trigonometric relations

sinϕa =a2

‖a‖ , sinϕb =b2

‖b‖cosϕa =

a1

‖a‖ , cosϕb =b1

‖b‖tanϕa =

a2

a1, tanϕb =

b2

b2.

If ϕ is the angle between a and b, then ϕ = ϕa −ϕb. Thus,

cosϕ = cos(ϕa − cosϕb) = cosϕa cosϕb + sinϕa sinϕb

=a1

‖a‖b1

‖b‖ +a2

‖a‖b2

‖b‖=a1b1 + a2b2

‖a‖‖b‖ ,

where the numerator in the last expression is 〈a,b〉. Note that this analysis holds inhigher dimensions, as well. Thus, the angle ϕ between vectors a,b ∈Rn is determined by

cosϕ =〈a,b〉‖a‖‖b‖ , (.)

where ϕ ∈ (0,π). We can just as well let ϕ be in the closed interval [0,π], which includesthe possibility that a,b lie on the same line. If the angle between a and b is 0 or π, a andb are called parallel.


Proof of Cauchy-Schwarz inequality. The Cauchy-Schwarz inequality is a directconsequence of the cosine identity (.). By (.), for vectors a,b ∈Rn we have

cosϕ =〈a,b〉‖a‖‖b‖ .

Since −1 ≤ cosϕ ≤ 1, this yields

−1 ≤ 〈a,b〉‖a‖‖b‖ ≤ 1,

hence−‖a‖‖b‖ ≤ 〈a,b〉 ≤ ‖a‖‖b‖

which is equivalent to|〈a,b〉| ≤ ‖a‖‖b‖.

Definition .. Two vectors a,b ∈Rn are called orthogonal (or perpendicular) if〈a,b〉 = 0. Note that, in this case, by (.)

cosϕ =〈a,b〉‖a‖‖b‖ = 0 =⇒ ϕ =

π2.

If a and b are orthogonal, we denote this by a⊥ b.

Example . Let’s compute the angle ϕ between a = (1,1) and

b =(−1

2+

√3

2,12

+

√3

2

).

We have norms‖a‖ =

√12 + 12 =

√2

and

‖b‖ =

√(−1

2+

√3

2

)2

+(

12

+

√3

2

)2

=

√(14−√

32

+34

)+(

14

+

√3

2+

34

)

=√

2.

The inner product is

〈a,b〉 = 1(−1

2+

√3

2

)+ 1

(12

+

√3

2

)=√

3.


Thus, the angle ϕ between a and b is determined by

cosϕ =〈a,b〉‖a‖‖b‖ =

√3√

2√

2=

√3

2.

Keeping in mind that ϕ must be in the interval (0,π), we have ϕ = π6 . ®

Proposition . (Parallelogram Law).

‖a + b‖2 + ‖a−b‖2 = 2‖a‖2 + 2‖b‖2 (.)

Proof. We will give a proof of the parallelogram law using the law of cosines. For atriangle with angles A,B,C with opposing sides of length a,b,c, respectively, such as thetriangle depicted in Figure ., the law of cosines states

a2 = b2 + c2 − 2bccosA

b2 = a2 + c2 − 2accosB

c2 = a2 + b2 − 2abcosC.

(.)

The geometric relationship between the vectors a,b,a + b and a−b is depicted in Figure..

First, let ϕ denote the angle between a and a + b. By (.),

cosϕ =〈a,a + b〉‖a‖‖a + b‖ . (.)

By the law of cosines, we have

‖b‖2 = ‖a‖2 + ‖a + b‖2 − 2‖a‖‖a + b‖cosϕ

= ‖a‖2 + ‖a + b‖2 − 2‖a‖‖a + b‖( 〈a,a + b〉‖a‖‖a + b‖

)

= ‖a‖2 + ‖a + b‖2 − 2〈a,a + b〉= ‖a‖2 + ‖a + b‖2 − 2(‖a‖2 + 〈a,b〉). (.)

Thus, we have‖a‖2 + ‖b‖2 = ‖a + b‖2 − 2〈a,b〉. (.)

Now, let θ denote the angle between a and b. By (.),

cosθ =〈a,b〉‖a‖‖b‖ . (.)


c

b

a

A

B

C

Figure .: Relationship between A,B,C and a,b,c for the law of cosines.

a

b

a+b

a−b

Figure .: The parallelogram generated by a,b with diagonals a + b and a−b.

By the law of cosines, we have

‖a−b‖2 = ‖a‖2 + ‖b‖2 − 2‖a‖‖b‖cosθ

= ‖a‖2 + ‖b‖2 − 2‖a‖‖b‖( 〈a,b〉‖a‖‖b‖

)

= ‖a‖2 + ‖b‖2 − 2〈a,b〉. (.)

Thus, we have‖a‖2 + ‖b‖2 = ‖a−b‖2 + 2〈a,b〉. (.)

Adding equation (.) and (.) we arrive at the parallelogram law

2‖a‖2 + 2‖b‖2 = ‖a + b‖2 + ‖a−b‖2.

Proposition .. The vectors a,b are orthogonal if and only if

‖a + b‖2 = ‖a‖2 + ‖b‖2.


Proof. By the parallelogram law (.)

‖a + b‖2 + ‖a−b‖2 = 2‖a‖2 + 2‖b‖2.If a,b are orthogonal 〈a,b〉 = 0. If we expand the term ‖a−b‖2 in terms of innerproducts, we have

‖a−b‖2 = 〈a−b,a−b〉= 〈a,a−b〉 − 〈b,a−b〉= 〈a,a〉 − 〈a,b〉 − 〈b,a〉+ 〈b,b〉= ‖a‖2 + ‖b‖2,

since 〈a,b〉 = 0 = 〈b,a〉. Since ‖a−b‖2 = ‖a‖2 + ‖b‖2, the parallelogram law reduces to

‖a + b‖2 = ‖a‖2 + ‖b‖2.

§.. Linear Independence, Bases, other definitions

Definitions.

..) Given a collection of vectors {v1, . . . ,vk} in Rn, a linear combination of these

vectors is an expression of the form

k∑

j=1

αjvj = α1v1 + · · ·+αkvk ,

where αj ∈R for j = 1, . . . , k.

..) A collection of vectors {v1, . . . ,vk} in Rn is called linearly independent if

k∑

j=1

αjvj = 0 implies α1 = . . . = αk = 0.

Otherwise, the collection is called linearly dependent. Geometrically, any two vectorsare linearly independent if they do not lie on the same line. Suppose a,b ∈Rn arelinearly dependent. This means that we can find α,β ∈R which are not both 0 such thatαa + βb = 0. Then, by rearranging the previous equality,

a = −βα

b,


or, equivalently,b = −α

βb.

Thus, the vectors a,b are scalar multiples of each other, so they lie on the same line.

..) The span of {v1, . . . ,vk} in Rn is the set of all linear combinations of {v1, . . . ,vk},

span{v1, . . . ,vk} = {α1v1 + · · ·+αkvk |αj ∈R, j = 1, . . . , k}.We say R

n (or a subspace of Rn; see below) is spanned by {v1, . . . ,vk} if any vector x ∈Rncan be expressed as a linear combination of v1, . . . ,vk; that is,

x =k∑

j=1

αjvj

for some scalars α1, . . . ,αk ∈R.

The span of a single vector a ∈Rn is simply the (infinite) line on which a lies. Sincespan{a} = {αa |α ∈R} is the set of all scalar multiples of a and scalar multiples of a vectorlie on the same line.

Let’s look at the span of two linearly independent vectors. First, why do we want them tobe linearly independent? Well, if we take two linearly dependent vectors, they lie on thesame line. That is, if a,b ∈Rn are linearly dependent, then

span{a,b} = span{a} = span{b}.Geometrically, we do not gain any new information by considering a linearly dependentcollection. Now, if a,b ∈Rn are linearly independent, then span{a,b} generates a plane.

..) A basis for Rn is a collection of linearly independent vectors that span Rn. In

other words, a collection of vectors {v1,v2, . . . ,vn} is a basis for Rn if every vector a ∈Rn isuniquely expressible as a linear combination of the basis vectors. This means, given anyvector a ∈Rn we can find unique scalars α1,α2, . . . ,αn such that

a = α1v1 +α2v2 + · · ·+αnvn. (.)

In this case, the scalars (α1,α2, . . . ,αn) are called the coordinates of a relative to the basis{v1, . . . ,vn} (or simply coordinates).

It is a basic theorem of linear algebra that any basis for Rn consists of exactly n vectors.We often deal with the canonical basis {e1, . . . ,en}, where ej ∈Rn has a 1 in the j-thcomponent and 0 in all other components. For R2, this is {e1,e2} where e1 = (1,0) and


e2 = (0,1). For R3, the canonical basis is {e1,e2,e3}, where e1 = (1,0,0), e2 = (0,1,0) ande3 = (0,0,1). The canonical basis {e1,e2,e3} consists of unit vectors directed along the x,yand z axes, respectively. The notation {i, j,k} for the canonical basis of R3 (in that order)is quite common in vector calculus texts, as well as other texts that rely heavily on vectorcalculus (e.g. electrodynamics and other physics texts).

..) A collection of vectors {v1, . . . ,vk} in Rn is called orthonormal if 〈vi ,vj〉 = 0 for

i , j and 〈vj ,vj〉 = ‖vj‖2 = 1 for all j = 1, · · · , k. Equivalently, this collection is orthonormalif 〈vi ,vj〉 = δij , where

δij ={

1, i = j0, i , j.

In other words, an orthonormal collection is a collection of mutually orthogonal unitvectors. The canonical bases for R2 and R

3 are both orthonormal.

To any collection {v1, . . . ,vk} of vectors in Rn there is an associated n× k matrix whose

columns are the vectorsA =

(v1 v2 · · · vk

).

For instance, the canonical basis {e1,e2,e3} for R3 corresonds to the matrix

A =(e1 e2 e3

)=

1 0 00 1 00 0 1

.

Given this correspondence, numerous statements about collections of vectors can betranslated into statements about the corresponding matrix (and vice versa).

Theorem .. Let {v1, . . . ,vn} be a collection of vectors in Rn. Let A ∈Rn×n be the matrix

associated with this collectionA =

(v1 · · · vn

).

The following statements are equivalent.

. The collection {v1, . . . ,vn} is a basis for Rn.

. The matrix A is invertible.

. det(A) , 0

. For any b ∈Rn, the nonhomogeneous system Ax = b has a solution. Moreover, thesolution x ∈Rn is unique.

. The homogeneous system Ax = 0 has only the trivial solution x = 0.


§.. Projection

Suppose a,b ∈R3 are linearly independent (hence, not on the same line). Since a,b arelinearly independent, they span (or generate) a plane in R

3. Let

M = span{a,b} = {αa + βb |α,β ∈R}

denote this plane. By definition, {a,b} is a basis for M (the vectors are lin earlyindependent and span M); however, as we will see, it is often convenient to have anorthonormal basis. Another linear algebra fact, that we will not cover in detail at thispoint, is that we can always determine an orthonormal basis. Let {v1,v2} be anorthonormal basis for M, where v1 = b/‖b‖. At this point, we do not need to know what v2is; we can use the fact the {v1,v2} is orthonormal to determine v2! Since {v1,v2} is a basisfor M

a = α1v1 +α2v2, (.)

for some α1,α2 ∈R. First, take the inner product of the previous equation ( both sides)with v1. Then,

〈a,v1〉 = α1〈v1,v1〉+α1〈v2,v1〉 (.)= α1, (.)

which follows since v1 is a unit vector (‖v1‖ = 1) and v1,v2 are orthogonal to each other(〈v1,v2〉 = 0). Similarly, if we take the inner product of (.) with v2, we have

〈a,v2〉 = α1〈v1,v2〉+α2〈v2,v2〉 (.)= α2, (.)

which also follows by the orthonormality of {v1,v2}. Now, if we go back to the equationa = α1v1 +α2v2, we know the values of the scalars α1,α2 and the vector v1. We can nowsolve for v2. But first, let’s rewrite α1 as

α1 = 〈a,v1〉 = 〈a, b‖b‖〉 =

〈a,b〉‖b‖ .

Then, solving for v2, we have

v2 =1α2

a− α1

α2v1 =

a−α1v1

α2(.)

=a− 〈a,b〉‖b‖ b

‖b‖‖a− 〈a,b〉b〈b,b〉 ‖

(.)


We have done several things here. First of all, we have determined formulae for anorthonormal basis of the plane spanned by {a,b}, namely

v1 =b‖b‖ , v2 =

a− 〈a,b〉‖b‖ b‖b‖

‖a− 〈a,b〉b〈b,b〉 ‖.

Secondly, α1 is the component of the vector a parallel to the vector b, while

α1v1 =〈a,b〉〈b,b〉b (.)

is the orthogonal projection of a onto b. The orthogonal projection of a onto b may bedenoted projb a.

Example . Let a = (1,1) and b = (2,1). We have

〈a,b〉 = 1 · 2 + 1 · 1 = 3

and〈b,b〉 = 2 · 2 + 1 · 1 = 5.

So the projection of a onto b is

〈a,b〉〈b,b〉b =

35

(2,1) =(65,35

),

as depicted in Figure ..

®

The projection of one vector onto another yields another geometric interpretation of theinner product, as depicted in Figure .. It is straightforward to verify that

〈a,b〉 = ‖projb a‖‖b‖ (.)= ‖proja b‖‖a‖. (.)

§ A little about matrices

Although we will not cover the details (yet), at first glance a matrix may seem a mere“bookkeeping” strategy, yet matrices are much more. At this point, it is perfectlyacceptable to think of a matrix as a collection of numbers, organized by rows and


x

y

a b

projb a

Figure .: The orthogonal projection of a onto b.

x

y

a

b

`

Figure .: The inner product 〈a,b〉 equals ` times the length of b. The length ` is pre-cisely ‖projb a‖.

columns (as with a spreadsheet). As with vectors, we denote a matrix by a single letter,


often a capital letter. For instance,

A =

2 −37 10 −4

is what we call a real 3× 2 matrix; real since each entry is a real number and 3× 2 todenote that A has 3 rows and 2 columns. A real m×n matrix has m rows and n columns.We denote the (vector) space of all real m×n matrices by R

m×n. When speaking of anarbitrary (non-specific) matrix A, we denote the corresponding entries by the lower casecounterpart with indices. For instance, if A ∈Rm×n, we may express this in the form

A =

a11 a12 a13 · · · a1na21 a22 a23 · · · a2n...

...... · · · ...

am1 am2 am3 · · · amn

.

That is, the entry in the ith row and jth column is denoted by aij . So, the first index isfor the row and the second index for the column. If we wish to be more concise, we maywrite A = (aij)i=1,...,m

j=1,...,n. A matrix is called square if it has an equal number of columns and

rows.

§.. Determinant Formulas

As with vectors, we would like to associate to any given matrix a scalar which tells ussomething about the matrix. For vectors, we have norms and inner products which yieldinformation in the form of a scalar number. We can, in fact, define norms and innerproducts for matrices, but these do not suffice for our purposes. What we need is thedeterminant of a square matrix. The determinant of a square matrix tells us when amatrix is invertible or not (see the discussion in the previous section). More importantlyfor our purposes, the determinant yields useful geometric information. Arguably, thedeterminant of a square 2× 2 matrix is the most important, as the formula for largermatrices relies on the 2× 2 case. So, we’ll start with 2× 2. An arbitrary real 2× 2 matrixtakes the form

A =(a bc d

),

where a,b,c,d ∈R. In this case, the determinant of A, denoted det(A) is

det(A) = ad − bc. (.)

For example, if

A =(2 −31 7

),


then det(A) = 2(7)− (−3)1 = 17.

Now, suppose A ∈R3×3 is the matrix

A =

a11 a12 a13a21 a22 a23a31 a32 a33

.

Then, the determinant of A is given by

det(A) = a11 det(a22 a23a32 a33

)− a12 det

(a21 a23a31 a33

)+ a13 det

(a21 a22a31 a32

)

= a11(a22a33 − a32a23)− a12(a21a33 − a31a23) + a13(a21a32 − a31a22).(.)

Let’s take a closer look at (.)

det(A) = a11 det

a11 a12 a13a21 a22 a23a31 a32 a33

− a12 det

a11 a12 a13a21 a22 a23a31 a32 a33

+ a13 det

a11 a12 a13a21 a22 a23a31 a32 a33

(.)

Formula (.) computes the determinant of A by expansion along the first row.Notice that there are three terms, each one corresponding to an entry in the first row ofA. For the first entry in row one, a11, we remove row 1 and column 1 to obtain thesubmatrix (

a22 a23a32 a33

).

Then, we multiply a11 by the determinant of this submatrix. For the next entry in rowone, a12, we remove row 1 and column 2 to obtain the submatrix

(a21 a23a31 a33

).

Now, we multiply a12 by the determinant of this submatrix. However, this time there is anegative thrown in front. Why?...

More importantly, what we should see is that the the indices for each aij coefficient tellus which row and column to remove.

So, for the last term in (.) corresponding to a13, we remove row 1 and column 3 toobtain (

a21 a22a31 a32

).

Again, we compute the determinant of this 2× 2 submatrix, then multiply by thecorresponding coefficient a13. Notice that this term does not have a negative in front.


Working our way from left to right along the first row, the sign in front of each termalternates between + and −. If there were more terms in the determinant formula, thisalternating behavior would continue. Indeed, this must be taken into account whencomputing the determinant of a matrix larger than 3× 3. However, knowing thedeterminant formula for a matrix no larger than 3× 3 is sufficient for our needs.

Example . Let

A =

1 2 80 1 −3−1 3 5

.

We will compute det(A) by (.). We have

det(A) = 1det(1 −33 5

)− 2det

(0 −3−1 5

)+ 8det

(0 1−1 3

)

= 1(5− (−9))− 2(0− 3) + 8(0− (−1))= 28.

®

§.. Determinant Geometry

So, what can the determinant tell us geometrically? We will first show that if a,b ∈R2,then the determinant of the matrix whose columns are a,b is the area of theparallelogram they generate. First, let’s assume one of the vectors is directed along acoordinate axis, namely, let a = (a1,0) and b = (b1,b2). Moreover, let’s assume a1,b1,b2 areall positive. We will compute the area of the parallelogram depicted in Figure ..

Without referring to a known area formula, there are several ways to determine the areaof the shaded region. One way is to decompose the parallelogram into a smallerrectangle and two right triangles (however, this only works if b1 < a1). One can alsodetermine the area of a larger rectangle and subtract the area of two right triangles,which works no matter how a1 and b1 compare. Using this method, as depicted in Figure., we find that the area of the large rectangle is (a1 + b1)b2, while each right trianglehas area 1

2b1b2. Thus, the area of the parallelogram is (a1 + b1)b2 − b1b2 = a1b2.

Of course, in general, such a parallelogram may be in any quadrant, such as theparallelogram depicted in Figure .. You should convince yourself that if the vectorsa,b were rotated so that a is on the positive x-axis, the area of the parallelogram wouldnot change. It turns out that if a = (a1, a2) and b = (b1,b2), the area of this parallelogram


x

y

a

b

Figure .: Any two linearly independent vectors a,b ∈R2 generate a parallelogram.

x

y

a1 + b1

b2

b1

Figure .: The area of the parallelogram can be computed by subtracting the area of theshaded triangular regions from the rectangle of area (a1 + b1)b2.

is given by ∣∣∣∣∣∣det(a1 b1a2 b2

)∣∣∣∣∣∣ = |a1b2 − a2b1|.


x

y

a

b

Figure .: An arbitrary parallelogram in the plane, generated by two linearly indepen-dent vectors a,b ∈R2.


Note that the absolute value is necessary, so that a positive value is yielded for the area,regardless of the sign of the vector components.

§.. Cross product, triple product

One of the primary reasons we need matrices is to define the determinant, which inturn, we need to define the cross product of vectors.

Definition .. Given two vectors a = (a1, a2, a3), b = (b1,b2,b3) in R3, we define the

cross product of a and b as the vector

a×b = (a2b3 − a3b2, a3b1 − a1b3, a1b2 − a2b1). (.)

We will see later in the semester, after introducing differential forms, that the crossproduct is a particular type of wedge product. For this reason, the cross product may alsobe denoted a∧b

A convenient method for computing the cross product is to compute

a×b = det

e1 e2 e3a1 a2 a3b1 b2 b3

. (.)

Caution must be exercised when using formula (.)! When expanding thedeterminant, the e1,e2,e3 are symbolic, in a sense; these vectors tell us which componentthe resulting scalars belong to.

Example . Let a = (1,−2,1) and b = (2,1,1). To compute the cross product a×b, wecompute

det

e1 e2 e31 −2 12 1 1

= (−2− 1)e1 − (1− 2)e2 + (1− (−4))e3

= −3e1 + e2 + 5e3

= (−3,1,5).

® The following properties of the cross product hold,

C1) b× a = −(a×b)C2) 〈a×b,a〉 = 0 = 〈a×b,b〉C3) a,b are linearly dependent if and only if a×b = 0C4) a× (b + c) = a×b + a× cC5) (a + b)× c = a× c + b× cC6) (αa)×b = α(a×b) = a× (αb), where α ∈R.


Property C) states that a×b is orthogonal to both a and b. Going back to Example .,one can directly verify that a×b = (−3,1,5) is orthogonal to both a and b. These vectorsare depicted in Figure ..

xy

z

a

b

a×b

Figure .: The vectors a,b and a×b.

The standard basis vectors {e1,e2,e3} for R3 satisfy

e1 × e2 = e3 (.)e3 × e1 = e2 (.)e2 × e3 = e1. (.)

Let ϕ denote the angle between a and b, then

‖a×b‖ = ‖a‖‖b‖sinϕ. (.)

Furthermore, if a,b are linearly independent, (.) yields the area of the parallelogramgenerated by a and b. Consider the parallelogram generated by two linearlyindependent vectors a,b in R

3, such as depicted in Figure .. As we did in the planarcase, we can decompose the parallelogram into a rectangular region and two righttriangular regions. Notice that the base of the right triangle corresponds precisely toproja b, the projection of b onto a, as depicted in Figure .. Moreover, if ϕ is the anglebetween a and b, the height of this triangle is ‖b‖sinϕ.

Definition .. Given vectors a,b,c ∈R3, the quantity 〈a,b× c〉 is called the scalartriple product.


x

y

z

a

b

projab

Figure .: The parallelogram.

ϕ

‖b‖

‖projab‖

Figure .: The right triangular portion of the parallelogram generated by a and b.

If a,b,c ∈R3 are linearly independent, then the volume of the parallelepiped generatedby a,b,c, as depicted in Figure ., is given by

|〈a,b× c〉|.Equivalently, the volume can be computed by the triple product |〈c,a×b〉|.

§.. Lines and Planes

In this section we will derive various equations of lines and planes in space. Whether ornot not we distinguish notationally, we must be careful to conceptually distinguishbetween points and vectors in R

3.


xy

z

a

b

c

Figure .: The parallelepiped generated by a,b,c.

Given two vectors a and b, the vector starting at the tip of a and ending at the tip of b isb− a. Recall that we can define a line in the plane by specifing two points, a point and aslope, or a slope and an intercept. For instance, the line through (a1, a2) and (b1,b2) is

y = a2 +b2 − a2

b1 − a1(x − a1). (.)

Notice that the graph of this line goes through the tips of the vectors a = (a1, a2) andb = (b1,b2). In addition to the form given above, we can express the same line inparametric form; that is, in terms of a real parameter t on which both coordinates maydepend. Rather than specifying a point and slope, we can specify a point and adirection. Since the line goes through the tips of a and b, the line is parallel to the vectorb− a (equivalently, a−b). The parametric equation

(x(t), y(t)) = (a1, a2) + t(b1 − a1,b2 − a2) (.)

where t ∈R defines the same line in terms of vectors. The algebraic equivalence of thisequation and (.) follows from setting t = x−a1

b1−a1. Making this substitution in (.)

yields y = a2 + t(b2 − a2), which is exactly the second coordinate of (.). Similarly, if

t =x − a1

b1 − a1

then x = a1 + t(b1 − a1). Since both the x and y coordinates depend on the parameter t,this dependence is denoted in the parametric equation (.).

We define lines in space by similar means.


Equation of line through a point in direction of a vector:The line ` through the point (x1,x2,x3) in the direction a = (a1, a2, a3) is given by theequation

`(t) = (x1,x2,x3) + t(a1, a2, a3) for t ∈R. (.)

In this case, the line ` may be expressed as `(t) = (x(t), y(t), z(t)), in terms of its coordinatefunctions. Thus, the coordinate functions x(t), y(t), z(t) are given by

x(t) = x1 + ta1, y(t) = x2 + ta2, z(t) = x3 + ta3.

Although the line ` goes through the point corresponding to the vector x = (x1,x2,x3), itis also common to express this line as

`(t) = x + ta, t ∈R, (.)

since the coordinate expressions are equivalent.

Now, if we want the line passing through the tips of two vectors a and b, we need theline pointed in the direction b− a (or a−b). This line passes through the pointscorresponding to the vectors a and b.

Equation of line through tips of vectors:The line ` through the tips of the vectors a = (a1, a2, a3) and b = (b1,b2,b3) is givenby the equation

`(t) = a + t(b− a) for t ∈R. (.)

In terms of the coordinate functions, ` has the form

`(t) = (x(t), y(t), z(t)) = (a1 + t(b1 − a1), a2 + t(b2 − a2), a3 + t(b3 − a3)). (.)

The same line is defined by any one of the following expressions

`(t) = a + t(a−b)= b + t(b− a)= b + t(a−b).

In terms of an infinite line, any one of these equations will typically suffice. However, ifwe want a line segment (that is, a finite line) we often want the segment directed as wesee fit. A line segment can be obtained by restricting the values of the parameter t. Forinstance, the line segment beginning at the tip of a and ending at the tip of b is given by`(t) = a + t(b−a), for t ∈ [0,1]. Notice that `(0) = a and `(1) = b. The line segment directedfrom b to a may be obtained by traversing `(t) = a + t(b− a) backwards. That is, take

˜̀(t) = `(1− t) = a + (1− t)(b− a).


Then, ˜̀(0) = b and ˜̀(1) = a.

There are various ways of describing any given plane in R3. For instance, the x,y-plane

may be expressed as the set of points in R3 for which the z-coordinate is 0,

{(x,y,0) |x,y ∈R}.More formally, this means the x,y-plane is the span of {e1,e2}. Similarly, if we fix z = α,where α is a fixed real number, the set

{(x,y,α) |x,y ∈R}is a plane in R

3, parallel to the x,y-plane and shifted vertically by α. This seems simpleenough, but what if we took the x,y-plane and tilted it about the origin. The resultwould still be a plane, but it would not be as straightforward to describe this plane bysimply fixing one of the coordinates. We need to know what distinguishingcharacteristics can be used to fully and unambiguously describe a plane in R

3. One suchcharacteristic is any vector which is orthogonal to the plane. For instance, any vector inthe x,y-plane is orthogonal to the z-axis (more particularly, orthogonal to the vector e3).Any vector which is parallel to e3 is also orthogonal to the x,y-plane. However, thischaracteristic alone is not enough to uniquely define the x,y-plane. The vector e3 (or anyparallel vector) is also parallel to all of the planes

{(x,y,α) |x,y ∈R},where α is fixed. We need more information to completely define a plane. Specifying apoint contained in the plane, along with an orthogonal vector, provides enoughinformation to completely define a plane in R

3. For instance, there is only one plane inR

3 which is orthogonal to e3 and contains the point (1,−2,0), the x,y-plane.

Suppose we want to determine an equation defining the plane in R3 which contains the

point (x1,x2,x3) and which is orthogonal to the vector a = (a1, a2, a3). Let (x,y,z) be anyother point in this plane. Then, the vector b = (x − x1, y − x2, z − x3), directed from onepoint to the other, must lie on this plane. Since the vector a is orthogonal to all vectorson this plane, we must have 〈a,b〉 = 0, hence

a1(x − x1) + a2(y − x2) + a3(z − x3) = 0. (.)

Any point (x,y,z) on this plane must satisfy equation (.).

Equation of plane orthogonal to given vector:The plane P in R

3 containing the point (x1,x2,x3) which is orthogonal to the vectora = (a1, a2, a3) satisfies the equation

a1(x − x1) + a2(y − x2) + a3(z − x3) = 0. (.)


Suppose (p1,p2,p3) is a point in space and let p denote the corresponding vector. Wewould like to determine the distance between this point and the plane P defined by(.). If (x1,x2,x3) is a point on P , let u = (x1 − p1,x2 − p2,x3 − p3) denote the vectordirected from (p1,p2,p3) to (x1,x2,x3). The distance from p to P is the norm of

proja u =〈u,a〉〈a,a〉a,

the projection of u onto a. We have

〈proja u,proja u〉 =〈x−p,a〉2〈a,a〉2 〈a,a〉

=〈x−p,a〉2‖a‖2 ,

where x is the vector with coordinates (x1,x2,x3). Hence, the distance from p = (p1,p2,p3)to the plane P is given by

|〈x−p,a〉|‖a‖ . (.)

Chapter

Multivariable functions

We are interested in defining functions of more than one variable, for the purpose ofcalculus in multiple variables (vector calculus). A real-valued function f in n variables,x1,x2, . . . ,xn, is a map which sends n real numbers (the inputs) to a single real number(the output), wherever f is defined. We denote the fact that f accepts n variables asinputs and outputs a single real number by f : Rn→R, which is read as “f maps Rn toR”. For example, f (x,y) = x2 + y2 is a real-valued function of 2 variables, so we wouldwrite f : R2→R.

Definition .. A real-valued function, f , from Rn to R is a mapping sending each

n-tuple of real numbers to a single real number, wherever defined. In this case, we writef : Rn→R. Equivalently, if f : Rn→R, f sends a vector x ∈Rn to a real number, whichwe denote as f (x). Thus, if x = (x1,x2, . . . ,xn), the output may be expressed asf (x1,x2, . . . ,xn) or f (x).

The fact is, we have already dealt with a function which maps vectors to scalars, namelythe norm. If we temporarily define f : R3→R by

f (a) = f (a1, a2, a3) =√a2

1 + a22 + a2

3,

then f (a) = ‖a‖ is the real valued function which determines the length of a given vectora.

We will also deal with maps which output multiple variables, in addition to acceptingmultiple inputs. If we write g : Rn→R

m, then g accepts n real numbers as inputs andoutputs m real numbers. When m = 1, g is exactly a real-valued function defined on R

n,as above.

Definition .. A vector-valued function, g, from Rn to R

m is a mapping sending eachn-tuple of real numbers to an m-tuple of real numbers, wherever defined. In this case,


we write g : Rn→Rm. Here, the inputs and the outputs can be regarded as vectors. If

x ∈Rn, then the output f (x) is a vector in Rm.

Remark .. Before discussing numerous examples, let’s address one technical detail.When we write g : Rn→R

m, we do not necessarily mean that the domain of g is all of Rn,nor do we mean that the range of g is all of Rm. Rather, if g : Rn→R

m, then the domainof g is a subset of Rn and the range of g is a subset of Rm. Also note that, whenevernecessary, we will denote the domain of a function g by dom(g).

The equation of a line in R3 is an example of a vector-valued function. Specifically, the

line ` : R→R3 defined by

`(t) = (x1 + ta1,x2 + ta2,x3 + ta3)= (x(t), y(t), z(t))

is a function which maps a single real number t to a vector (x(t), y(t), z(t)) in R3.

Before diving deeper into the subject of vector-valued functions, we will discussreal-valued functions in more detail. Eventually, we want to develop a calculus formultivariable functions, both real and vector valued. As in single variable calculus, wewant to develop limits, differentiation, extreme values, integration, etc. and see howthese ideas come up in applications. First, how can we visualize a function of more thanone variable? In particular, what does the graph of a function f : Rn→R look like?

Definition .. If f : Rn→R is a real-valued function, the graph of f is the subset ofRn+1 defined by

graph(f ) = {(x1,x2, . . . ,xn, f (x1,x2, . . . ,xn)) |x1, . . . ,xn ∈R}. (.)

Equivalently, if we express the input as a vector x = (x1, . . . ,xn), then the graph of f maybe expressed more concisely as

graph(f ) = {(x, f (x)) |x ∈Rn}. (.)

Since x ∈Rn and f (x) ∈R, graph(f ) is indeed a subset of Rn+1. Many of our exampleswill be functions from R

2 to R. If f : R2→R, then

graph(f ) = {(x,y, f (x,y)) |x,y ∈R}

is a subset of R3. In this particular case, a point (x,y,z) in R3 lies on the graph of f if

z = f (x,y). Thus, the function value f (x,y) determines the z component, or height.


§.. Level Sets

One particular method for determining graphical information about f is to determinethe level sets of f . Determining the level sets of f is analogous to determining x and yintercepts of a single variable function h : R→R.

Definition .. If f : Rn→R and α ∈R, the α-level set of f is the set of (x1, . . . ,xn) suchthat f (x1, . . . ,xn) = α. If we denote the α-level set of f by Lα(f ) then

Lα(f ) = {(x1, . . . ,xn) |f (x1, . . . ,xn) = α}.If f : R2→R, the α-level set of f is the set

Lα(f ) = {(x,y) ∈R2 |f (x,y) = α},which is often called a level curve in this particular case. If f : R3→R, the α-level set off

Lα(f ) = {(x,y,z) ∈R3 |f (x,y,z) = α}is called a level surface.

For a function f : R2→R, the α-level curve of f corresponds to setting z = α.

Example . Let z = f (x,y) =√x2 + y2. The 0-level curve, L0(f ), is the set of (x,y) ∈R2

for which√x2 + y2 = 0. The equation

√x2 + y2 = 0, equivalently x2 + y2 = 0, is only

satisfied for (x,y) = (0,0). This tells us that the point (0,0,0) (i.e., the origin) is on thegraph of f and, moreover, it is the only point on the graph with z coordinate 0.

®

§.. Sections

In addition to determining level sets of a multivariable function f , we can determinemore graphical information by determining cross-sections of the graph. A cross-section(or slice) of the graph of f is the intersection of a plane with graph(f ). For level sets, wedetermine the intersection of graph(f ) with a plane parallel with the xy-plane. Incontrast, for cross-sections we determine the intersection of graph(f ) with a ’vertical’plane. In this context, a vertical plane is any plane which is not parallel with thexy-plane. For example, the intersection of graph(f ) with the yz-plane is a cross-section.The yz-plane is the set {(0, y,z) |y,z ∈R} and the intersection with graph(f ) is the set

{(0, y, f (0, y)) |y ∈R}. (.)


−2 −1 0 1 2

−1

0

1

2

x

y

Figure .: The 0,1 and 2 level curves of f (x,y) =√x2 + y2.

1

2

xy

z

Figure .: The 0,1 and 2 level sets of f (x,y) =√x2 + y2 lifted to the corresponding z

values.

This cross-section is precisely the portion of the graph which lies on the yz-plane.

Another common cross-section is the intersection of graph(f ) with the xz-plane. Thexz-plane is the set {(x,0, z) |x,z ∈R} and the intersection with graph(f ) is the set

{(x,0, f (x,0)) |x ∈R}, (.)

which is the portion of the graph which lies on the xz-plane.

Typically, level sets give enough information that, along with one or two cross-sections,we can sketch the graph of f fairly accurately.

Example (. Continued). Returning to Example ., for which f (x,y) =√x2 + y2, let’s

determine the cross-sections of graph(f ) with the yz and xz planes. A point on the graph


of f is of the form (x,y,√x2 + y2). For the yz-plane cross-section, we simply substitute

x = 0 which yields the set of points

{(0, y,√y2) |y ∈R} = {(0, y, |y|) |y ∈R}.

This is the graph of the absolute value function, z = |y|, placed on the yz-plane. Similary,the xz-plane cross-section can be obtained by substituting y = 0, which yields the set ofpoints

{(x,0,√x2) |x ∈R} = {(x,0, |x|) |x ∈R}.

This is the graph of the absolute value function, z = |x|, placed on the xz-plane. Thesecross-sections, along with the previously determined level curves, are depicted in Figure.. A full surface plot of f (x,y) =

√x2 + y2 can be seen in Figure ..

1

2

xy

z

Figure .: The 0,1 and 2 level sets, along with the xz and yz cross-sections of graph(f ).The graph of f (x,y) =

√x2 + y2 is a cone.

2

xy

z

Figure .: A surface plot of z = f (x,y).


Example . Let g : R2→R be defined by g(x,y) = 1/3√

36− 4x2 − y2. The 0-level set,L0(g), is the set of (x,y) for which

0 =13

√36− 4x2 − y2,

hence 4x2 + y2 = 36. Equivalently,

19x2 +

136y2 = 1,

which is the equation of an ellipse with radii rx = 3 and ry = 6. The α-level set, Lα(g), ischaracterized by the equation

3α =√

36− 4x2 − y2,

hence4x2 + y2 = 36− 9α2.

Equivalently, for (x,y) to be in Lα(g) we must have

436− 9α2x

2 +1

36− 9α2y2 = 1.

Again, this is the equation of an ellipse, but with radii

rx =

√36− 9α2

4=

32

√4−α2, ry =

√36− 9α2 = 3

√4−α2.

From this we see that the we must have −2 ≤ α ≤ 2. For instance, if α = 1 thecorresponding ellipse has radii rx = 3/2

√3 and ry = 3

√3.

For the xz-plane cross-section, where y = 0, we have

z =13

√36− 4x2,

which becomes the ellipse19x2 +

14z2 = 1

with radii rx = 3 and rz = 2 in the xz-plane. Similarly, the yz-plane cross-section ischaracterized by

z =13

√36− y2,

corresponding to the ellipse1

36y2 +

14z2 = 1

with radii ry = 6 and rz = 2 in the yz-plane. The corresponding cross-sections, level setsand surface plot are depicted in Figures .-.. ®


−2−1

1

2

xy

z

Figure .: The xz and yz cross-sections of graph(g). The graph of g(x,y) =1/3

√36− 4x2 − y2 is an ellipsoid.

1

xy

z

Figure .: The 0,1 and 1.5 level sets of graph(g). The graph of g(x,y) = 1/3√

36− 4x2 − y2

is an ellipsoid.

§ Derivatives

Some of the techniques used for visualizing multivariable functions can help us makesense of derivatives/rates of change for multivariable functions. In particular, iff : R2→R is a real-valued function, if we slice the graph of f we obtain a single variablefunction. The intersection of graph(f ) with a plane parallel to the yz-plane yields acurve depending only on z and y. Equivalently, taking x to be a fixed number yields thecorresponding single variable function. The intersection of graph(f ) with a planeparallel to the xz-plane yields a curve depending only on z and x, which corresponds tothe single variable function obtained by fixing y.

In the same manner, we may regard x as fixed and then compute a derivative as wewould in single variable calculus. That is, if x is not changing, f (x,y) is a functiondepending only on y, so we may compute the instantaneous rate of change of f withrespect to y by

limh→0

f (x,y + h)− f (x,y)h

. (.)


Figure .: A surface plot of z = g(x,y).

If the limit in (.) exists, we call this the partial derivative of f with respect to y, at thepoint (x,y), which we denote by ∂f

∂y (x,y). Whenever the point at which a partialderivative is evaluated need not be specified, it is conventional to denote the partialderivative as simply ∂f

∂y . Similarly, if we regard y as fixed, f (x,y) depends only onchanges in x. The instantaneous rate of change of f with respect to x is given by

limh→0

f (x+ h,y)− f (x,y)h

. (.)

If the limit in (.) exists, this is the partial derivative of f with respect to x, at the point(x,y), which is denoted by ∂f

∂x (x,y) or ∂f∂x whenever it is unnecessary to specify the point.

Example . Let f : R2→R be the real-valued function f (x,y) = x2 cosy. By definition,


we have

∂f

∂x= limh→0

(x+ h)2 cosy − x2 cosyh

= limh→0

(2xh+ h2)cosyh

= limh→0

(2x+ h)cosy

= 2xcosy.

Notice that the same result is obtained by y as a constant and differentiating as usual,using the power rule in this case.

In the same way, we may compute ∂f∂y without resorting to limits. Thus,

∂f

∂y= −x2 siny.

®

The definition of partial derivatives is analogous for any real-valued functionf : Rn→R.

Definition .. Let f : Rn→R be a real-valued function. The partial derivative of fwith respect to xj , at the point (x1, . . . ,xn), is defined by

∂f

∂xj= limh→0

f (x1, . . . ,xj + h, . . . ,xn)− f (x1, . . . ,xn)

h, (.)

which is the result of differentiating f with respect to xj , regarding all other variables asfixed. Whenever it is necessary to denote the point at which the partial derivative isevaluated, we denote (.) by

∂f

∂xj(x1, . . . ,xn).

The limit definition of ∂f∂xj

can also be expressed as

∂f

∂xj= limh→0

f (x + hej)− f (x)

h, (.)

where ej = (0, . . . ,1↑

jth comp.

, . . . ,0) is the jth standard basis vector for Rn. We denote the operation

of partial differentiation with respect to the jth variable by ∂∂xj

.


By only considering the rate of change of f as the variable xj changes, we are computingthe rate of change in a direction parallel to the respective coordinate axis. For example,if f : R2→R, then ∂f

∂x and ∂f∂y are rates of change parallel to the x-axis and y-axis,

respectively. That is, if we consider a particle moving along the graph of f , then we onlyknow how much the height (z-coordinate) of the particle changes as the particle movesparallel to the x and y axes.

Example . Let f : R3→R be defined by f (x,y,z) = a−√z2−xy where a > 0 is a constant.

Recall the exponential derivative formula

ddtat = ln(a)at.

Then,

∂f

∂x= ln(a)a−

√z2−xy ∂

∂x(−

√z2 − xy)

= ln(a)a−√z2−xy

−

1

2√z2 − xy

∂∂x

(z2 − xy)


−

1

2√z2 − xy

(−y)

=ln(a)ya−

√z2−xy

2√z2 − xy

.

Similarly, we have

∂f

∂y= ln(a)a−

√z2−xy ∂

∂y(−

√z2 − xy)


−

1

2√z2 − xy

∂∂y

(z2 − xy)

=ln(a)xa−

√z2−xy

2√z2 − xy

and∂f

∂z= ln(a)a−

√z2−xy ∂

∂z(−

√z2 − xy)


−

1

2√z2 − xy

∂∂z

(z2 − xy)

= − ln(a)za−√z2−xy

√z2 − xy

.


®

Of course, we can take multiple partial derivatives of a function f , whenever thecorresponding limits are defined, in the same way that we compute multiple derivativesof a single variable function. For instance, if f : Rn→R we can differentiate with respectto xj , then differentiate with respect to xk. If the corresponding limit exists, this is asecond order partial derivative of f , denoted by

∂∂xk

(∂f

∂xj

)=

∂2f

∂xk∂xj.

Notice that the order in which we differentiate is from right to left, or inside to outside.If k = j, we denote this as

∂2f

∂x2j

,

which is the second partial derivative of f with respect to xj . If f : R2→R, there are 4second order partial derivatives, namely

∂2f

∂x2 ,∂2f

∂y∂x,

∂2f

∂x∂y,

∂2f

∂y2 .

Remark .. Depending on the properties of the real-valued function f , the order inwhich we take partial derivatives may matter! For a function f : R2→R, we may have

∂2f

∂y∂x,∂2f

∂x∂y,

in general. However, we will discuss a particular class of functions for which these mixedpartial derivatives are equal.

Example (. continued). Consider f (x,y) = x2 cosy. In Example . we computed thefirst partials

∂f

∂x= 2xcosy,

∂f

∂y= −x2 siny.

The mixed second derivatives are

∂2f

∂y∂x=∂∂y

(2xcosy) = −2x siny

and∂2f

∂x∂y=∂∂x

(−x2 siny) = −2x siny,


so∂2f

∂y∂x=∂2f

∂x∂y

in this case. The remaining second derivatives are

∂2f

∂x2 = 2cosy, and∂2f

∂y2 = −x2 cosy.

Now, if f : Rn→Rm is a vector-valued function, we can define partial derivatives

similarly. Since the output f (x) is a vector in Rm for each x ∈Rn, it is convenient to

introduce the notationf (x) = (f 1(x), f 2(x), . . . , f m(x)),

where each f i : Rn→R is a real-valued function, i = 1, . . . ,m. We call f i the ith coordinatefunction of f . Rather than defining derivatives of f directly, we can use existingderivative definitions for each of the m coordinate functions. Since each coordinatefunction maps Rn to R, there are n partial derivatives for each of the m coordinatefunctions.

Remark .. The coordinate functions can also be denoted using subscripts, fi , ratherthan superscripts, f i . However, the use of subscripts can be confusing when we opt todenote partial derivatives with subscripts. For example, we will represent the partialderivative of the 2nd coordinate function with respect to x3 by

∂f 2

∂x3or f 2

3 .

Definition .. If f : Rn→Rm is a vector-valued function, the Jacobian of f at the

point x is the m×n matrix

Df (x) =

∂f 1

∂x1

∂f 1

∂x2· · · ∂f 1

∂xn

∂f 2

∂x1

∂f 2

∂x2· · · ∂f 2

∂xn

.... . .

...

∂f m

∂x1

∂f m

∂x2· · · ∂f m

∂xn

, (.)

where each partial derivative is evaluated at the point x. The (i, j)-entry in Df (x) is

∂f i

∂xj.

The Jacobian of f is also called, simply, the matrix of partial derivatives of f .


The Jacobian of a real-valued function f : Rn→R has a special name, the gradient of f ,denoted gradf or ∇f . In accordance with the above definition, the gradient of f is therow vector

gradf =(∂f

∂x1,∂f

∂x2, . . . ,

∂f

∂xn

). (.)

If p is a point, we denote the gradient of f evaluated at p by gradp f , or ∇f (p).

Example . If r : R3→R is defined by r(x,y,z) =√x2 + y2 + z2 then

∂r∂x

=x√

x2 + y2 + z2,

∂r∂y

=y√

x2 + y2 + z2,

∂r∂z

=z√

x2 + y2 + z2,

hence

gradr =

x√x2 + y2 + z2

,y√

x2 + y2 + z2,

z√x2 + y2 + z2

.

If p = (1,2,−1), then gradp r = 1/√

6(1,2,−1). ®

Example . Let f : R2→R2 be the vector-valued function defined by

f (x,y) = (x2 cosy,y2 sinx).

In this case, we have two coordinate functions

f 1(x,y) = x2 cosy and f 2(x,y) = y2 sinx.

The corresponding partial derivatives are

∂f 1

∂x= 2xcosy,

∂f 1

∂y= −x2 siny

and∂f 2

∂x= y2 cosx,

∂f 2

∂y= 2y sinx.


Hence, the Jacobian of f is the 2× 2 matrix

Df =

∂f 1

∂x∂f 1

∂y

∂f 2

∂x∂f 2

∂y

=

2xcosy −x2 siny

y2 cosx 2y sinx

.

Notice that the rows of the Jacobian are precisely the gradients of f 1 and f 2,respectively. ®

§.. What’s wrong with partial derivatives?

Computing partial derivatives is relatively straightforward, albeit time consuming. Asmentioned previously, the partial derivatives yield rates of change parallel with thecoordinate axes. This straightforward process gives us some useful information aboutrates of change, but not quite enough. In single variable calculus, a derivative at a pointyields a rate of change which corresponds to the slope of the tangent line at that point.In the sense of slopes, partial derivatives only yield slopes in directions parallel to thecoordinate axes. We would like to be able to determine slopes/rates of change in anydirection, as rates of change can certainly depend on a particular direction. Moreover,we would like to generalize the local linear approximation (tangent line approximation)

g(x) ≈ g(c) + g ′(c)(x − c)of a single variable function g : R→R to a local approximation for f : Rn→R.

In what follows, we will give a definition of differentiability which is consisten withthese goals. But first, let’s briefly revisit differentiability for a single variable functiong : R→R. If the function g is differentiable at x ∈R, then

g ′(x) = limh→0

g(x+ h)− g(x)h

.

Equivalently, g is differentiable at x if and only if there exists a number L and a functionr(h) such that

g(x+ h)− g(x) = Lh+ r(h) (.)

and

limh→0

r(h)h

= 0. (.)


In this case,

limh→0

g(x+ h)− g(x)h

= limh→0

L+r(h)h

= L,

so L = g ′(x). To see this equivalent definition in action, take g(x) = x2. Corresponding to(.), we have

(x+ h)2 − x2 = 2xh+ h2 = (2x)h+ h2.

For this example, r(h) = h2 and we do have

limh→0

r(h)h

= 0,

and L = 2x = g ′(x).

Definition .. If f : Rn→Rm is a function (possibly vector-valued), then we say f is

differentiable at x ∈O ⊂Rn if there exists a matrix L and a function r(h) such that

f (x + h)− f (x) = Lh + r(h) (.)

and

limh→0

r(h)‖h‖ = 0. (.)

Equivalently,

limh→0

f (x + h)− f (x)−Lh‖h‖ = 0. (.)

If f is differentiable at x, the matrix L is unique and we denote it by f ′(x) or Df (x). If fis differentiable at every x ∈O, then we simply say f is differentiable (or f isdifferentiable on O). When f is differentiable, we refer to Df as the total derivative of f .

In the previous definition, the set O is an open subset contained in the domain of f .Since x is in the domain of f , we can at least be assured that f (x) is defined. The fact thatO is open has to do with convergence and the limits being defined, but this is beyond thescope of our course, so we need not worry about this technical detail. However, do takenote that h is a vector, so we divide by ‖h‖, whereas h is a scalar in (.)-(.).

Theorem .. If f : Rn→Rm is differentiable at x ∈ dom(f ), then f is continuous at x.

Example . To illustrate how definition . yields the appropriate derivative,consider f : R2→R

2 defined by f (x,y) = (x2 − y2,2xy). Let h = (h1,h2) ∈R2 and denote


by x the vector with components x and y. The corresponding difference in the left-handside of (.) is

f (x + h)− f (x) = ((x+ h1)2 − (y + h2)2,2(x+ h1)(y + h2))− (x2 − y2,2xy)

= (2xh1 + h21 − 2yh2 − h2

2,2xh2 + 2yh1 + 2h1h2)

= (2xh1 − 2yh2,2xh2 + 2yh1) + (h21 − h2

2,2h1h2)

=(2x −2y2y 2x

)(h1h2

)+ r(h), (.)

where r(h) = (h21 − h2

2,2h1h2). Although we have not formalized the notion limits formultivariable functions, it is indeed the case that

limh→0

r(h)‖h‖ = 0.

Thus, the total derivative of f is the 2× 2 matrix

Df =(2x −2y2y 2x

).

Notice that Df is precisely the Jacobian of f . For those with some background in linearalgebra, note there is an abuse of notation in (.), the last line in the computation ofDf . Namely, the matrix vector product should be transposed to be consistent with theprevious line and consistent with our convention of expressing vectors in row form.However, the end result will be the same. ®

Theorem .. Let f : Rn→Rm, where dom(f ) =O is a subset of Rn. If the partial

derivatives∂f i

∂xjexist and are continuous, for each i = 1, . . . ,m and j = 1, . . . ,n, near x ∈O,

then f is differentiable at x. In this case, the total derivative at x, Df (x), is the Jacobianevaluated at x.

According to this theorem, for a function f to be differentiable, not only should eachpartial derivative exists, but they must all be continuous. Moreover, it is possible for all(first-order) partial derivatives of a function to exist and the function benon-differentiable. This is possible if, for example, if one of the partial derivatives has adiscontinuity.

Definition .. We say a function f : Rn→Rm is of class C1 if all first order partial

derivatives ∂f i

∂xjexist and are continuous on some open subset O ⊂R

n. In some

circumstances, we say f is of class C1(O) to be clear about what the set O is. If f is ofclass C1, we may simply say f is C1 or that f is continuously differentiable.

Based on the previous definition, Theorem . can be stated more concisely.

Theorem .. If f : Rn→Rm is C1, then f is differentiable.


§.. Directional derivatives

Suppose f : Rn→R. We want a quantity representing the instantaneous rate of changeof f in a particular direction. For instance, if f : R2→R represents the temperature of arectangular plate at the (x,y) location. If we specify a direction in the form of a vectorv ∈R2, we want to know how much the temperature changes as a particle moves parallelto v. In the general case, f : Rn→R, let’s assume v ∈Rn is a unit vector. We will justifythis choice soon.

Definition .. The directional derivative of f at x in the direction v, which we willdenote by Dvf (x), is defined by

Dvf (x) = limλ→0

f (x +λv)− f (x)λ

. (.)

Analogous to the notation ∇f for the gradient of f , the notation ∇vf (x) is also used todenote the directional derivative.

Thankfully, by the following theorem, we do not need to use the previous limitdefinition for computing directional derivatives.

Theorem .. If f : Rn→R is differentiable, then all directional derivatives exist.Furthermore, the directional derivative of f in the direction v at x can be computed by

Dvf (x) = 〈gradx f ,v〉. (.)

Theorem .. Suppose gradx f , 0, then gradx f points in the direction along which fincreases the fastest.

Proof. Let v ∈Rn be a unit vector. Then, by (.) we have

Dvf (x) = 〈gradx f ,v〉 (.)

and hence

Dvf (x) = ‖gradx f ‖cosϕ (.)

by (.), where ϕ is the angle between v and gradx f . Thus, Dvf (x) is maximal whenϕ = 0, that is, when v is parallel to gradx f . By the same reasoning, f decreases mostrapidly in the direction −gradx f , for then cosϕ = −1.


§ Tangent vectors and planes

Consider a single-variable function f : R→R, for which

graph(f ) = {(x,f (x)) |x ∈R}is a subset of the plane. If f is differentiable at a point a ∈R, the tangent line at a is givenby the equation

y = f (a) + f ′(a)(x − a).Just as when we derived (.), the tangent line equation can be put in parametric formby setting t = x − a. Then y = f (a) + tf ′(a) and the tangent line can be parametrized by

`(t) = (a+ t, f (a) + tf ′(a))= (a,f (a)) + t(1, f ′(a)).

Comparing this parametric form with (.), we see that the tangent line passes throughthe point (a,f (a)) and is in the direction of the vector (1, f ′(a)). The vector (1, f ′(a)) istangent to the graph of f at a, hence we call it a tangent vector. Let τ(a) = (1, f ′(a))denote this tangent vector.

A point (x,y) is on the graph of f if y = f (x). Let g : R2→R be defined byg(x,y) = f (x)− y, then (x,y) is on the graph of f if g(x,y) = 0. Moreover,gradg = (f ′(x),−1) and grada g = (f ′(a),−1). Notice that

〈grada g,τ(a)〉 = f ′(a)− f ′(a) = 0.

This shows that a vector which is tangent to the graph of f at a is orthogonal to thegradient at a.

§.. Surfaces in R3

Now, suppose f : R2→R and consider the surface corresponding to the graph of f ,where as usual, we set z = f (x,y). Let p = (x0, y0, z0) be a point on graph(f ), so

z0 = f (x0, y0). Then ∂f∂x

∣∣∣∣(x0,y0)

represents the rate of change, with respect to x, parallel to

the xz-plane. In particular, we imagine slicing graph(f ) with the plane y = y0. Theintersection of y = y0 and graph(f ) (which is a cross-section) is a curve. This curve is thegraph of a single-variable function. Define g(x) = f (x,y0). Then g ′(x) = ∂f

∂x and, in

particular, g ′(x0) = ∂f∂x

∣∣∣∣(x0,y0)

. Thus, the slope of this curve’s tangent line at x0 is

g ′(x0) = ∂f∂x

∣∣∣∣(x0,y0)

. The tangent line is defined by the two conditions

z = g(x0) + g ′(x0)(x − x0), y = y0.


In parametric form, the tangent line is given by

`x(t) = (x0 + t,y0, g(x0) + tg ′(x0))

=(x0, y0, f (x0, y0)

)+ t

(1,0,

∂f

∂x

∣∣∣∣(x0,y0)

). (.)

Hence, the corresponding tangent vector is

u =(1,0,

∂f

∂x

∣∣∣∣(x0,y0)

). (.)

Similarly, ∂f∂y

∣∣∣∣(x0,y0)

represents the rate of change, with respect to y, parallel to the

yz-plane. In this case, we slice graph(f ) with the plane x = x0. The intersection of x = x0and graph(f ) is also a curve, which is the graph of a single-variable function. Define

h(y) = f (x0, y). Then h′(y) = ∂f∂y and h′(y0) = ∂f

∂y

∣∣∣∣(x0,y0)

. So, the slope of this curve’s tangent

line is h′(y0) = ∂f∂y

∣∣∣∣(x0,y0)

. In parametric form, the tangent line is given by

`y(t) = (x0, y0 + t,h(y0) + th′(y0))

=(x0, y0, f (x0, y0)

)+ t

(0,1,

∂f

∂y

∣∣∣∣(x0,y0)

)(.)

Hence, the corresponding tangent vector is

w =(0,1,

∂f

∂y

∣∣∣∣(x0,y0)

). (.)

The point p = (x0, y0, z0) on graph(f ) is on both tangent lines, (.) and (.). Wedefine the tangent plane to graph(f ) at p to be the plane containing both tangent vectors(equivalently, containing both tangent lines). In linear algebra lingo, the tangent plane isspanned by u and w, the tangent vectors given by (.) and (.). Since u and w areboth on the tangent plane, their cross product u×w must be orthogonal to the tangentplane. We have

u×w =(−∂f∂x,−∂f∂y,1

), (.)

where the partial derivatives are evaluated at (x0, y0). Now, a point (x,y,z) is on graph(f )if z = f (x,y) or, equivalently, if f (x,y)− z = 0. Let F(x,y,z) = f (x,y)− z, then

gradF =(∂f

∂x,∂f

∂y,−1

).


Thus, the gradient of F (evaluated at (x0, y0)) is orthogonal to the tangent plane! We alsoknow that the tangent plane passes through the point p, since both tangent lines passthrough p. Thus, the tangent plane is characterized as the plane orthogonal to gradFand passing through p = (x0, y0, f (x0, y0)). According to (.), the equation of thetangent plane is determined by 〈gradF,p〉 = 0. Thus, we have the following:

Equation of tangent plane to graph(f ) at (x0, y0, z0):

If f : R2→R, the tangent plane at (x0, y0, z0), where z0 = f (x0, y0), satisfies

∂f

∂x(x − x0) +

∂f

∂y(y − y0)− (z − f (x0, y0)) = 0, (.)

where both partial derivatives are evaluated at (x0, y0).

§ Coordinates

When specifying a point (or vector) in the plane, R2, we typically use use rectangularcoordinates (or Cartesian coordinates). If p is a point in R

2, we specify how to get fromthe origin to p in terms of how many units to move along the x-axis and y-axis.

Another natural choice for specifyin the location of a point in R2 is given via the

identification between x,y and cosθ,sinθ. If (x,y) is on the unit circle, then (x,y) is at aradial distance of 1 from the origin. The radial line connecting (0,0) to (x,y) forms anangle θ with the positive x-axis. For an arbitrary point in the plane, we can specify aradial distance r from the origin and an angle θ formed between the positive x-axis andthe radial line from (0,0) to (x,y).

Polar coordinates in R2:

The location of a point is specified as (r,θ) in polar coordinates, in terms of theradius and angle. We can determine the polar coordinates of a point from its Carte-sian coordinates (x,y) and vice-versa via:

x = r cosθ r =√x2 + y2

y = r sinθ θ = tan−1(yx

).

When specifying (x,y,z) coordinates of a point in R3, we are using rectangular/Cartesian


x

y

r

x

y

θ

Figure .: Cartesian coordinates (x,y) versus polar coordinates (r,θ).

coordinates. One alternative is to use polar coordinates for x,y and stick with Cartesianfor z.

Cylindrical coordinates in R3:

The location of a point is specified as (r,θ,z) in cylindrical coordinates, in termsof the radius, angle and Cartesian z coordinate. We can determine the cylindricalcoordinates of a point from its Cartesian coordinates (x,y,z) and vice-versa, in thesame way as for polar, since z remains unchanged:

x = r cosθ r =√x2 + y2

y = r sinθ θ = tan−1(yx

)

z = z z = z.

The third coordinate system for R3 that we will use often, requires a radius and twoangles.


x

y

z

r

z

θ

(r,θ,z)

Figure .: Cylindrical coordinates

Spherical coordinates in R3:

The location of a point, with Cartesian coordinates (x,y,z), is specified as (r,θ,ϕ) inspherical coordinates. Here, r is the straight line distance from (0,0,0) to the point;θ is the angle between the projection of (x,y,z) onto the xy-plane and the positivex-axis; ϕ is the angle between the positive z-axis and the radial line from (0,0,0) to(x,y,z). We can determine the spherical coordinates of a point from its Cartesiancoordinates (x,y,z), and vice-versa,via the coordinate transformations:

x = r sinϕ cosθ r =√x2 + y2 + z2

y = r sinϕ sinθ θ = tan−1(yx

)

z = r cosϕ ϕ = tan−1

√x2 + y2

z2

.


x

y

z

r

p

θ

ϕ

Figure .: Spherical coordinates

§ Parametric Curves

A path is a continuous mapping γ : [a,b]→Rn. The image γ([a,b]) is a curve in R

n. Thatis, as t varies over the interval [a,b], a curve is traced out by the values γ(t) ∈Rn. If wedenote this curve by C, then γ : [a,b]→R

n is also called a parametrization of C. Notethat our textbook assumes γ is differentiable, which means the associated curve issmooth. A parametrization of a curve is simply a vector-valued function. As with othervector-valued functions, we sometimes express γ in terms of its coordinate functions

γ(t) = (x1(t),x2(t), . . . ,xn(t)).

If γ is differentiable, then the derivative is

γ ′(t) = (x′1(t),x′2(t), . . . ,x′n(t)),

which is a tangent vector to the curve at the point γ(t). If γ(t) represents the position ofa particle (displacement) at time t, then γ ′(t) may be called the velocity vector at time t.In this case, the quantity ‖γ ′(t)‖ represents the speed of the particle at time t. Moreover,γ ′′(t) would be the acceleration vector at time t.

Examples.

..) The mapping γ : [0,1]→R3 defined by

γ(t) = a + t(b− a)


is a parametrization of the line segment joining a and b.

..) The mapping γ : [0,2π]→R2 defined by

γ(t) = (cos t,sin t)

is a parametrization of the unit circle. In terms of the terminology defined above, theunit circle is the curve, while the function γ is the parametrization or path.

..) Given a function y = f (x), whose graph is in the plane, we may always define aparametrization of the corresponding curve by defining

γ(t) = (t, f (t)).

Notice that γ ′(t) = (1, f ′(t)), which is precisely the tangent vector we defined in §.

®

§ Parametric surfaces

A parametrization of a surface is a map ψ : R2→R3 of the form

ψ(u,v) = (x(u,v), y(u,v), z(u,v)) .

In general, each component of a point on the surface depends on both u and v. Each ofthe vectors

∂ψ

∂u=

(∂x∂u,∂y

∂u,∂z∂u

)

and

∂ψ

∂v=

(∂x∂v,∂y

∂v,∂z∂v

)

is a tangent vector to the surface at the point ψ(u,v). Hence, the vector

∂ψ

∂u× ∂ψ∂v

(.)

is orthogonal to the tangent plane, as depicted in Figure ..


∂ψ

∂u

∂ψ

∂v

∂ψ

∂u× ∂ψ∂v

Figure .: The cross product (.) is orthogonal to the surface parametrized by ψ.

Examples.

..) If f : R2→R is a real-valued function of 2 variables, a point (x,y,z) is on thegraph of f if z = f (x,y). The graph of f may be parametrized by x = u,y = v,z = f (u,v).The corresponding parametrization would be

ψ(u,v) = (u,v,f (u,v)) .

Notice that the tangent vectors

∂ψ

∂u=

(1,0,

∂f

∂u

)

and

∂ψ

∂v=

(0,1,

∂f

∂v

)

are precisely what we found in (.) and (.), since u = x and v = y.

..) The surface determined by

x2

9+y2

12+z2

9= 1

is an ellipsoid. The upper half of the ellipsoid can be parametrized by

ψ(u,v) =(u,v,

√9−u2 − 3/4v2

).


..) Let’s determine the tangent plane to the cone 2z2 = x2 + y2 at (1,1,1). The conemay be parametrized as in the previous examples. However, in many cases,parametrizing a surface (just like a curve) via non-rectangular coordinates can simplifymatters. Using cylindrical coordinates, since x = r cosθ and y = r sinθ, we have z = r/

√2.

So, we will use the parametrization

ψ(r,θ) =(r cosθ,r sinθ,

r√2

).

The derivatives are

∂ψ

∂r=

(cosθ,sinθ,

1√2

)

and

∂ψ

∂θ= (−r sinθ,r cosθ,0) .

Now, since we want the tangent plane at the point (1,1,1), we must determine r and θsuch that ψ(r,θ) = (1,1,1). Equating components, we must have

r cosθ = 1,r sinθ = 1,

andr√2

= 1.

Clearly, r =√

2 so cosθ = 1/√

2 = sinθ, hence θ = π/4. Evaluating∂ψ

∂rand

∂ψ

∂θat r =

√2 and

θ = π/4, we have

∂ψ

∂r

∣∣∣∣(r,θ)

=(√

22,

√2

2,

1√2

)

and

∂ψ

∂θ

∣∣∣∣(r,θ)

= (−1,1,0) .

The cross product of the previous two tangent vectors is(− 1√

2,− 1√

2,√

2).


Thus, the equation of the tangent plane is

− 1√2

(x − 1)− 1√2

(y − 1) +√

2(z − 1) = 0,

which simplifies tox+ y − 2z = 0.

®

z

xx

yy

Figure .: The portion of the sphere ρ = 2 above the xy-plane and below the graph ofz = r.

§ Practice problems

P -.

For each of the following functions, determine at least 3 level sets and at least 2cross-sections. Sketch the graph as best you can, based on the level sets andcross-sections. Check your answer with Mathematica or other software.

a) f (x,y) = x2 − y2


b) f (x,y) = x2 + xy

c) f (x,y) = cos(xy)

d) f (x,y,z) =1√

x2 + y2 + z2

P -. For each of the following functions determine the gradient.

a) f (x,y) = x2y sin(x2y)

b) f (x,y) =x2 − 2xyx3 + y3

c) f (x,y,z) =xyz

x2 + y2 + z2

P -. Sketch the following surfaces.

a) r = 4sinθ

b) ρ sinϕ = 2

c) ρ = cos2θ

d) ρ = 1 + 2cos2ϕ

e) ρ = 1 + sin5θ5

P -. Determine an equation of the plane z = x in cylindrical coordinates and sphericalcoordinates.

P -. Determine the surface parametrized by

ψ(θ,ϕ) = (cosϕ cosθ,cosϕ sinθ,sinϕ),


with 0 ≤ θ ≤ 2π and −π/4 ≤ ϕ ≤ π/4.

P -. Determine a parametrization for the portion of the plane 2x+ 3y − z − 3 = 0 thatlies above the square

{(x,y) |0 ≤ x ≤ 1,0 ≤ y ≤ 1} .In other words, the portion of the plane 2x+ 3y − z − 3 = 0 for which the projection ontothe xy-plane is the above square.

P -. Determine a parametrization of the graph of z = x2 + y2 that lies above therectangle

{(x,y) |0 ≤ x ≤ 1,0 ≤ y ≤ 2} .

P -. Consider the parametrization

ψ(u,v) = (2u,v,u2 + v3),

where 0 ≤ u ≤ 1,0 ≤ v ≤ 2. Determine a function f (x,y) whose graph is the parametrizedsurface.


Chapter

Exterior Forms

A tangent vector at a point p gives us directional information about rates of change,locally, near p. Due to this fact (and many others that will become clear as we proceed),it makes more sense to have a tangent vector emanate from p rather than the origin, asdepicted in Figure .. Moreover, since a tangent vector lies on the tangent line, thischoice is much more natural than having tangent vectors emanate from the origin.

x

y

z

Figure .: Rather than the tangent vector emanating from the origin in xyz-space, thetangent vector emanates from the point γ(t) on the curve.

Let’s fix some notation. Let M be a geometric object in Rn, e.g. a curve, a surface, a region.

All of these geometric objects are particular types of manifolds, a topic which is beyondthe scope of our course. However, the machinery we will develop in Chapters andbeyond will extend to more arbitrary manifolds in a straightforward way. If p is a pointin M, we will denote the tangent space to M at p by TpM, which is the subspace of Rn

spanned by all the tangent vectors at p. For example, if M is a curve, then TpM is a line.If M is a surface (graph of a function), then TpM is a plane.


Now, if p ∈M and X is a tangent vector at p, we write X ∈ TpM and we imagine Xemanating from p. More formally, if M is a subset of Rn, we have another copy of Rn

which is translated so that the origin is at p, as depicted in Figure .. Since tangentvectors technically live in a different copy of Rn, we will represent tangent vectors bycapital letters. Whenever it is irrelevant, we may refer to tangent vectors in TpRn to beclear that these tangent vectors are in the ambient space R

n.

x

y

p

X

Figure .: A second copy of R2 with its origin at p, where all tangent vectors at p emanatefrom.

§ Constant Forms

In this section, we will discuss constant forms, which is precisely what Chapter 3 is allabout in our textbook. Constant forms are a special case of differential forms, which isthe machinery we will be using for vector calculus. The concept of a form can be a bitstrange at first, as it is likely quite different than any mathematical structure we haveseen before. However, forms are an incredibly powerful tool and they are usedthroughout mathematics and physics. For this reason, we will become acquainted withconstant forms first, as they are easier to deal with and this will give us the opportunityto slowly adjust to their nuances.

§.. 1-Forms

Definition .. An exterior 1-form, or simply a 1-form, is a linear function ω : Rn→R

which maps a (tangent) vector to a real number. To say that ω is a linear function is to


say that the following conditions are satisfied:

ω(x + y) = ω(x) +ω(y) (L)ω(αx) = αω(x) (L)

for all x,y ∈Rn and α ∈R. Note that the two conditions (L)-(L) are equivalent to thesingle condition

ω(αx + βy) = αω(x) + βω(y) (L’)

for all x,y ∈Rn and α,β ∈R.

If ω and η are two 1-forms, we can add them

(ω+ η)(x) = ω(x) + η(x),

and we can multiply 1-forms by constants

(αω)(x) = αω(x).

Any vector x ∈Rn induces a canonical 1-form

ωx(y) = 〈x,y〉 (.)

obtained by fixing one entry in the inner product and allowing the other entry to vary. Inparticular, to each standard basis vector ej , j = 1, . . . ,n, corresponds the canonical 1-form

ωej (y) = 〈ej ,y〉. (.)

Since ej = (0, . . . ,1↑

jth comp.

, . . . ,0), the canonical 1-form ωej simply returns the jth component of

the evaluated vector, ωej (y) = yj .

Definition .. For reasons to be discussed later, we denote the 1-form defined by (.)as dxj , which we call the jth coordinate 1-form. That is, the coordinate 1-formdxj : Rn→R is defined by

dxj(y) = yj . (.)

The coordinate 1-forms are also referred to as elementary 1-forms. Any 1-form ω on Rn

can be expressed asω = a1dx1 + a2dx2 + · · ·+ andxn,

where a1, . . . , an are real numbers.


On TpR2, in order to agree with earlier notation, we denote the coordinate 1-forms by dxand dy. Similarly, the coordinate 1-forms on TpR3 will be denoted by dx, dy and dz.

Example . Suppose ω = a1dx+ a2dy is a 1-form on TpR2. Then, the 1-form ω can beidentified with the vector a = (a1, a2) in the following way. If c = (c1, c2) is an arbitraryvector in TpR2 then

ω(c) = a1dx(c) + a2dy(c)= a1c1 + a2c2,

which is precisely 〈a,c〉. Hence, the action of the 1-form ω on an arbitrary vector c canbe interpreted in terms of the inner product. Recall that, by (.)-(.),〈c,a〉 = ‖proja c‖‖a‖. Thus, the value of the inner product 〈c,a〉 is proportional to thelength of the projection of c onto a and, in turn, the value of ω(c) is proportional to thelength of this projection. ®

Example . Experimental evidence shows that the work, W , done by a constant force Fin moving an object through a distance d is W = Fd, assuming the force is directed alongthe line of motion of the object. If the force is constant, but otherwise directed, we thinkof force as a vector F. Suppose an object is at position x and a force F, directed along u,applied to the object moves it from x to v. The displacement of the object is ‖v− x‖, themagnitude of the vector from x to v. If we let d = v− x be the displacement vector, thework done is

W = ‖d‖‖projd F‖= 〈F,d〉

The work done in moving an object by a displacement d,

W (d) = 〈F,d〉,

is a particular example of a 1-form. Here, we are regarding the force vector F as fixed.The 1-form W yields a scalar (the work) when given a displacement vector d.

®


F (force)

d (displacement)

W (d) = 〈F,d〉

Figure .: The work of a force F is a 1-form acting on the displacement vector.

§ 2-Forms

Definition .. An exterior 2-form, or simply a 2-form, is a function ω : Rn ×Rn→R onpairs of vectors which is linear in each component (bilinear) and skew-symmetric:

ω(αx + βy,v) = αω(x,v) + βω(y,v) (linear in 1st component) (F)ω(x,αy + βv) = αω(x,y) + βω(x,v) (linear in 2nd component) (F)

ω(x,y) = −ω(y,x) (skew-symmetric) (F)

Example . The signed area of the projection of the parallelogram with sides a,b ∈R3

on the xy-plane is a 2-form. When we speak of a signed area (sometimes referred to asan oriented area), we allow for the area to be negative. If a = (a1, a2, a3) and b = (b1,b2,b3),then the projected vectors have components (a1, a2,0) and (b1,b2,0), respectively. Thesigned area is a1b2 − a2b1 (whereas the unsigned area would be |a1b2 − a2b1|). ®

Example . Suppose v ∈R3 is a fixed vector. The 2-form ω defined by

ω(a,b) = 〈v,a×b〉

corresponds to the scalar triple product. This 2-form comes up when v is a uniformvelocity vector for the flow of fluid over the area of the parallelogram spanned bya,b. ®

§.. Wedge (Exterior) product

Many of the forms we are interested in are the result of combining 1-forms.


Definition .. Let ω,η be 1-forms on Rn. The value of ω∧ η (pronounced ω wedge η)

on the pair of vectors a,b ∈Rn is the signed area of the image of the parallelogram withsides (ω(a),η(a)) and (ω(b),η(b)) on the ωη-plane. That is,

(ω∧ η)(a,b) = det(ω(a) η(a)ω(b) η(b)

). (.)

By its very definition, ω∧ η is a 2-form on Rn. Forming the wedge product of two

1-forms yields a 2-form.

Example . Let ω be the 2-form defined by ω = dx∧ dy + 3dx∧ dz. If a = (a1, a2, a3) andb = (b1,b2,b3), then

(dx∧ dy)(a,b) = det(dx(a) dy(a)dx(b) dy(b)

)

= det(a1 a2b1 b2

)

= a1b2 − a2b1.

Similarly,

3(dx∧ dz)(a,b) = det(dx(a) dz(a)dx(b) dz(b)

)

= det(a1 a3b1 b3

)

= a1b3 − a3b1.

Combining the results, we have

ω(a,b) = (a1b2 − a2b1) + 3(a1b3 − a3b1).

®

Lemma .. Every (constant) 2-form ω on R3 can be expressed as

ω =ω1dx∧ dy +ω2dy ∧ dz+ω3dz∧ dx,

for some real numbers ω1,ω2,ω3.


§ k-forms

A k-form on Rn is a mapping which accepts k vectors, outputs a scalar, and is

multilinear and alternating:

ω(X1, . . . ,Xj + X̃j , . . . ,Xk) = ω(X1, . . . ,Xj , . . . ,Xk) +ω(X1, . . . , X̃j , . . . ,Xk), (kF)

ω(X1, . . . ,αXj , . . . ,Xk) = αω(X1, . . . ,Xj , . . . ,Xk), (kF)

ω(X1, . . . ,Xi , . . . ,Xj , . . . ,Xk) = −ω(X1, . . . ,Xj , . . . ,Xi , . . . ,Xk), (kF)

for j = 1, . . . , k. The conditions (kF) and (kF) state that ω is linear in the jth component.Since j = 1, . . . , k, ω is linear in each component. In other words, ω is multilinear. Thecondition (kF) states that ω is alternating. If we swap the order of two vectors, weintroduce a negative.

Remark .. If ω is a k-form on Rn and k > n, then ω ≡ 0. For instance, any 3-form on

R2 is 0 (meaning its action on every vector in R

2 results in 0). This also tells us that weneed only concern ourselves with 1,2 and 3-forms on R

3.

In general, a k-form is a machine that accepts any k (tangent) vectors X1, . . . ,Xk at a pointand returns a number ω(X1, . . . ,Xk) which can be thought of as the signed volume of theparallelepiped spanned by the vectors, according to a particular scale.

§.. Wedge product again

Just as we can take the wedge product of two 1-forms to yields a 2-form, we cancompute the wedge product of arbitrary forms in a simliar fashion. To be perfectlyhonest, the formal definition of the wedge product for arbitrary forms can appear a bitdaunting at first. Rather than giving the formular right away, we will proceed with amore constructive approach. Let’s look at some of the properties of the wedge product.


Properties of the wedge product:

Let ω be a k-form on Rn, let η be an `-form on R

n, and let ν be an s-form on Rn. Let

a,b ∈R be scalars.

W. The wedge product ω∧ η is a (k + `)-form.

W. (aω+ bη)∧ ν = a(ω∧ ν) + b(η ∧ ν)

W. ν ∧ (aω+ bη) = a(ν ∧ω) + b(ν ∧ η)

W. ω∧ (η ∧ ν) = (ω∧ η)∧ νW. ω∧ η = (−1)k`η ∧ωW. If ω1, . . . ,ωk are 1-forms and v1, . . . ,vk are vectors, then

ω1 ∧ · · · ∧ωk(v1, . . . ,vk) = det

ω1(v1) ω2(v1) · · · ωk(v1)ω1(v2) ω2(v2) · · · ωk(v2)...

.... . .

...ω1(vk) ω2(vk) · · · ωk(vk)

(.)

The formula (.) is particularly important, as it tells us how to compute the action of ak-form produced from k 1-forms on k vectors. In this way, we avoid the technical detailsof defining a k-form as the wedge product of 1-forms, directly, and focus on the action ofthe resulting k-form on vectors. Also, this allows us to progress from 1-forms to 2-formsto 3-forms, and so on, via the wedge product.

Lemma .. If ω is a k-form and k is odd, then ω∧ω = 0.

Proof. By W., we know ω∧ω = (−1)k2ω∧ω. Since k is odd, k2 is also odd, hence

(−1)k2

= −1. Thus, ω∧ω = −ω∧ω, which implies 2(ω∧ω) = 0. Hence, ω∧ω = 0.

Examples.

..) Let ω = 3dx∧ dy ∧ dz be a 3-form on R3 and let

X1 = (1,−3,4), X2 = (2,5,−2), X3 = (−1,3,6).


According to (.),

ω(X1,X2,X3) = 3det

dx(X1) dy(X1) dz(X1)



= 3det

1 −3 42 5 −2−1 3 6

= 3(36 + 30 + 44)= 330.

Note that we have used formula (.) for the determinant of a 3× 3 matrix.

..) Let ω = 2dx − dy + 4dz, η = −dx+ 2dy − dz, and ν = 2dx+ 3dy + 2dz. Givenvectors X1,X2,X3, we may determine the action of ω∧ η ∧ ν on these vectors by using(.). However, we may also determine the wedge product ω∧ η ∧ ν directly. By Lemma., the wedge product of any 1-form with itself is 0. In particular, dx∧ dx = 0 ,dy ∧ dy = 0 and dz∧ dz = 0. Using properties W.-W., we have

ω∧ η =(2dx − dy + 4dz)∧ (−dx+ 2dy − dz)= − 2(dx∧ dx) + 4(dx∧ dy)− 2(dx∧ dz)

+ (dy ∧ dx)− 2(dy ∧ dy) + (dy ∧ dz)− 4(dz∧ dx) + 8(dz∧ dy)− (dz∧ dz)

(by W. and W.)

=4(dx∧ dy)− 2(dx∧ dz) + (dy ∧ dx) + (dy ∧ dz)− 4(dz∧ dx) + 8(dz∧ dy)

(by Lemma .)

=4(dx∧ dy)− (dx∧ dy) + 2(dz∧ dx)− 4(dz∧ dx)+ (dy ∧ dz)− 8(dy ∧ dz) (by W.)

=3(dx∧ dy)− 2(dz∧ dx)− 7(dy ∧ dz).

Thus, ω∧ η is the 2-form

ω∧ η = 3(dx∧ dy)− 2(dz∧ dx)− 7(dy ∧ dz).


Similarly, we can compute (ω∧ η)∧ ν using properties of the wedge product,

(ω∧ η)∧ ν =(3(dx∧ dy)− 2(dz∧ dx)− 7(dy ∧ dz))∧ (2dx+ 3dy + 2dz)

=6(dx∧ dy ∧ dx) + 9(dx∧ dy ∧ dy) + 6(dx∧ dy ∧ dz)− 4(dz∧ dx∧ dx)− 6(dz∧ dx∧ dy)− 4(dz∧ dx∧ dz)− 14(dy ∧ dz∧ dx)− 21(dy ∧ dz∧ dy)− 14(dy ∧ dz∧ dz)

(by W. and W.)

=6(dx∧ dy ∧ dz)− 6(dz∧ dx∧ dy)− 14(dy ∧ dz∧ dx) (by Lemma .)=6(dx∧ dy ∧ dz)− 6(dx∧ dy ∧ dz)− 14(dx∧ dy ∧ dz) (by W.)= − 14(dx∧ dy ∧ dz).

Thus, the wedge product ω∧ η ∧ ν is the 3-form given by

ω∧ η ∧ ν = −14(dx∧ dy ∧ dz). (.)

Now, let’s consider an alternate method for determining (.). Let’s compute the actionof ω∧ η ∧ ν on the standard basis vectors e1,e2,e3. By (.), we have

(ω∧ η ∧ ν)(e1,e2,e3) = det

ω(e1) η(e1) ν(e1)

ω(e2) η(e2) ν(e2)

ω(e3) η(e3) ν(e3)

= det

2 −1 2−1 2 34 −1 2

= −14.

Notice that the resulting scalar is precisely the scalar appearing in (.)! ®

§ Vector fields

A vector field on Rn is a continuous mapping X :U ⊂R

n→Rn, where U is an open

subset of Rn. The vector field X assigns to each point p a vector in Rn, which we will

denote by Xp. Vector fields can be visualized as attaching an arrow (vector) to each pointin U . A vector field X can be expressed in terms of component functions

X =(X1,X2, . . . ,Xn

),

where each Xi :U →R, i = 1, . . . ,n. We let X(M) denote the vector space of all vectorfields defined on M. For instance, if X is a vector field on R

3, we may write X ∈ X(R3).


Remark .. If M is a geometric object, we may want vectors which are tangent to M atsome point. In this context, if p is a point in M, the vector Xp is a tangent vector to M atp, hence we write Xp ∈ TpM. The set of all tangent vectors to M at all points of M isdenoted by TM. In this sense, a vector field X can be regarded as a map from M to TM,which we express by X :M→ TM. More often, when chances of confusion are at aminimum, we will write X ∈ TpM, in which case it should be understood that X isevaluated at p.

Examples.

..) An example of a vector field on R2 is

X = (sinx,ex−y) .

If p = (π/2,0), then Xp = (1, eπ/2).

..) An example of a vector field on R3 is

X = (x2z,x,2yz).

If p = (1,1,1), then Xp = (1,1,2) and if q = (2,0,3) then Xq = (12,2,0).

®

In fact, we have seen lots of vector fields, we use them all the time. If f : Rn→R is areal-valued function, then gradf is a vector field, which is often reffered to as, simply,the gradient field of f . At any point p, the gradient vector at p is denoted gradp f .

Example . If f (x,y) =√x2 + y2, then

gradf =

x√x2 + y2

,y√

x2 + y2

.

The vector field gradf is visualized by attaching the vector gradp f to each point p, asdepicted in Figure .. In this particular figure, two level sets of f are also depicted.Notice that each gradient vector is orthogonal to the level curves!

®

If γ : R→R3 is a parametrization of a curve, then γ ′(t) is a vector field. For each point

γ(t) on the curve, there corresponds a tangent vector γ ′(t). For example, if


−2 −1 0 1 2

−2

−1

0

1

2

Figure .: The gradient field for f (x,y) =√x2 + y2 depicted with two level curves.

Figure .: The tangent vector field for the parametrization γ(t) = (cos t,sin t).

γ(t) = (cos t,sin t) is the counterclockwise parametrization of the unit circle, then thetangent vector field is γ ′(t) = (−sin t,cos t), which is depicted in Figure ..

Definitions.

..) Let X = (X1, . . . ,Xn) be a vector field on Rn. The divergence of X, denoted divX,

is the scalar function defined by

divX =n∑

i=1

∂Xi

∂xi. (.)


..) Let X = (X1,X2,X3) be a vector field on R3. The curl of X, denoted curlX, is the

vector field defined by

curlX = det

e1 e2 e3

∂∂x

∂∂y

∂∂z

X1 X2 X3

. (.)

As with the cross product, the determinant in (.) should be understood as a formalexpression. By expanding (.) along the first row and treating the second row entries aspartial differential operators, we have

curlX =(∂X3

∂y− ∂X

2

∂z,∂X1

∂z− ∂X

3

∂x,∂X2

∂x− ∂X

1

∂y

)(.)

Both the divergence and curl can be expressed in terms of the del operator

∇ =(∂∂x,∂∂y,∂∂z

). (.)

The divergence of X can be expressed by

divX = 〈∇,X〉, (.)

while the curl of X can be expressed by

curlX = ∇×X. (.)

The curl of a vector field only makes sense in R3, but divergence can be defined on R

n.On R

n, the del operator is

∇ =(∂∂x1

,∂∂x2

, . . . ,∂∂xn

)(.)

and divX can still be defined by (.) for a vector field on Rn, by using the correct del

operator.

Now, if X is a vector field on Rn, it is also commonplace to use the notation

X =n∑

i=1

Xi∂∂xi

(.)

as an alternative to X = (X1, . . . ,Xn). The notation (.) is particularly useful in thefollowing context. If f : Rn→R is a continuously differentiable (smooth) function, thenX maps f to a new function Xf defined by

Xf (p) = Xpf .


At this point, we will briefly discuss the meaning of divergence and curl of a vector field.The (physical) interpretation of the divergence and curl of a vector field should becomemore clear after studying Stokes’ Theorem and its consequences. The divergence of avector field measures how much of the field is flowing out of a region. As the namemight suggest, the curl of a vector field tells us if a particle placed in the field will rotate,or curl. More particularly, at a point p, the vector curlpX defines an axis around whichparticles in the vicinity of p will rotate. Moreover, ‖curlpX‖measures how quicklyparticles rotate around this axis.

§ Differential forms

In this section, we extend the constant forms of the last section to families of forms. If Mis a geometric object (manifold) and p is a point on M, we want to define forms that varyas a tangent vector X ∈ TpM varies. Recall that TpM denotes the set of all tangent vectorsat p. If ω is a 1-form, in addition to the criteria (L)-(L), we want ω(X) to vary smoothlyas the tangent vector X varies. In short, what this requires is that the coefficients of ω besmooth functions. The definition of smooth can vary depending on the context, but afunction must be at least differentiable to be considered smooth. Unless otherwisestated, we will assume smooth functions are infinitely differentiable, thus we can take asmany derivatives as necessary.

Thus, a differential 1-form on Rn, or simply a 1-form, is a linear function ω : Rn→R

which maps a (tangent) vector to a real number and which can be expressed as

ω =ω1dx1 +ω2dx2 + · · ·+ωndxn, (.)

where each ωi is a real-valued, continuously differentiable function on Rn, i = 1, . . . ,n.

The coordinate 1-forms are defined, as before, by (.). Differential 1-forms are alsoreferred to as covector fields, since the input of a 1-form is really a vector field, ingeneral. The vector space of all 1-forms on R

n will be denoted by Λ1(Rn). If ω is a1-form and X ∈ TpM, then

ω(X) = ω1(p)dx1(X) +ω2(p)dx2(X) + · · ·+ωn(p)dxn(X).

Notice that each component function ωi is evaluated at the point p, while the coordinate1-forms are evaluated at the vector X.

Example . Suppose γ(t) = (t, t2, t3) and let M be the corresponding curve. Then,

γ ′(t) = (1,2t,3t2)


is the vector field which defines the tangent vector at each point on M. Letω = 2xyzdx − y2dy − (yz − 30)dz be a covector field (1-form) on R

3. At t = 2, the pointp = γ(2) = (2,4,8) has tangent vector γ ′(2) = (1,4,12). Hence,

ω(γ ′(2)) = 2(2 · 4 · 8)dx(γ ′(2))− (4)2dy(γ ′(2))− (4 · 8− 30)dz(γ ′(2))= 128(1)− 16(4)− 2(12)= 40.

Again, notice that the component function 2xyz was evaluated by using the x,y,zcoordinates of p = γ(2), respectively. ®

A differential k-form on Rn, or simply a k-form, is a mapping which accepts k vectors,

outputs a scalar, and is multilinear and alternating, for which the component functionsare continuously differentiable. A k-form on R

n can be expressed

ω =∑

I

ωI dxI ,

where we are summing over all indices I = {i1, i2, . . . , ik} with i1 < i2 < · · · < ik and the ωI

are component functions. The notation dxI is an abbreviation for dxi1 ∧ · · · ∧ dxik , whichis an elementary k-form. There are

(nk

)=

n!k!(n− k)!

many elementary k-forms on Rn, where n! denotes n factorial. The vector space of all

k-forms on Rn is denoted Λk(Rn).

Remark .. A 0-form on Rn is simply a real-valued function defined on R

n.

Example . Let ω = y dx − xdy and η = x2y dy − xz2dz be 1-forms on R3. Then,

ω∧ η = x2y2dx∧ dy − xyz2dx∧ dz − x3y dy ∧ dy + x2z2dy ∧ dz= x2z2dy ∧ dz+ xyz2dz∧ dx+ x2y2dx∧ dy.

®

§.. Exterior derivative

Let f : Rn→R be a real-valued function. The differential of f is the covector field,denoted df , which is defined by

df =n∑

i=1

∂f

∂xidxi . (.)


Example . If f (x,y) = cos(xy), then df is given by

df =∂f

∂xdx+

∂f

∂ydy

= −y sin(xy)dx − x sin(xy)dy.

®

Proposition . (Properties of the differential). Let f ,g be continuously differen-tiable functions on R

n.

. For any a,b ∈R, d(af + bg) = adf + bdg.

. d(f g) = f dg + g df .

. d(f /g) = (g df − f dg)/g2 whenever g , 0.

Example . Let f (x,y) = cos(xy) and g(x,y) = xy. By virtue of ..., we maycompute the differential of (f g)(x,y) = xy cos(xy) by a product rule. In the previousexample, we found that

df = −y sin(xy)dx − x sin(xy)dy.

Since dg = y dx+ xdy, we have

d(f g) = f dg + g df= cos(xy) (y dx+ xdy) + xy (−y sin(xy)dx − x sin(xy)dy)

=(y cos(xy)− xy2 sin(xy)

)dx+

(xcos(xy)− x2y sin(xy)

)dy.

One can readily verify that computing the differential of xy cos(xy), directly, producesthe same result. ®

Definition .. If ω is a k-form on Rn given by

ω =∑

I

ωI dxI ,

then the exterior derivative of ω is the (k + 1)-form dω defined by

dω =∑

I

dωI ∧ dxI . (.)


The differential of a real-valued function f is a special case of the exterior derivative.Since the exterior derivative of a k-form ω is a (k + 1)-form, d maps k-forms to(k + 1)-forms; that is,

d :Λk(Rn)→Λk+1(Rn).

Examples.

..) If η = (x2 + y2)dz then

dη = d(x2 + y2)∧ dz= (2xdx+ 2y dy)∧ dz= 2xdx∧ dz+ 2y dy ∧ dz.

..) Let ω = 2xyzdx, then

dω = d(2xyz)∧ dx= (2yzdx+ 2xzdy + 2xy dz)∧ dx= 2yzdx∧ dx+ 2xzdy ∧ dx+ 2xy dz∧ dx= −2xzdx∧ dy + 2xy dz∧ dx.

..) An arbitrary differential 1-form on R3 can be expressed as

ω =ω1dx+ω2dy +ω3dz,


where the component functions ωi depend on x,y and z. According to (.), we have

dω = dω1 ∧ dx+ dω2 ∧ dy + dω3 ∧ dz

=(∂ω1

∂xdx+

∂ω1

∂ydy +

∂ω1

∂zdz

)∧ dx

(∂ω2

∂xdx+

∂ω2

∂ydy +

∂ω2

∂zdz

)∧ dy

(∂ω3

∂xdx+

∂ω3

∂ydy +

∂ω3

∂zdz

)∧ dz

=∂ω1

∂ydy ∧ dx+

∂ω1

∂zdz∧ dx+

∂ω2

∂xdx∧ dy

+∂ω2

∂zdz∧ dy +

∂ω3

∂xdx∧ dz+

∂ω3

∂ydy ∧ dz

=(∂ω3

∂y− ∂ω

2

∂z

)dy ∧ dz+

(∂ω1

∂z− ∂ω

3

∂x

)dz∧ dx+

(∂ω2

∂x− ∂ω

1

∂y

)dx∧ dy.

..) An arbitrary differential 2-form on R3 can be expressed as

ω =ω1dy ∧ dz+ω2dz∧ dx+ω3dx∧ dy,where the component functions ωi depend on x,y and z. According to (.), we have

dω = dω1 ∧ dy ∧ dz+ dω2 ∧ dz∧ dx+ dω3 ∧ dx∧ dy

=(∂ω1

∂xdx+

∂ω1

∂ydy +

∂ω1

∂zdz

)∧ dy ∧ dz

(∂ω2

∂xdx+

∂ω2

∂ydy +

∂ω2

∂zdz

)∧ dz∧ dx

(∂ω3

∂xdx+

∂ω3

∂ydy +

∂ω3

∂zdz

)∧ dx∧ dy

=∂ω1

∂xdx∧ dy ∧ dz+

∂ω2

∂ydy ∧ dz∧ dx+

∂ω3

∂zdz∧ dx∧ dy

=(∂ω1

∂x+∂ω2

∂y+∂ω3

∂z

)dx∧ dy ∧ dz.

®


Theorem .. For an arbitrary k-form ω

d(dω) = 0.

Proposition . (Properties of the exterior derivative).

. If f is a smooth function on Rn and X is a smooth vector field on R

n, then

df (X) = Xf .

. If ω ∈Λk(Rn), η ∈Λ`(Rn), then

d(ω∧ η) = dω∧ η + (−1)kω∧ dη. (.)

. d ◦ d = 0.

Proposition . (Exterior derivative of 1-forms). For any 1-form ω and vector fields Xand Y ,

dω(X,Y ) = X(ω(Y ))−Y (ω(X))−ω(XY −YX). (.)

Definitions.

..) A 1-form ω is called exact if there exists a real-valued function f : Rn→R suchthat df =ω, in which case f is called a potential for ω. In general, a k-form ω is calledexact if there exists a k − 1-form η such that ω = dη. Note that the general case agreeswith the definition for a 1-form, since a real-valued function f is a 0-form.

..) On the other hand, a vector field X is called conservative if there exists areal-valued function f : Rn→R such that gradf = X, in which case f is called apotential for X.

Proposition ..

a) If the vector field X ∈ X(R3) is conservative, then curlX = 0.

b) If a k-form ω is exact, then dω = 0.

Proof.


a) Since X = gradf ,

curlX = curl(gradf ) = 0,

since curl(gradf ) = 0 for any f : R3→R.

b) If ω is a 1-form, then ω = df for some function f , hence

dω = d(df ) = 0,

by Theorem ..

The same logic holds if ω is an exact k-form, since d ◦ d = 0.

Definition .. A differential k-form η is called closed if dη = 0.

Thus, the statement of Proposition .-b) takes the following form.

Proposition .. Every exact differential form is closed.

§.. Pullbacks/Change of coordinates

Let F : Rm→Rn be a smooth, vector-valued mapping, which we will consider as a

change of coordinates (or change of variables). If ω is k-form on Rn, we want to

determine how ω changes under this change of coordinates. More specifically, themapping F sends coordinates in R

m to coordinates in Rn (the forward direction). The

k-form ω is defined in terms of Rn-coordinates and we want to represent ω in terms ofRm-coordinates, thus we have to convert back to R

m-coordinates. In this context, we arepulling back coordinates in R

n to those in Rm.

Example . Consider the following example of changing coordinates (thesubstitution rule) from single-variable calculus. To compute the integral

∫x

x2 + 4dx,

we may set u = x2 + 4. In order to appropriately represent the integrand in terms of thenew variable u, we compute du = 2xdx. In this way, we are pulling back from theu-coordinate to the x-coordinate, to give an expression for the differential form du interms of x-coordinates. From this, we find

∫x

x2 + 4dx, =

∫1

2udu

=12

ln(u).


®

Examples. change of coordinate mappings

..) Let F : R2→R2 be defined by F(r,θ) = (r cosθ︸︷︷︸

x

, r sinθ︸︷︷︸y

), which changes

coordinates from polar to Cartesian. In this context, we think of F as mapping(r,θ)-coordinates to (x,y)-coordinates (the forward direction). Given a k-form defined in(x,y)-coordinates, we want to pull back to (r,θ)-coordinates.

..) If F : R3→R2 is defined by F(x,y,z) = ( x2y

︸︷︷︸u

, y sinz︸︷︷︸

v

), then u = x2y, v = y sinz and

we want to pull back a k-form from (u,v)-coordinates to (x,y,z)-coordinates.

®


Proposition . (Computing the pullback).

Let F : Rm→ Rn be a smooth mapping with component functions Fi , so F = (F1, . . . ,Fn).

Let ω be a differential form on Rn. The pullback of ω by F, denoted F∗ω, satisfies the

following properties.

. If ω = dxi , the ith coordinate 1-form, then

F∗dxi = dFi , (.)

the differential of the ith component of F.

. If ω =n∑i=1ωi dxi , then

F∗ω =n∑

i=1

(ωi ◦F)dFi . (.)

. If g : Rn→R is continuous and ω is a covector field (1-form), then

F∗(gω) = (g ◦F)F∗ω. (.)

If g is also differentiable, then

F∗dg = d(g ◦F). (.)

. If F : Rm→Rn, G : Rn→R

p and ω is a differential form on Rp, then

(G ◦F)∗ω = F∗(G∗ω). (.)

. If ω is a k-form and η is an `-form, then

F∗(ω∧ η) = (F∗ω)∧ (F∗η). (.)

Example . (Polar coordinates.) Consider the elementary 2-form on R2 ω = dx∧ dy.

Let F(r,θ) be as in Example .. and let F1(r,θ) = r cosθ, F2(r,θ) = r sinθ be thecoordinate functions. The pullback of ω by F can be determined by using the previous


properties. We have

F∗ω = F∗(dx∧ dy)= (F∗dx)∧ (F∗dy) by (.)= (dF1)∧ (dF2) by (.)= d(r cosθ)∧ d(r sinθ)= (cosθdr − r sinθdθ)∧ (sinθdr + r cosθdθ)

= r cos2θdr ∧ dθ − r sin2θdθ ∧ dr= r(cos2θ + sin2θ)dr ∧ dθ= r dr ∧ dθ.

This yields an expression for ω in polar coordinates. Since ω is an elementary form,computing the pullback F∗ω simply involves making the substitutions

x = r cosθ, y = r sinθ

and computing the exterior derivative (differential) of each component, in terms of thevariables r and θ. ®

Example . Consider the 1-form on R2 ω = v du +udv. Let F(x,y,z) be as in Example

... The pullback F∗ω can be computed as follows,

F∗ω = y sinzd(x2y) + x2y d(y sinz)

= y sinz(2xy dx+ x2dy) + x2y(sinzdy + y coszdz)

= 2xy2 sinzdx+ 2x2y sinzdy + x2y2 coszdz.

Notice that, since ω is not constant, we also must substitute for the componentfunctions, according to (.).

®


P -. For the given forms ω,η compute ω∧ η and simplify.

a) ω = 2dx − 7dy, η = −3dx+ 5dy

b) ω = y dx − xdy, η = xy dx+ y dy


c) ω = dx − 4dz, η = 6dy ∧ dz − dz∧ dx+ 2dx∧ dy

d) ω = xzdx+ yzdy, η = y dy ∧ dz+ z2dz∧ dx − x3dx∧ dy

e) ω = cosy dx − x siny dy, η = siny dx+ xcosy dy

P -. Let ω = η ∧ψ, where η,ψ are the (constant) 1-forms

η = n1dx+n2dy +n3dz

ψ = p1dx+ p2dy + p3dz.

a) Compute the wedge product η ∧ψ algebraically to determine a simplified expressionfor ω of the form

ω = ady ∧ dz+ bdz∧ dx+ cdx∧ dy.

b) Let n = (n1,n2,n3) and p = (p1,p2,p3). Show that (a,b,c) = n×p.

P -. For each of the following functions, determine the gradient vector field andsketch the vector field.

a) f (x,y,z) = x2 + y2 + z2

b) f (x,y,z) = 1√x2+y2+z2

c) f (x,y,z) = log(x2 + y2)

P -. Compute the divergence and curl of each vector field from the previous problem.Note that the gradient field resulting from the function in P-c should be defined onR

3, even though f is independent of z.

P -. Compute the divergence and curl of the vector field

X = (x,y,z).


P -. Compute the divergence and curl of the vector field

X = (x2 + 1,xyz,sin(x+ y)).

P -. For each of the following functions determine the differential.

a) f (x,y) = x2y sin(x2y)

b) f (x,y) =x2 − 2xyx3 + y3

c) f (x,y,z) =xyz

x2 + y2 + z2

d) f (x,y,z) = exy + cos(xyz)

e) f (x,y,z) = x2y3z − 2xyz2

P -. Show that d2f = d(df ) = 0 for an arbitrary function f : R3→R (0-form).

P -. Compute the exterior derivative of the following 1-forms, which are defined onR

3 unless otherwise stated.

a) ω = 2x2dx+ (x+ y)dy on R2

b) η = −xdx+ (x − 2y)dy on R2

c) ω = x3dx+ yzdy + (x2 + y2 + z2)dz

d) ψ = y2zdx − xzdy + (2x+ 1)dz

e) η = xdx+ y2dy + z3dz

f) ψ = 2dx − dy + 4dz


g) ω = x2yzdx − 2xy3zdy + 3xyz4dz

h) η = yz2dx+ (x+ z2)dy + (x2 − y)dz

P -. Show that d2ω = d(dω) = 0 for an arbitrary 1-form ω =ω1dx+ω2dy +ω3dz onR

3. If the superscripts bother you, let ω = f dx+ g dy +hdz, where f ,g,h are all functionsfrom R

3 to R.

P -. Compute the exterior derivative of the following 2-forms on R3.

a) ω = xdx∧ dy + zdy ∧ dz+ y dz∧ dx

b) η = (x2 + y2)dy ∧ dz

c) ω = (x2 − y2)dy ∧ dz+ (x − y2)dz∧ dx+ 3zdx∧ dy

d) ψ = (x2 + y2)dy ∧ dz+ (x+ y − z2)dz∧ dx − 6xy dx∧ dy

e) η = (9xz2 − 4xy)dx∧ dy + (x+ 2y − 3z)dy ∧ dz − (3x+ 4yz2)dz∧ dx

P -. Let ψ and ω be as in P-d and P-a, respectively. Compute the following:

dψ, ψ ∧ dψ, dω, ψ ∧ω

P -. Let ω = dx − zdy and ν = (x2 + y2 + z2)dz∧ dx+ (xyz)dy ∧ dz. Compute thefollowing:

dω, ω∧ dω, dν, ω∧ ν

P -. Let ω = xdy ∧ dz+ y dz∧ dx+ zdx∧ dy.

a) Determine an expression for ω in spherical coordinates. In other words, ifF : R3→R

3 is defined by

F(ρ,θ,ϕ) = (ρ sinϕ cosθ︸︷︷︸

x

,ρ sinϕ sinθ︸︷︷︸

y

,ρcosϕ︸︷︷︸

z

),


compute the pullback, F∗ω, of ω by F.

b) Compute dω in both Cartesian and spherical coordinates, then check that theyrepresent the same 2-form.


Chapter

Integration and the fundamentalcorrespondence

§ The correspondence between vector fields anddifferential forms

If f : Rn→R, we have already noted similarities between the gradient vector field

gradf =(∂f

∂x1, . . . ,

∂f

∂xn

)

and the differential

df =∂f

∂x1dx1 + · · ·+ ∂f

∂xndxn.

Likewise, if X = (P ,Q,R) is a vector field on R3, there seems to be a correspondence

between curlX and the exterior derivative of

ω = P dx+Qdy +Rdz.

In this section, we discuss this correspondence in detail and show how we can use thiscorrespondence in practice for certain types of integrals. Moreover, this correspondencewill aid in understanding and deciphering the many faces of Stokes Theorem. In orderto make the correspondence clear, we will introduce three operators.

Definitions.

..) Let X = (X1,X2,X3) be a vector field on R3 and let X[ (“X flat”) denote the

corresponding covector field (1-form)

X[ = X1dx+X2dy +X3dz. (.)


..) On the other hand, if ω =ω1dx+ω2dy +ω3dz is a covector field on R3, let ω] (“ω

sharp”) denote the corresponding vector field

ω] = (ω1,ω2,ω3). (.)

For example, if f : R3→R, then

gradf =(∂f

∂x,∂f

∂y,∂f

∂z

)

and

(gradf )[ =∂f

∂xdx+

∂f

∂ydy +

∂f

∂zdz.

That is, (gradf )[ = df . On the other hand, (df )] = gradf .

fgrad−−−−−→ gradf∥∥∥∥

y[f −−−−−→

ddf

Definition .. The Hodge star operator, ∗, on R3 takes k-forms to (3− k)-forms. If η is a

k-form on R3, then ∗η is a (3− k)-form on R

3. In terms of elementary forms on R3, we

have

∗1 = dx∧ dy ∧ dz,∗dx = dy ∧ dz, ∗(dy ∧ dz) = dx,∗dy = dz∧ dx, ∗(dz∧ dx) = dy,∗dz = dx∧ dy, ∗(dx∧ dy) = dz,

∗(dx∧ dy ∧ dz) = 1.

For example, if ω is an arbitrary 1-form on R3, given by

ω =ω1dx+ω2dy +ω3dz,

where the component functions ωi depend on x,y and z, then ∗ω is the 2-form

∗ω =ω1 ∗ dx+ω2 ∗ dy +ω3 ∗ dz=ω1dy ∧ dz+ω2dz∧ dx+ω3dx∧ dy.


On the other hand, if η is an arbitrary 2-form on R3, given by

η = η1dy ∧ dz+ η2dz∧ dx+ η3dx∧ dy,

then ∗η is the 1-form

∗η = η1 ∗ (dy ∧ dz) + η2 ∗ (dz∧ dx) + η3 ∗ (dx∧ dy)

= η1dx+ η2dy + η3dz.

Remark .. Notice the similarity between the relations

∗(dy ∧ dz) = dx, e2 × e3 = e1,

∗(dz∧ dx) = dy, and e3 × e1 = e2,

∗(dx∧ dy) = dz, e1 × e2 = e3.

If f : R3→R is a real-valued function (0-form), then ∗f is the 3-form ∗f = f dx∧ dy ∧ dz.On the other hand, if ψ is an arbitrary 3-form on R

3, given by

ψ = g dx∧ dy ∧ dz,

then ∗ψ is the 0-form ∗ψ = g, where g : R3→R is a real-valued function.

Proposition .. If X = (P ,Q,R) is a vector field on R3, where P ,Q,R are functions of x,y,z,

then

curlX =(∗d(X[)

)](.)

or, equivalently,

(curlX)[ = ∗d(X[). (.)

The divergence of X satisfies∗ d(∗X[) = divX. (.)

§ Flux integrals

Suppose X ∈ X(R3) is a vector field on R3 and let M be a surface. If the vector field X

represents the velocity vectors of a fluid, we want to determine the rate at which thefluid moves across the surface. This type of calculation corresponds to the physicalquantity of flux, which is the amount of fluid that crosses the surface per unit of time.


Let the surface M be parametrized by a map ψ :D ⊂R2→R

3,

ψ(u,v) = (x(u,v), y(u,v), z(u,v)),

where D is the subset of R2 on which ψ is defined. In this context, D may be referred toas the domain of integration. Then each of the vectors

∂ψ

∂u=

(∂x∂u,∂y

∂u,∂z∂u

)

and

∂ψ

∂v=

(∂x∂v,∂y

∂v,∂z∂v

)

is a tangent vector to the surface at the point ψ(u,v). The vector

∂ψ

∂u× ∂ψ∂v

is orthogonal (recall that normal and orthogonal are synonymous) to the tangent plane atthis point. Define the unit normal vector to M at the point ψ(u,v) by

N (u,v) =∂ψ∂u ×

∂ψ∂v

‖∂ψ∂u ×∂ψ∂v ‖

, (.)

where all the partial derivatives are evaluated at (u,v). The flux of the vector field Xacross the surface M, in the direction of the unit normal vector N , is defined by theintegral �

D

〈X(ψ),N 〉∥∥∥∥∥∂ψ

∂u× ∂ψ∂v

∥∥∥∥∥ dudv. (.)

If D is the rectangleD = {(u,v) |a ≤ u ≤ b,c ≤ v ≤ d},

then (.) can be expressed as the double integral∫ d

c

∫ b

a〈X(ψ),N 〉

∥∥∥∥∥∂ψ

∂u× ∂ψ∂v

∥∥∥∥∥ dudv.

Integrating over more general regions will be discussed in the sequel.

Example . Determine the flux of the vector field X = (x,y,2z) across the surface M,which is the portion of the unit sphere in the first octant. In terms of Cartesiancoordinates, the first octant is characterized by x ≥ 0, y ≥ 0 and z ≥ 0. The mapping

ψ(θ,ϕ) = (sinϕ cosθ,sinϕ sinθ,cosϕ)


parametrizes the unit sphere in spherical coordinates. The first octant is obtained by therestriction 0 ≤ ϕ ≤ π/2 and 0 ≤ θ ≤ π/2. The tangent vectors are

∂ψ

∂θ= (−sinϕ sinθ,sinϕ cosθ,0)

and∂ψ

∂ϕ= (cosϕ cosθ,cosϕ sinθ,2cosϕ).

The unit normal vector is

N =1

sinϕ(−sin2ϕ cosθ,−sin2ϕ sinθ,−sinϕ cosϕ)

= (−sinϕ cosθ,−sinϕ sinθ,−cosϕ).

We have X(ψ) = (sinϕ cosθ,sinϕ sinθ,2cosϕ), hence

〈X(ψ),N 〉 = −sin2ϕ cos2θ − sin2ϕ sin2θ − 2cos2ϕ

= −sin2ϕ − 2cos2ϕ

= −(1 + cos2ϕ).

Thus, to determine the flux, we compute

=∫ π/2

0

∫ π/2

0−sinϕ(1 + cos2ϕ)dϕdθ

=∫ π/2

0(cosϕ + 1/3 cos3ϕ)

∣∣∣∣π/2

0dθ

=∫ π/2

0−4

3dθ

= −2π3.

The amount of fluid crossing the surface per unit of time is −2π3 . Note that the chosen

normal vector N is inward pointing, so our calculation tells us how much fluid is enteringthe surface, but the value is negative which tells us that fluid is, in fact, exiting thesurface.

The same calculation can be performed by integrating a differential form. According tothe previous section, the vector field X = (x,y,2z) corresponds to the differential 2-form

ω = ∗X[ = xdy ∧ dz+ y dz∧ dx+ 2zdx∧ dy.


The flux integral of X over M is precisely the same as the integral of the pullback ψ∗ω.Note that

ψ∗(xdy ∧ dz) = (sinϕ cosθ)d(sinϕ sinθ)∧ d(cosθ)= (sinϕ cosθ)((cosϕ sinθdϕ + sinϕ cosθdθ)∧ (−sinϕdϕ))

= −sin3ϕ cos2θdθ ∧ dϕ,ψ∗(y dz∧ dx) = (sinϕ sinθ)d(cosϕ)∧ d(sinϕ cosθ)

= (sinϕ sinθ)((−sinϕdϕ)∧ (cosϕ cosθdϕ − sinϕ sinθdθ))

= −sin3ϕ sin2θdθ ∧ dϕ,ψ∗(2zdx∧ dy) = (2cosϕ)d(sinϕ cosθ)∧ d(sinϕ sinθ)

= 2cosϕ((cosϕ cosθdϕ − sinϕ sinθdθ)∧ (cosϕ sinθdϕ + sinϕ cosθdθ))

= (−2sinϕ cos2ϕ sin2θ − 2sinϕ cos2ϕ cos2θ)dθ ∧ dϕ= −2sinϕ cos2ϕdθ ∧ dϕ.

Thus, the pullback is given by

ψ∗ω = (−sin3ϕ − 2sinϕ cos2ϕ)dθ ∧ dϕ= −sinϕ(sin2ϕ + cos2ϕ + cos2ϕ)dθ ∧ dϕ= −sinϕ(1 + cos2ϕ)dθ ∧ dϕ,

hence we integrate

∫ψ∗ω =

∫ π/2

0

∫ π/2

0−sinϕ(1 + cos2ϕ)dθdϕ.

®

Surface area and integrating over simple domains

Let M be a surface parametrized by a map ψ :D ⊂R2→R

3,

ψ(u,v) = (x(u,v), y(u,v), z(u,v)),

where D is the subset of R2 on which ψ is defined. The surface area of M os given by�D

∥∥∥∥∥∂ψ

∂u× ∂ψ∂v

∥∥∥∥∥ dudv. (.)


Example . Let M be the surface parametrized by

ψ(r,θ) = (r cosθ,2r cosθ,θ)

for 0 ≤ r ≤ 1 and 0 ≤ θ ≤ 2π. Then, the tangent vectors are

∂ψ

∂r= (cosθ,2cosθ,0)

∂ψ

∂θ= (−r sinθ,−2r sinθ,1)

and the normal vector is∂ψ

∂r× ∂ψ∂θ

= (2cosθ,−cosθ,0).

The norm of the normal vector, ∂ψ∂r ×∂ψ∂θ , is

√5cos2θ =

√5|cosθ| =

√5

cosθ, 0 ≤ θ ≤ π

2 and 3π2 ≤ θ ≤ 2π,

−cosθ, π2 ≤ θ ≤ 3π

2 .

The absolute value is important here, as without it, we would get the wrong answer.Now, the first case is equivalent to cosθ for −π/2 ≤ θ ≤ π/2, so the area is

√5∫ 2π

0

∫ 1

0|cosθ|dr dθ =

√5∫ π/2

−π/2cosθdθ +

√5∫ 3π/2

π/2−cosθdθ

=√

5(sinθ

∣∣∣∣π2

−π2

− sinθ∣∣∣∣

3π2

π2

)

= 4√

5.

®

Suppose f : R2→R is a real-valued function and we want to compute∫

Df (x,y)dx∧ dy, (.)

where D is a domain of integration in R2. The integral (.) represents the volume in R

3

bounded by the graph of f and the region D in the xy-plane. So far, we have onlyencountered the case when D is a rectangular region, corresponding to the intervalsa ≤ x ≤ b and c ≤ y ≤ d. We want to allow for more general domains of integration. First,consider the case when

D = {(x,y) |0 ≤ y ≤ g(x), a ≤ x ≤ b},where g is a real-valued function of x. The idea is to parametrize the domain D, thencompute the pullback of the differential form f (x,y)dx∧ dy. In this case, the region D


can be parametrized by the map ψ(u,v) = (u,vg(u)), where a ≤ u ≤ b and 0 ≤ v ≤ 1. Thepullback is given by

ψ∗(f (x,y)dx∧ dy) = f (u,vg(u))d(u)∧ d(vg(u))= f (u,vg(u))du ∧ (vg ′(u)du + g(u)dv)= f (u,vg(u))g(u)du ∧ dv.

Hence, the integral of f over D becomes∫

Df =

∫ψ∗(f (x,y)dx∧ dy)

=∫ 1

0

∫ b

af (u,vg(u))g(u)dudv.

On the other hand, ifD = {(x,y) |0 ≤ x ≤ h(y), c ≤ y ≤ d},

then the region D can be parametrized by the map ψ(u,v) = (uh(v),v), where 0 ≤ u ≤ 1and c ≤ v ≤ d. In this case, the integral of f over D becomes

∫

Df =

∫ψ∗(f (x,y)dx∧ dy)

=∫ d

c

∫ 1

0f (uh(v),v)h(v)dudv.

Example . The portion of the cone x2 + y2 = z2 above z = 0 and inside the spherex2 + y2 + z2 = 4ax, where a > 0, can be parametrized by

F(r,θ) = (r cosθ,r sinθ,r),

for −π/2 ≤ θ ≤ π/2 and 0 ≤ r ≤ 2acosθ. If M is the portion of the cone and η = zdx∧ dy,let’s compute

∫Mη.

Method ) First, we compute the pullback of η by F,

F∗η = rd(r cosθ)∧ d(r sinθ)

= r2dr ∧ dθ.The domain of integration is

D = {(r,θ) | − π/2 ≤ θ ≤ π/2 and 0 ≤ r ≤ 2acosθ},


which can be parametrized by ψ(u,v) = (2au cosv,v), where 0 ≤ u ≤ 1 and−π/2 ≤ v ≤ π/2. Using this parametrization, we can compute the pullback of thepullback,

ψ∗(F∗η) = ψ∗(r2dr ∧ dθ)

= (2au cosv)2d(2au cosv)∧ dv= 8a3u2 cos3 v du ∧ dv.

Hence, the integral of η over M becomes

∫ψ∗(F∗η) =

∫ π/2

−π/2

∫ 1

08a3u2 cos3 v du dv

=8a3

3

∫ π/2

−π/2cos3 v dv

=8a3

3

(sinv − 1

3sin3 v

)∣∣∣∣π2

−π2

=32a3

9.

Method ) The same answer can be found by directly integrating F∗η over D, makingsure to first integrate with respect to the variable bounded by functions. In thiscase,

∫

DF∗η =

∫ π/2

−π/2

∫ 2acosθ

0r2dr dθ

=∫ π/2

−π/2

(13r3

∣∣∣∣2acosθ

0

)dθ

=8a3

3

∫ π/2

−π/2cos3θdθ

=32a3

9.

®

Example . Let’s compute the area of the surface in Example .. Given theparametrization

F(r,θ) = (r cosθ,r sinθ,r),


we have tangent vectors

∂F∂r

= (cosθ,sinθ,1)

and

∂F∂θ

= (−r sinθ,r cosθ,0).

The normal vector∂F∂r× ∂F∂θ

= (−r cosθ,−r sinθ,r)

has norm ∥∥∥∥∥∂F∂r× ∂F∂θ

∥∥∥∥∥ =√r2 cos2θ + r2 sin2θ + r2 =

√2r.

Given the bounds for r and θ from the previous example, we have∫ π/2

−π/2

∫ 2acosθ

0

√2r dr dθ =

∫ π/2

−π/2

√2

2r2

∣∣∣∣2acosθ

0dθ

=∫ π/2

−π/22√

2a2 cos2θdθ

=√

2a2∫ π/2

−π/2(1 + cos(2θ))dθ

=√

2a2(θ +

12

sin(2θ))∣∣∣∣

π/2

−π/2

=√

2πa2.

®

Example . Compute the area of the portion of 4z2 = x2 + y2 inside the cylinder(x − 1)2 + y2 = 1. The cylinder has radius 1 and is centered at (1,0,0), so x ≥ 0. Choosingspherical coordinates, the equation 4z2 = x2 + y2 becomes 4ρ2 cos2ϕ = ρ2 sin2ϕ, hencetan2ϕ = 4, which implies tanϕ = ±2. Since x ≥ 0, we only need to consider tanϕ = 2 orϕ = arctan2. Before proceeding, the problem can be restated: Determine the surface areaof the portion of the cone ϕ = arctan2 which is inside the given cylinder. Converting theequation of the cylinder to spherical coordinates, we have

(ρ sinϕ cosθ − 1)2 + ρ2 sin2ϕ sin2θ = 1.

Simplifying and solving this equation for ρ yields two solutions

ρ = 0 and ρ =2cosθsinϕ

.


Since ϕ = arctan2 is fixed, the denominator of the second solution is sin(arctan2) =2√5

,

henceρ =√

5cosθ

is the second solution and 0 ≤ ρ ≤ √5cosθ. Now, sinϕ =2√5

and cosϕ =1√5

, so the

portion of the cone inside the cylinder can be parametrized by

F(ρ,θ) =(

2ρ√5

cosθ,2ρ√

5sinθ,

ρ√5

),

where 0 ≤ ρ ≤ √5cosθ and −π/2 ≤ θ ≤ π/2. Notice that the bounds for θ correspond tox ≥ 0.

With the parametrization taken care of, we are now ready to compute the surface area,starting with the tangent vectors

∂F∂ρ

=(

2√5

cosθ,2√5

sinθ,1√5

)

and

∂F∂θ

=(− 2ρ√

5sinθ,

2ρ√5

cosθ,0).

The corresponding normal vector is

∂F∂ρ× ∂F∂θ

=(−2ρ

5cosθ,−2ρ

5sinθ,

4ρ5

),

with norm ∥∥∥∥∥∂F∂ρ× ∂F∂θ

∥∥∥∥∥ =2√5ρ.

Hence, the surface area is

∫ π/2

−π/2

∫ √5cosθ

0

2√5ρdρdθ =

∫ π/2

−π/25√5

cos2θdθ

=

√5

2

(θ +

12

sin(2θ))∣∣∣∣π2

−π2

=

√5

2π.

®


§.. Line integrals and work

If γ : [a,b]→Rn parametrizes a curve, where [a,b] ⊂R, and ω is a covector field on R

n,then the line integral (or path integral) of ω over the curve (parametrized by) γ is

∫

γω =

b∫

a

γ∗ω, (.)

where γ∗ω is the pullback of ω by γ .

Example . Consider the covector field ω = xdx − y dy on R2 and let

γ(t) = (cos t,sin t), 0 ≤ t ≤ 2π, parametrize the unit circle, oriented counterclockwise. Tocompute the line integral ∫

γω,

we first compute the pullback

γ∗ω = cos t d(cos t)− sin t d(sin t)= cos t(−sin t dt)− sin t(cos t dt)= −2sin t cos t dt.

Thus, we have∫

γω =

∫ 2π

0γ∗ω

= −2∫ 2π

0sin t cos t dt

= −sin2 t∣∣∣∣2π

0

= 0.

®

Now, suppose ω = P dx+Qdy +Rdz is a covector field on R3, where P ,Q,R are

real-valued functions of x,y,z. The covector field ω corresponds to the vector fieldω] = (P ,Q,R). If γ(t) = (x(t), y(t), z(t)) parametrizes a curve in R

3, then

γ∗ω = P (γ(t))d(x(t)) +Q(γ(t))d(y(t)) +R(γ(t))d(z(t))= P (γ(t))x′(t)dt +Q(γ(t))y′(t)dt +R(γ(t))z′(t)dt= (P (γ(t))x′(t) +Q(γ(t))y′(t) +R(γ(t))z′(t)) dt

= 〈ω](γ(t)),γ ′(t)〉dt.

(.)


What this tells us is that computing the pullback of the covector field ω by aparametrized curve γ is the same as evaluating the corresponding vector field at γ(t) andtaking the inner product with the tangent vector γ ′(t). In general, if X is a vector field on

Rn and γ : [a,b]→R

n, the line integral of X over γ , denoted∫

γX · ds, is defined by

∫

γX · ds =

∫ b

a〈X(γ(t)),γ ′(t)〉dt. (.)

By (.), the line integral of a covector field is the same as the line integral of thecorresponding vector field and vice versa. If X is a vector field, then

∫

γX · ds =

∫

γX[. (.)

Equivalently, if ω is covector field, then∫

γω =

∫

γω] · ds. (.)

If the vector field X is a force field and γ(t) is the position of a particle at time t, the line

integral∫

γX · ds represents the work done on the particle as it traverses from γ(a) to

γ(b).

Example . Let’s compute the work done by the force field X = (x,y,z) in moving aparticle along y = x2, z = 0, from x = −1 to x = 2. Whether we prefer to use the vectorfield X or the covector field X[ = xdx+ y dy + zdz, we must first determine aparametrization for the curve. Take γ(t) = (t, t2,0), with −1 ≤ t ≤ 2. The pullback of X[

by γ isγ∗(X[) = t dt + t2(2t dt) + 0,

hence the work done is∫ 2

−1γ∗(X[) =

∫ 2

−1(t + 2t3)dt

=(12t2 +

12t4)∣∣∣∣

2

−1

= (2 + 8)− (1/2 + 1/2)= 9.

®


M

γ

Figure .: The region M is bounded by the simple closed curve parametrized by γ .

Definitions.

..) A curve parametrized by γ : [a,b]→Rn is called simple if it does not cross, or

intersect, itself. In other words, the curve is simple if γ is one-to-one (i.e., γ(t) = γ(s) ifand only if t = s), except possibly at the endpoints of [a,b].

..) A curve parametrized by γ : [a,b]→Rn is called closed if γ(a) = γ(b).

Remark .. If γ parametrizes a closed curve, the notations γω,

�γω

are sometimes used to denote the line integrals of ω with respect to the clockwise andcounterclockwise orientations, respectively.

For example, the unit circle, parametrized by γ(t) = (cos t,sin t), 0 ≤ t ≤ 2π, is simple andclosed. The helix γ(t) = (cos t,sin t, t), 0 ≤ t ≤ 2π, is simple but not closed. There arecertainly examples of closed curves which are not simple, such as a figure eight in theplane.

Suppose M is a region in the plane bounded by a simple closed curve, which isparametrized by γ(t) = (x(t), y(t)), where a ≤ t ≤ b. A region of this type is depicted inFigure ..

The area of M can be found by integrating the 2-form dx∧ dy over M. That is, if A(M)denotes the area of M, then

A(M) =∫

Mdx∧ dy. (.)


Since the boundary of M is the image of γ , the solid region M may be parametrized byF(r, t) = (rx(t), ry(t)), where 0 ≤ r ≤ 1 and a ≤ t ≤ b. Thus, the area of M satisfies

A(M) =∫

Mdx∧ dy =

∫F∗(dx∧ dy). (.)

The pullback of dx∧ dy by F is

F∗(dx∧ dy) = d(rx(t))∧ d(ry(t))= (x(t)dr + rx′(t)dt)∧ (y(t)dr + ry′(t)dt)= (rx(t)y′(t)− rx′(t)y(t))dr ∧ dt.

Hence, the area of M is given by∫F∗(dx∧ dy) =

∫ b

a

∫ 1

0r(x(t)y′(t)− x′(t)y(t))dr dt

=12

∫ b

a(x(t)y′(t)− x′(t)y(t))dt.

This proves the first part of the following proposition.

Proposition .. Let γ(t) = (x(t), y(t)) be a parametrization of a simple, closed plane curve,where a ≤ t ≤ b. Assume the curve is positively oriented (counterclockwise).

. The area bounded by γ is given by

12

∫ b

a(x(t)y′(t)− x′(t)y(t))dt. (.)

. The area bounded by γ is given by the line integral∫

γω, (.)

where ω is the covector field

ω =12

(−y dx+ xdy). (.)

Proof. To show that (.) holds, we first compute the pullback

γ∗ω =12

(−y(t)d(x(t)) + x(t)d(y(t)))

=12

(−y(t)x′(t)dt + x(t)y′(t)dt).


By definition, ∫

γω =

∫ b

aγ∗ω,

which is precisely the same as (.).

We will soon see that the above formulas for the area bounded by a simple, closed planecurve can also be seen as a consequence of Stokes Theorem.

Examples.

..) Let M be the disc of radius a > 0 in the xy-plane, which can be parametrized byF(r,θ) = (r cosθ,r sinθ) for 0 ≤ r ≤ a and 0 ≤ θ ≤ 2π. According to (.), sinceF∗(dx∧ dy) = r dr ∧ dθ, the area of the disc is

∫ 2π

0

∫ a

0r dr dθ =

∫ 2π

0

a2

2dθ

= πa2.

..) Let’s compute the area bounded by the plane curve defined by the polarequation r = 1 + sinθ. Since x = r cosθ and y = sinθ, we can parametrize the curve withγ(θ) = ((1 + sinθ)cosθ, (1 + sinθ)sinθ), where 0 ≤ θ ≤ 2π. If ω is given by (.), thepullback by γ is

γ∗ω =12

((cosθ + sinθ cosθ)(cosθ + 2sinθ cosθ)dθ

−(sinθ + sin2θ)(−sinθ + cos2θ − sin2θ)dθ)

=12

(1 + 2sinθ cos2θ + sin2θ cos2θ + 2sin3θ + sin4θ)dθ

=12

(1 + 2sinθ cos2θ + sin2θ + 2sin3θ)dθ

=12

(1 + 2sinθ + sin2θ)dθ. (.)

After applying the identity sin2θ = 1/2− 1/2 cos2θ, we integrate

∫ 2π

0

(34

+ sinθ − 14

cos2θ)dθ =

(34θ − cosθ − 1

8sin2θ

)∣∣∣∣2π

0

=3π2.


®

Consider the special case of a simple, closed plane curve which can be expressed interms of a polar equation r = f (θ), where a ≤ θ ≤ b. The curve can be parametrized byγ(θ) = (f (θ)cosθ,f (θ)sinθ), where a ≤ θ ≤ b. Then,

γ ′(θ) = (f ′(θ)cosθ − f (θ)sinθ,f ′(θ)sinθ + f (θ)cosθ) .

hence, we have

x(θ)y′(θ)− x′(θ)y(θ) =f (θ)cosθ (f ′(θ)sinθ + f (θ)cosθ)− (f ′(θ)cosθ − f (θ)sinθ)f (θ)sinθ

=f (θ)f ′(θ)sinθ cosθ + f (θ)2 cos2θ

− f (θ)f ′(θ)sinθ cosθ + f (θ)2 sin2θ

=f (θ)2(sin2θ + cos2θ)

=f (θ)2.

Applying (.), the area bounded by the curve is given by

12

∫ b

a(x(θ)y′(θ)− x′(θ)y(θ))dθ =

12

∫ b

af (θ)2dθ. (.)

The area bounded by the curve of Example .. can be computed by (.). Here,f (θ) = 1 + sinθ, where 0 ≤ θ ≤ 2π. The area is

12

∫ 2π

0f (θ)2dθ =

12

∫ 2π

0(1 + sinθ)2dθ

=12

∫ 2π

0(1 + 2sinθ + sin2θ)dθ,

which is the integral of the differential form obtained in (.).

Arc length

Let γ : [a,b]→Rn be a parametrization of a curve, where [a,b] ⊂R. The arc length of γ ,

denoted L(γ), is defined by

L(γ) =∫ b

a‖γ ′(t)‖dt. (.)

Recall that ‖γ ′(t)‖ is the speed. If ‖γ ′(t)‖ ≡ 1 (for all t), then γ is called a unit speedcurve, or, more precisely, a unit speed parametrization of a curve. If γ is a unit speedcurve, then

L(γ) =∫ b

adt = b − a.


Since the arc length equals the length of the interval [a,b], a unit speed curve is alsocalled parametrized by arc length.

Examples.

..) Let γ(t) = (acos t,asin t), where a > 0, which parametrizes the circle of radius a,centered at the origin. Then γ ′(t) = (−asin t,acos t) and ‖γ ′(t)‖ =

√a2 sin2 t + a2 cos2 t = a.

Hence,

L(γ) =∫ 2π

0adt = 2πa.

..) Let γ(t) = (et cos t, et sin t) for 0 ≤ t ≤ 1. The tangent (velocity) vector isγ ′(t) = (et cos t − et sin t, et sin t + et cos t) and

‖γ ′(t)‖ =(e2t(cos t − sin t)2 + e2t(sin t + cos t)2

)1/2

=(e2t(cos2 t − 2cos t sin t + sin2 t + sin2 t + 2cos t sin t + cos2 t)

)1/2

=√

2e2t

=√

2et.

So, the length of the curve is

L(γ) =∫ 1

0

√2et dt =

√2et

∣∣∣∣1

0=√

2(e − 1).

®

§.. Orientations

In many cases, in order to properly integrate differential forms over surfaces andparametrized curves, we need to discuss how to orient them, when possible. Inparticular, when we are calculating a physical quantity, such as flux or work, the wronganswer can cause misinterpretation. Luckily, if the opposite orientation is chosen and allother information is correct, our calculation will only differ from the true answer by anegative. For instance, if γ(t) = (cos t,sin t) is a parametrization of the unit circle,representing the position of a particle, then the particle travels counterclockwise. On theother hand, we can just as well parametrize the unit circle by γ̃(t) = (cos t,−sin t), for


which a particle would travel clockwise. By choosing γ or γ̃ , we are orienting the unitcircle which, among other things, will allow us to appropriately define a path integralwith respect to the oriented unit circle.

For our purposes, we are mainly interested in orienting curves,surfaces and othergeometric objects in R

2 or R3. However, we may benefit from a brief discussion oforienting vector spaces. If V is a vector space, such as R2 or R3, then specifying anordered basis of V determines an orientation of V . For instance, {e1,e2} and {e2,e1} aretwo, distinct ordered bases for R2 and, as it turns out, these two bases determinedifferent orientations of R2. An orientation of R2 (or a 2-dimensional subspace) isdetermined by the direction of rotation from the first basis vector to the second. Byorienting R

2 in this way, we are assigning a positive direction for the measurement ofangles from a fixed axis. For the ordered basis {e1,e2}, the direction of rotation iscounterclockwise, which agrees with our convention that the positive direction forangles in the plane is counterclockwise. On the other hand, the ordered basis {e2,e1}specifies a clockwise direction of rotation, as depicted in Figure .. A geometric objectin R

2 whose orientation agrees with the ordered basis {e1,e2} will be called positivelyoriented. For example, the parametrization γ(t) = (cos t,sin t) is positively oriented, as aparticle would travel in the counterclockwise direction. A geometric object in R

2 whoseorientation agrees with the ordered basis {e2,e1} will be called negatively oriented. Forexample, the parametrization γ̃(t) = (cos t,−sin t) is negatively oriented, as a particlewould travel in the clockwise direction.

x

y

e1

e2 e1

e2{e2,e1}

{e1,e2}

Figure .: The positive orientation of R2 is specified by the ordered basis {e1,e2}, whilethe negative orientation is specified by the ordered basis {e2,e1}.

In general, to orient a curve, we specify a direction of traversal. To specify this direction,imagine the tangent line to the curve. If we ignore the zero tangent vector, at each pointon the curve, the tangent line consists of two half-lines. At each point, the tangentvectors on these half-lines are pointed in opposing directions, as depicted for the unit


circle in Figure .. By picking one of these half-lines, we specify a direction of traversal,and thus, an orientation.

x

y

Figure .: Orientations specified by two tangent half-lines along the unit circle.

An orientation of a surface M in R3 is a continuous assignment of a direction of rotation,

to the tangent space TpM at each point p of M. In other words, we choose a direction ofrotation between the basis vectors for TpM and this choice varies continuously over thesurface. This choice varies continuously in the sense that it does not change abruptly. Ifsuch a continuous assignment can be made, the surface is called orientable. If a surfaceconsists of more than one piece (e.g. a cylinder with a top), it is orientable if each piececan be assigned an orientation. A surface consisting of multiple smooth pieces is calledpiecewise smooth. One way to specify such a direction of rotation, is to specify a vectorwhich is orthogonal (or normal) to the tangent space. If a surface M is parametrized by

F : R2→R3, F(u,v) = (x(u,v), y(u,v), z(u,v)), then

∂F∂u

and∂F∂v

are tangent vectors. These

two vectors form a basis for the tangent space at each point. Just as with R2, if we


specify an ordered basis for TpM, we specify a direction of rotation. Now, the vector∂F∂u× ∂F∂v

is orthogonal to the tangent space, but so is

−∂F∂u× ∂F∂v

=∂F∂v× ∂F∂u.

When we specify a direction of rotation between the basis vectors for TpM, we areimplicity choosing

∂F∂u× ∂F∂v

or∂F∂v× ∂F∂u.

So, which one do we choose? We choose the normal vector, so that when taken with thechosen ordered basis for TpM, we obtain a basis for R3 that agrees with the right-handrule. For example, if we choose the ordered basis

{∂F∂u,∂F∂v

}

for TpM, then we should choose∂F∂u× ∂F∂v

since

{∂F∂u,∂F∂v,∂F∂u× ∂F∂v

}

is a right-handed basis of R3. If we had chosen the ordered basis

{∂F∂v,∂F∂u

}

for TpM, then we should choose∂F∂v× ∂F∂u

since

{∂F∂v,∂F∂u,∂F∂v× ∂F∂u

}

is a right-handed basis of R3. A surface M whose orientation agrees with the right-handrule is called positively oriented. Likewise, any right-handed basis of R3 corresponds tothe positive orientation of R3. This discussion relies on properties of the cross product,which are summed up in the following proposition.

Proposition .. If v1 and v2 are linearly independent vectors in R3, then the ordered basis

{v1,v2,v1 × v2} is positively oriented.


Based on this proposition and (.)-(.) in § §.., the ordered bases {e1,e2,e3},{e2,e3,e1} and {e3,e1,e2} are right-handed, or positively oriented. On the other hand, thebases {e2,e1,e3}, {e3,e2,e1} and {e1,e3,e2} are left-handed, or negatively oriented.

In textbooks, it is conventional to specify an orientation by choosing a normal vector tothe surface. For a surface, the choice of normal vector is specified as upward ordownward, whenever it is clear which one is which. If a surface is closed, such as asphere, the choice is betwen inward or outward normal. If a normal vector is specifiedfirst, we imagine that it points out of one side of the surface. When the surface is viewedfrom this side, the direction of rotation between tangent vectors is counterclockwise.Below, we outline a simple procedure for ensuring the correct answer is obtained whenintegrating over oriented surfaces.

When integrating over a surface with respect to a given orientation, or normal vec-tor:

. Choose a parametrization for the surface.

. If the parametrization agrees with the specified orientation (or normal vector)then integrate as usual.

. If the parametrization does not agree, then choose a new parametrization thatdoes agree or simply integrate as usual and negate the final answer!

Example . Let X = (xzey ,−xzey , z). Determine the flux across the portion of the planex+ y + z = 1 in the first octant, oriented according to the downward normal. Theprojection of the plane x+ y + z = 1, in the first octant, onto the xy-plane is the triangularregion where 0 ≤ x ≤ 1 and 0 ≤ y ≤ 1− x. The most straightforward parametrization isgiven by

ψ(u,v) = (u,v,1−u − v),

where 0 ≤ u ≤ 1 and 0 ≤ v ≤ 1−u. The tangent vectors are

∂ψ

∂u= (1,0,−1)

and

∂ψ

∂v= (0,1,−1).

The cross product∂ψ

∂u× ∂ψ∂v

= (1,1,1)


is normal to the surface, but it is upward pointing. Using this parametrization, the flux is∫ 1

0

∫ 1−u

01−u − v dv du =

∫ 1

0

((1−u)v − 1

2v2

)∣∣∣∣1−u

0du

=∫ 1

0

12

(1−u)2du

=16.

But, since the chosen parametrization has opposite orientation, the flux according to the

downward normal is −16

. ®

Now, suppose M is a solid body in R3. For example, the unit ball, defined by

{(x,y,z) |x2 + y2 + z2 ≤ 1}in Cartesian coordinates, is a solid body. The boundary of M is denoted ∂M. If M is theunit ball, then ∂M is the unit sphere. If M is an oriented solid body, with boundary ∂M,then the orientation of M induces an orientation on ∂M. The induced orientation on∂M is such that the outward pointing normal to M at ∂M, followed by the orientation of∂M, agrees with the orientation of M. For instance, consider the top of a cylinder (adisc), the boundary of which is the outer circle. The induced orientation on theboundary is such that rotation is counterclockwise, when viewed from the side thenormal vector points out of. In this way, the induced orientation followed by the chosennormal vector agrees with the right hand rule. Now, for piecewise smooth surfaces, theboundaries which are common to two pieces are oriented in opposing directions. Forexample, see the oriented cylinder depicted in Figure ..

Example . Let the surface M be the graph of z = x2 − y2 over the region x2 + y2 ≤ 1,y ≥ 0. Determine the flux of X = (x,y,−2y2) across M with respect to the upwardpointing normal. In cylindrical coordinates, z = r2 cos2θ − r2 sin2θ = r2 cos(2θ), hence

F(r,θ) = (r cosθ,r sinθ,r2 cos(2θ))

parametrizes M. Since x2 + y2 ≤ 1, we must have 0 ≤ r ≤ 1 and since y ≥ 0, we need0 ≤ θ ≤ π. The tangent vectors are

∂F∂u

= (cosθ,sinθ,2r cos(2θ))

and

∂F∂v

=(−r sinθ,r cosθ,−2r2 sin(2θ)

).


n

n

n

Figure .: The common boundaries of the cylinder are oriented in opposite directionswith respect to the corresponding pieces of the surface.

Taking {∂F∂u,∂F∂v

}

as an ordered basis for the tangent space, TpM, the basis

{∂F∂u,∂F∂v,∂F∂u× ∂F∂v

}

for R3 is right-handed, where

∂F∂u× ∂F∂v

=(−2r2(sinθ sin(2θ) + cosθ cos(2θ)),−2r2(sinθ cos(2θ)− cosθ sin(2θ)), r

)

=(−2r2 cosθ,2r2 sinθ,r

).


Since r > 0, the z-component is positive, hence∂F∂u× ∂F∂v

is an upward pointing normal.Now,

X(F) = (r cosθ,r sinθ,−2r2 sin2θ),

so ⟨X(F),

∂F∂u× ∂F∂v

⟩= −2r3 cos2θ + 2r3 sin2θ − 2r3 sin2θ = −2r3 cos2θ.

So, the flux is given by∫ π

0

∫ 1

0−2r3 cos2θdr dθ =

∫ π

0

(−1

2r4 cos2θ

)∣∣∣∣1

0dθ

= −12

∫ π

0cos2θdθ

= −12

∫ π

0

12

(1 + cos(2θ))

= −14

(θ +

12

sin(2θ))∣∣∣∣π

0

= −π4.

Since the chosen parametrization for M agrees with the specified orientation, the flux ofX across M in the direction of the upward pointing normal is −π/4. Since this answer isnegative, the net flow is actually in the opposite direction; in the direction of thedownward pointing normal.

®

Suppose, in the last example, we had been asked to determine the flux of X across M inthe direction of the downward pointing normal. Using the same parametrization for M,we would have to negate the final answer, since ∂F

∂u × ∂F∂v is upward pointing. The numericanswer would be different, but the result would be the same. Indeed, we would arrive atan answer of π/4. Since the answer is positive, the flow is in the direction of thedownward normal.

§.. Integration of 3-forms

Suppose M is a (solid) 3-dimensional region and ω is a 3-form on R3. If F :D ⊂R

3→R3

is a parametrization of M, then the integral of ω over M can be computed by the samegeneral procedure used to integrate 1-forms and 2-forms. The integral of ω over M is

∫

Mω =

∫

DF∗ω. (.)


Recall that any 3-form ω on R3 can be expressed as ω = f dx∧ dy ∧ dz, where f : R3→R

is a real-valued function of x,y,z. In particular, if V (M) denotes the volume of M, then

V (M) =∫

Mdx∧ dy ∧ dz. (.)

For this reason, the elementary 3-form dx∧ dy ∧ dz is referred to as the volume form.

Example .

®

Example . Let M be the unit ball (the solid sphere of radius 1) and let

ω =1√

2 + x2 + y2 + z2dx∧ dy ∧ dz.

Let’s compute ∫

Mω.

For the unit ball, M, the radius satisfies 0 ≤ ρ ≤ 1. We take the standard sphericalcoordinate parametrization

F(ρ,θ,ϕ) = (ρ sinϕ cosθ,ρ sinϕ sinθ,ρcosϕ),

with 0 ≤ ρ ≤ 1, 0 ≤ θ ≤ 2π and 0 ≤ ϕ ≤ π. As with integrating 1-forms and 2-forms, thefirst step is to compute the pullback of ω by F. First, we compute F∗(dx∧dy ∧dz), wherewe use (.) and (.). We have

F∗dx = sinϕ cosθdρ+ ρcosϕ cosθdϕ − ρ sinϕ sinθdθF∗dy = sinϕ sinθdρ+ ρcosϕ sinθdϕ + ρ sinϕ cosθdθF∗dz = cosϕdρ − ρ sinϕdϕ.

Taking the wedge product of F∗dx with F∗dy, we have

F∗(dx∧ dy) = (F∗dx)∧ (F∗dy)

= (ρ sinϕ cosϕ sinθ cosθ − ρ sinϕ cosϕ sinθ cosθ) dρ∧ dϕ+(ρ sin2ϕ cos2θ + ρ sin2ϕ sin2θ

)dρ∧ dθ

+(ρ2 sinϕ cosϕ cos2θ + ρ2 sinϕ cosϕ sin2θ

)dϕ ∧ dθ

= ρ sin2ϕdρ∧ dθ + ρ2 sinϕ cosϕdϕ ∧ dθ.


Taking the wedge product of F∗(dx∧ dy) with F∗dz, we have

F∗(dx∧ dy ∧ dz) =(ρ sin2ϕdρ∧ dθ + ρ2 sinϕ cosϕdϕ ∧ dθ

)∧ (cosϕdρ − ρ sinϕdϕ)

= (ρ2 sin3ϕ + ρ2 sinϕ cos2ϕ)dρ∧ dϕ ∧ dθ= ρ2 sinϕdρ∧ dϕ ∧ dθ.

Finally, using (.), we have

F∗ω =1√

2 + ρ2ρ2 sinϕdρ∧ dϕ ∧ dθ.

Thus, to compute the integral of ω over M, we compute the triple integral∫ 2π

0

∫ π

0

∫ 1

0

ρ2 sinϕ√2 + ρ2

dρdϕdθ. (.)

The integral with respect to ρ is given by the formula

∫ρ2

√2 + ρ2

dρ =ρ√

2 + ρ2

2− log

(ρ+

√2 + ρ2

),

hence, the innermost integral evaluated between 0 and 1 is(√

32− log(1 +

√3)

)+ log(

√2).

Let

C =(√

32− log(1 +

√3)

)+ log(

√2).

Then, the triple integral (.) becomes∫ 2π

0

∫ π

0C sinϕdϕdθ =

∫ 2π

0(−C cosϕ)

∣∣∣∣π

0dθ

= 2C∫ 2π

0dθ

= 4πC.

Hence, the final answer is

4π((√

32− log(1 +

√3)

)+ log(

√2)

).

®



P -. Compute the surface area of the surface described in Example ..

P -. Let ω = xdx − y dy and γ(t) = (cos t,sin t) for 0 ≤ t ≤ 2π. Compute∫

γω.

P -. Let ω = yzdx+ xzdy + xy dz and let γ(t) be a parametrization of the path from(1,0,0) to (0,1,0) to (0,0,1). Compute

∫

γω.

P -. Let γ : [0,π/2]→R2 be defined by γ(t) = (sin4 t,cos4 t).

a) Determine whether the curve is oriented counterclockwise or clockwise.

b) Compute the area bounded by γ .

P -. Determine the area of the following surfaces.

a) The part of the sphere x2 + y2 + z2 = 4z that lies inside the paraboloid z = x2 + y2.

b) The part of the sphere x2 + y2 + z2 = a2 that lies inside the cylinder x2 + y2 = ax, wherea > 0.

c) The part of the cone z =√x2 + y2 that lies between the plane y = x and the cylinder

y = x2.

d) The part of the paraboloid y = x2 + z2 that lies within the cylinder x2 + z2 = 16.

e) The part of the paraboloid z = x2 + y2 between z = 0 and z = 1, including the top( theintersection of z = 0 and z = x2 + y2).


f) The surface described by z = xy, where x2 + y2 ≤ 2.

P -. Let ω = e−(x2+y2)dx∧ dy and let M be the half-disc in the plane described byx2 + y2 ≤ 1, y ≤ 0. Compute ∫

Mω.

P -. Determine the area bounded by the lemniscate (x2 + y2)2 = 2a2(x2 − y2), a > 0.

P -. Determine the area bounded by the curve described by the polar equationr = 1 + sinθ.

P -. Determine the flux of curlX across the surface x2 + y2 + 3z2 = 1, z ≤ 0, withrespect to the upward pointing normal, where X = (y,−x,zx3y2).


Chapter

Stokes’ Theorem

§ Surfaces with boundary

In order to specialize the generalized Stokes’ theorem to particular cases, we will unifysome of the terminology and notation for the geometric objects we have been studying.Let M be such a geometric object. We will refer to M as an n-dimensional surface, ingeneral, or an n-surface, for short. Here, the word surface can be replaced with manifold.We are primarily interested in the types of n-surfaces that are subsets of R3. In R

3, thereare 4 types of n-surfaces that we discuss.

Types of n-surfaces in R3:

. (n = 0) A 0-surface in R3 is simply a point, or collection of points.

. (n = 1) A 1-surface is a curve, such as a line segment or any parametrizedcurve.

. (n = 2) A 2-surface is, up until this point, what we have called, simply, a sur-face. Examples of 2-surfaces include planes, cones and cylinders.

. (n = 3) A 3-surface is a solid region, e.g. the unit ball.

Some surfaces have a boundary. If M is an n-surface with boundary,where n ≥ 1, thenthe boundary of M, ∂M, is an (n− 1)-dimensional surface. For example, suppose M is acurve (a 1-surface) which can be parametrized by γ(t), where a ≤ t ≤ b. Then∂M = {γ(a),γ(b)}. That is, the boundary of the 1-surface M is the 0-surface consisting ofjust the end points γ(a) and γ(b). As another example, suppose M is the upper


hemisphere of radius 1, centered at the origin. Then M is a 2-surface and ∂M is theequator; the unit circle in the xy-plane, which is a 1-surface.

§ The generalized Stokes’ Theorem

Theorem . (Stokes’ Theorem). Suppose M is an oriented surface of dimension n with an(n− 1)-dimensional boundary ∂M. If ω is a smooth (n− 1)-form on M, then

∫

Mdω =

∫

∂Mω. (.)

Remark .. A few comments are in order before we delve into the different cases ofStokes’ Theorem. The boundary ∂M should be understood to have the inducedorientation. The differential form ω in the right hand side of (.) is really ω

∣∣∣∂M

, therestriction of ω to the boundary. If M does not have a boundary (∂M = ∅), then the righthand side is 0. If M is 1-dimensional, then the right hand side of (.) is a finite sum.

§.. Stokes’ Theorem for 1-surfaces

Let us first consider the case when M is a 1-surface in R3 and ω is a 0-form on M. Recall

that a 0-form is just a function, which we will denote by f instead of ω. Supposeγ : [a,b]→R

3 parametrizes M, so M = γ([a,b]). The left hand side of (.) is

∫

γdf =

∫ b

aγ∗(df ),

which is just the line integral of the 1-form df . Now, as we mentioned in the last section,the boundary of M = γ([a,b]) consists of the endpoints, ∂M = {γ(a),γ(b)}. Then, Stokes’Theorem says ∫

γdf =

∫ b

aγ∗(df ) =

∫

∂Mf = f (γ(b))− f (γ(a)).

This version of Stokes’ Theorem is called the Fundamental Theorem for line integrals.

Theorem . (Fundamental Theorem for line integrals). Suppose f : Rn→R is a smoothreal-valued function and γ : [a,b]→R

n is smooth parametrization of a curve in Rn. Then

∫

γdf = f (γ(b))− f (γ(a)). (.)


Corollary .. If ω is an exact covector field on Rn with potential function f , then

∫

γω = f (γ(b))− f (γ(a)). (.)

Example . Let γ(t) = (cos4 t,sin2 t + cos3 t, t), 0 ≤ t ≤ π, and ω = xdx+ y dy + zdz. Wewant to compute the line integral ∫

γω.

The covector field ω is exact, since the function

f (x,y,z) =12x2 +

12y2 +

12z2

is a potential for ω. That is,

df = xdx+ y dy + zdz =ω.

The end points of γ (the boundary) are

γ(0) = (1,1,0)

and

γ(π) = (1,−1,π).

By Theorem .,∫

γω = f (γ(π))− f (γ(0))

=12

(12 + 12 + 02

)− 1

2

(12 + (−1)2 +π2

)

=12π2.

®

Example . As another example, consider the covector fieldω = 2xyzdx+ x2zdy + x2y dz and the curve γ(t) = (t2 + 1, t2 − 1, t5 + 1) for 1 ≤ t ≤ 2.Computing the line integral ∫

γω


directly, we first compute the pullback

γ∗ω = 2(t2 + 1)(t2 − 1)(t5 + 1)2t dt + (t2 + 1)2(t5 + 1)2t dt + (t2 + 1)2(t2 − 1)5t4dt

=(4t(t4 − 1)(t5 + 1) + 2t(t4 + 2t + 1)(t5 + 1) + 5t4(t4 − 1)(t2 + 1)

)dt

=(4t10 + 4t5 − 4t6 − 4t + 2t10 + 2t5 + 4t8 + 4t3 + 2t6 + 2t + 5t10 + 5t8 − 5t6 − 5t4

)dt

=(11t10 + 9t8 − 7t6 + 6t5 − 5t4 + 4t3 − 2t

)dt.

Hence, the line integral is∫

γω =

∫ 2

1

(11t10 + 9t8 − 7t6 + 6t5 − 5t4 + 4t3 − 2t

)dt

=(t11 + t9 − t7 + t6 − t5 + t4 − t2

)∣∣∣∣2

1

=(211 + 29 − 27 + 26 − 25 + 24 − 22

)−(111 + 19 − 17 + 16 − 15 + 14 − 12

)= 2476− 1

= 2475.

Clearly, if ω is exact, it should take less effort to compute the line integral. If f is apotential for ω, then df =ω, so

∂f

∂x= 2xyz

∂f

∂y= x2z

∂f

∂z= x2y.

Integrating with respect to x,y and z, or possibly by inspection, we find a candidatef (x,y,z) = x2yz. Indeed, we can quickly verify that df =ω, so f is a potential for ω. With

γ(1) = (2,0,2)

and

γ(2) = (5,3,33),

by the fundamental theorem for line integrals, we have∫

γω = f (γ(2))− f (γ(1)) = 52(3)(33)− 22(0)(2) = 2475.

®

The following corollary explains why the line integral in Example . is 0.


Corollary .. If ω is an exact covector field on Rn and γ : [a,b]→R

n is a closed curve, then∫

γω = 0. (.)

Proof. Since ω is exact, ω = df for some potential function f : Rn→R. By Theorem .,∫

γω =

∫

γdf = f (γ(b))− f (γ(a)),

but γ(b) = γ(a) since the curve is closed. Thus∫

γω = 0.

Although our attention is focused on R3, Theorem . is valid on R

n. Consider thespecial case when γ : [a,b]→R is given by γ(t) = t. If f : R→R is a real-valued function(0-form), df = f ′(x)dx. Moreover, γ∗(df ) = f ′(t)dt. Hence, in this specific case, Stokes’Theorem reduces to

∫

γdf =

∫ b

aγ∗(df )

=∫ b

af ′(t)dt

= f (γ(b))− f (γ(a))= f (b)− f (a).

Stokes’ theorem, and hence the fundamental theorem for line integrals, reduces to theFundamental Theorem of Calculus!

For vector fields, the conclusion of Theorem . is∫ b

a

⟨gradγ(t) f ,γ

′(t)⟩dt = f (γ(b))− f (γ(a)). (.)

In general, if X is a conservative vector field with potential f , then∫ b

a

⟨X(γ(t)),γ ′(t)

⟩dt = f (γ(b))− f (γ(a)). (.)

Both (.) and (.) are analogous to (.) and (.), hence direct consequences of Stokes’theorem. Note that the vector field version of the fundamental theorem for line integrals


can be seen as a direct consequence of the single-variable fundamental theorem ofcalculus. Indeed, if f : R3→R is a differentiable function and γ : [a,b]→R

3 is a smoothpath, it follows from the chain rule that

ddtf (γ(t)) =

⟨gradγ(t) f ,γ

′(t)⟩.

Hence,∫ b

a

⟨gradγ(t) f ,γ

′(t)⟩dt =

∫ b

a

ddtf (γ(t)) dt

= f (γ(t))∣∣∣∣b

a

= f (γ(b))− f (γ(a)),

by the fundamental theorem of calculus.

Now, when a vector field X is conservative, hence the corresponding covector field isexact, we have seen that the line integral over any closed path is 0, by Corollary .. Infact, we can conclude more than this.

Theorem .. Let ω be a continuous covector field on Rn. The following statements are

equivalent.

a) ω is an exact covector field.

b) ∫

γω = 0

for any closed path γ : [a,b]→Rn.

c) If γ1 and γ2 are any two smooth paths with the same endpoints, then∫

γ1

ω =∫

γ2

ω.

The statements analogous to a)-c) for a continuous vector field X are also equivalent. A1-form satisfying c) of Theorem . is called path independent (or independent ofpath). Analogously, a vector field X satisfying

∫

γ1

X · ds =∫

γ2

X · ds,

for any two paths with common endpoints, is called path independent. Theorem .states that a vector field (covector field) is conservative (exact) if and only if it is pathindependent.


γ1

a)γ2

γ1

b)γ̃2

ξ

c)

Figure .: a) Two smooth paths, γ1 and γ2, with the same endpoints. b) The path γ̃2 hasthe opposite orientation as γ2. c) The closed path ξ, which is the union of γ1 and γ̃2.

Proof. The proof of Corollary . shows that a) implies b). We will show that b) impliesc). Suppose

∫

γω = 0

for any smooth, closed path. Let γ1 : [a,b]→Rn and γ2 : [c,d]→R

n be smooth pathswith common endpoints; that is, γ1(a) = γ2(c) and γ1(b) = γ2(d). Since these paths startand end at the same points, both paths are oriented from the initial point to the finalpoint. Let γ̃2 denote a parametrization of the path γ2 with the opposite orientation, asdepicted in Figure .. Then, the path ξ : [a,b]∪ [c,d]→R

n defined by

ξ(t) =

γ1(t), a ≤ t ≤ b,γ̃2(t), c ≤ t ≤ d

is a closed curve. Hence,∫

ξω = 0.


Moreover, since γ2 and γ̃2 have opposite orientations, we have

0 =∫

ξω

=∫

γ1

ω+∫

γ̃2

ω

=∫

γ1

ω −∫

γ2

ω,

thus ∫

γ1

ω =∫

γ2

ω.

Conservation of Energy

Suppose X is a vector force field on Rn acting on a particle traversing a curve

parametrized by γ : [a,b]→Rn. We now know that the work done by X on the particle

as it moves from γ(a) to γ(b) is given by the line integral

∫

γX =

∫ b

a〈X(γ),γ ′〉dt.

If the particle has mass m, then the kinetic energy of the particle, Ek(t), is given by

Ek(t) =12m‖γ ′(t)‖2. (.)

If f : Rn→R is a potential for X, then the potential energy of the particle, Ep(t), is givenby

Ep(t) = −f (γ(t)). (.)

The total energy, E(t), is the sum of the kinetic and potential energies

E(t) = Ek(t) +Ep(t). (.)

Consider that the rate of change in the kinetic energy is

ddtEk(t) =

12m (〈γ ′′(t),γ ′(t)〉+ 〈γ ′(t),γ ′′(t)〉)

=m〈γ ′′(t),γ ′(t)〉.


By Newton’s Second Law (force=mass×acceleration), X(γ(t)) =mγ ′′(t), hence

ddtEk(t) = 〈X(γ(t)),γ ′(t)〉. (.)

Integrating the left-hand side of (.), we have

∫ b

a

ddtEk(t)dt = Ek(b)−Ek(a),

by the fundamental theorem of calculus. Integrating the right-hand side of (.), wehave

∫ b

a〈X(γ(t)),γ ′(t)〉dt = f (γ(b))− f (γ(a)) by (.)

= Ep(a)−Ep(b) by (.).

Thus, by (.), we have

Ek(b)−Ek(a) = Ep(a)−Ep(b)

which implies

Ek(b) +Ep(b) = Ek(a) +Ep(a),

orE(b) = E(a).

This shows that the total energy remains the same, that is, energy is conserved. This isprecisely the reason X is called a conservative vector field.


Suppose M is a 2-surface parametrized by F : R2→R3,

F(u,v) = (x(u,v), y(u,v), z(u,v)) .

Let ∂M be parametrized by γ : [a,b]→R3, unless M does not have a boundary. Let ω be

the 1-form ω = f dx+ g dy + hdz, where f ,g,h are real-valued functions of x,y and z. Bythe correspondence between differential forms and vector fields, ω] = X, whereX = (f ,g,h). Moreover, the exterior derivative of ω,

dω =(∂h∂y− ∂g∂z

)dy ∧ dz+

(∂f

∂z− ∂h∂x

)dz∧ dx+

(∂g

∂x− ∂f∂y

)dx∧ dy,


corresponds to the curl of X; specifically,

curlX = (∗dω)].

According to Stokes’ Theorem ., we have∫

Mdω =

∫

∂Mω. (.)

Since dω is a 2-form, the left hand side of (.) is equivalent to the flux integral of thecorresponding vector field � ⟨

curlX,∂F∂u× ∂F∂v

⟩dudv, (.)

where curlX is evaluated on F. Now the right hand side of (.) is the integral of a1-form over a 1-surface, that is, a line integral. The line integral of ω is equivalent to theline integral of the corresponding vector field

∫ b

a

⟨X(γ),γ ′

⟩dt, (.)

where γ([a,b]) = ∂M. The equality of (.) and (.) is the original form of Stokes’Theorem. Since the following version of Stokes’ theorem involves the curl of a vectorfield, it is unique to R

3.

Theorem . (Classical Stokes’ theorem). Let M be a smooth surface in R3 (a 2-surface),

parametrized by F : R2→R3, with boundary ∂M, parametrized by γ : [a,b]→R

3. If X is asmooth vector field on M, then�

M

⟨curlX,

∂F∂u× ∂F∂v

⟩dudv =

∫ b

a

⟨X(γ),γ ′

⟩dt. (.)

The classical version of Stokes’ theorem is often stated in the following manner. SupposeM is a smooth surface, with boundary, in R

3 which is oriented according to the unitnormal vector N . If X is a smooth vector field on M, then�

M〈curlX,N 〉 =

∫

∂M〈X,T 〉 , (.)

where T is the unit tangent to ∂M compatible with the induced orientation of ∂M.

Example . (Problem P-) Let M be the surface x2 + y2 + 3z2 = 1, z ≤ 0, and letX = (y,−x,zx3y2). In problem P- we were asked to compute the flux of curlX across M


with respect to the upward pointing normal. That is, if F : R2→R3 is a parametrization

of M compatible with the upward normal, we computed�M

⟨curlX,

∂F∂u× ∂F∂v

⟩dudv.

Since z ≤ 0, M is the lower half of an ellipsoid, hence it has a boundary. The boundary ofM is

∂M = {(x,y,0) |x2 + y2 = 1},which is the unit circle in the plane z = 0. For a parametrization of ∂M to be compatiblewith the induced orientation, when viewed from above (since the normal to M isupward) the circle should be oriented counterclockwise. Thus, γ(t) = (cos t,sin t,0),0 ≤ t ≤ 2π, is a parametrization of ∂M compatible with the induced orientation. Since

X(γ) = (sin t,−cos t,0)

and

γ ′(t) = (−sin t,cos t,0),

according to Theorem . we have�M

⟨curlX,

∂F∂u× ∂F∂v

⟩dudv =

∫ 2π

0〈X(γ),γ ′〉dt

=∫ 2π

0−sin2 t − cos2 t dt

= −2π.

®

The following theorem is a special case of Stokes’ theorem, applied to regions in R2.

Theorem . (Green’s Theorem). If M is a simple plane region, where the boundary, ∂M,is a simple closed curve, oriented counterclockwise, then

∫

∂MP dx+Qdy =

�M

(∂Q∂x− ∂P∂y

)dxdy (.)

for any real-valued differentiable functions P and Q.

Proof. Any 1-form, ω, on R2 can be expressed as ω = P dx+Qdy, where P and Q are

functions from R2 to R. Since

dω =(∂P∂x

dx+∂P∂y

dy

)∧ dx+

(∂Q∂x

dx+∂Q∂y

dy

)∧ dy

=(∂Q∂x− ∂P∂y

)dx∧ dy,


Stokes’ Theorem . says

∫

∂MP dx+Qdy =

∫

∂Mω =

∫

Mdω =

�M

(∂Q∂x− ∂P∂y

)dxdy.

Suppose ω is a closed covector field on R2, that is, dω = 0. Since any covector field on R

2

can be expressed as ω = P dx+Qdy, we have

dω =(∂Q∂x− ∂P∂y

)dx∧ dy = 0,

which implies∂Q∂x− ∂P∂y

= 0,

or∂Q∂x

=∂P∂y.

Now, suppose ω is closed and exact. Since ω is exact,

ω = df =∂f

∂xdx+

∂f

∂ydy,

for some function f : R2→R. Since ω is closed, dω = 0, thus

∂∂x

∂f

∂y=∂∂y

∂f

∂x.

So, the mixed partial derivatives of f are equal

∂2f

∂x∂y=∂2f

∂y∂x.


Let M be a 3-surface in R3. If F : R3→R

3 parametrizes M, let dVM = F∗(dx∧ dy ∧ dz)denote the volume form on M. That is, dVM is the 3-form such that the volume of Mequals ∫

MdVM .


For example, if M can be parametrized by the spherical change of coordinates

F(ρ,ϕ,θ) = (ρ sinϕ cosθ,ρ sinϕ sinθ,ρcosϕ),

then dVM = ρ2 sinϕdρ∧ dϕ ∧ dθ. If M has a boundary, ∂M, let dVM̃ denote the inducedvolume form on ∂M. For example, the ball of radius a has volume form

dVM = ρ2 sinϕdρ∧ dϕ ∧ dθ.The boundary is the sphere of radius a. Since the radius is now fixed, ρ = a, the volumeform collapses to

dVM̃ = a2 sinϕdϕ ∧ dθ.

Now, if ω = P dy ∧ dz+Qdz∧ dx+Rdx∧ dy is a 3-form on R3, where P ,Q,R are

functions of x,y,z, recall that

dω =(∂P∂x

+∂Q∂y

+∂R∂z

)dx∧ dy ∧ dz.

As detailed in § of Chapter , ω corresponds to the vector field X = (P ,Q,R) and dωcorresponds to the function divX. In terms of the Hodge star operator and musicalnotation, (∗ω)] = X and ∗(dω) = divX. Since ω and dω correspond to X and divX,respectively, Stokes’ Theorem . relates the integral of X to the integral of divX.

Theorem . (Divergence Theorem). If X is a smooth vector field on a 3-surface M, then∫

M(divX)dVM =

∫

∂M〈X,N 〉 dVM̃ , (.)

where N is the outward-pointing unit normal vector field along ∂M.

This statement of the Divergence Theorem is given in terms of 3-surfaces (volumes) inR

3. In this particular case, Theorem . is often called Gauss’ Theorem or Gauss’Divergence Theorem. Moreover, the equality (.) may take the form�

M(∇ ·X)dVM =

�∂M

(X ·N )dVM̃ . (.)

Recall, however, that divergence of a vector field on Rn can always be defined, in

contrast to the curl, which is only defined on R3. Hence, Theorem . holds when X is

a vector field on Rn and M is a smooth n-surface on R

n.

Example . Consider the radial vector field X = (x,y,z). First, let’s directly computethe flux of X across the sphere of radius a, centered at the origin. LetF(θ,ϕ) = (asinϕ cosθ,asinϕ sinθ,acosϕ) for 0 ≤ θ ≤ 2π and 0 ≤ ϕ ≤ π. Since

X(F) = (asinϕ cosθ,asinϕ sinθ,acosϕ)


and

∂F∂θ× ∂F∂ϕ

= (−a2 sin2ϕ cosθ,−a2 sin2ϕ sinθ,−a2 sinϕ cosϕ), (.)

we have ⟨X(F),

∂F∂θ× ∂F∂ϕ

⟩= −a3 sinϕ.

Because∫ 2π

0

∫ π

0−a3 sinϕdϕdθ =

∫ 2π

0a3 cosϕ

∣∣∣∣π

0dθ

=∫ 2π

0−2a3dθ

= −4πa3,

the flux is ±4πa3, depending on the orientation. Notice that the direction of the normalvector (.) depends on cosϕ, since a2 > 0 and sinϕ ≥ 0 for ϕ ∈ [0,π]. Now, cosϕ ≥ 0 forϕ ∈ [0,π/2], on the upper hemisphere, while cosϕ ≤ 0 for ϕ ∈ [π/2,π], on the lowerhemisphere. Due to the negative sign in the z-component of (.), the normal vectorwill point downward in the upper hemisphere and point upward in the lowerhemisphere. That is, we have used the inward pointing normal.

Now, let’s use the Divergence Theorem . to compute the flux. The sphere of radius ais the boundary of the ball of radius a. Our direct computation of the flux corresponds tothe right hand side of (.). Since divX = 3, we need to compute

∫

M3dVM ,

where M is the ball of radius a. Here, we must exercise caution. The volume form dVMdepends on the chosen coordinate system (parametrization) for M. Using sphericalcoordinates, dVM = ρ2 sinϕdρ∧ dϕ ∧ dθ, hence the flux is

∫ 2π

0

∫ π

0

∫ a

03ρ2 sinϕdρdϕdθ =

∫ 2π

0

∫ π

0ρ3

∣∣∣∣a

0sinϕdϕdθ

=∫ 2π

0−a3 cosϕ

∣∣∣∣π

0dθ

= 4πa3.

Stokes’ theorem assumes the outward-pointing normal, hence this answer agrees withthe direct computation of the flux above. ®


Suppose X = (P ,Q,R) is a smooth vector field on R3 and f : R3→R is differentiable.

Then

div(f X) =∂∂x

(f P ) +∂∂y

(f Q) +∂∂z

(f R)

=∂f

∂xP + f

∂P∂x

+∂f

∂yQ+ f

∂Q∂y

+∂f

∂zR+ f

∂R∂z

=(∂f

∂xP +

∂f

∂yQ+

∂f

∂zR

)+ f

(∂P∂x

+∂Q∂y

+∂R∂z

)

= 〈gradf ,X〉+ f divX.

Now, by the Divergence Theorem . we have∫

M(divf X)dVM =

∫

∂Mf 〈X,N 〉 dVM̃ ,

which becomes∫

M〈gradf ,X〉dVM +

∫

Mf divXdVM =

∫

∂Mf 〈X,N 〉 dVM̃ .

The last equality is usually expressed as∫

M〈gradf ,X〉dVM =

∫

∂Mf 〈X,N 〉 dVM̃ −

∫

Mf divXdVM . (.)

The formula (.), which holds on Rn, is sometimes called “integration by parts”. The

identity÷(f X) = 〈gradf ,X〉+ f divX

also holds on Rn.

Suppose f : Rn→R is smooth. The operator ∆, defined by

∆f = div(gradf ), (.)

is called the Laplace operator, or simply Laplacian. In Cartesian coordinates,

gradf =(∂f

∂x1, . . . ,

∂f

∂xn

),

hence

div(gradf ) =∂∂x1

∂f

∂x1+ . . .+

∂∂xn

∂f

∂xn

=∂2f

∂x21

+ . . .+∂2f

∂x2n.


So, the Laplacian of f in Cartesian coordinates is simply the sum of its second partialderivatives

∆f =n∑

i=1

∂2f

∂x2i

. (.)

In particular,

∆f =∂f

∂x+∂f

∂y

and

∆f =∂f

∂x+∂f

∂y+∂f

∂z

on R2 and R

3, respectively. A function f satisfying ∆f = 0 is called harmonic. Letg : Rn→R. Setting X = gradg in (.), we obtain

∫

M〈gradf ,gradg〉dVM +

∫

Mf ∆g dVM =

∫

∂Mf⟨gradg,N

⟩dVM̃ , (.)

which is one of Green’s identities. Swapping the roles of f and g, we have∫

M〈gradg,gradf 〉dVM +

∫

Mg∆f dVM =

∫

∂Mg⟨gradf ,N

⟩dVM̃ . (.)

The difference of (.) and (.) yields another of Green’s identities,∫

M(f ∆g − g∆f )dVM =

∫

∂M

⟨f gradg − g gradf ,N

⟩dVM̃ . (.)

Green’s Identities:Expressed in more compact form, we have

. ∫

M〈∇f ,∇g〉dVM +

∫

Mf ∆g dVM =

∫

∂Mf⟨∇g,N⟩

dVM̃

. ∫

M(f ∆g − g∆f )dVM =

∫

∂M

⟨f ∇g − g∇f ,N⟩

dVM̃

Appendix A

Coordinate representations

Polar coordinates:For standard polar coordinates on R

2,

x = r cosθ, r2 = x2 + y2,

y = r sinθ, θ = tan−1(yx

),

we have the following.

. dVM = r dr ∧ dθ.

gradf =(∂f

∂r,

1r2∂f

∂θ

),

where f : R2→R is a function of r and θ.

. If X = (P ,Q) is a vector field on R2, where P ,Q are functions of r and θ, then

divX =1rP +

∂P∂r

+∂Q∂θ

.


Cylindrical coordinates:For standard cylindrical coordinates on R

3,

x = r cosθ, r2 = x2 + y2,

y = r sinθ, θ = tan−1(yx

),

z = z, z = z,


. dVM = r dr ∧ dθ ∧ dz.

gradf =(∂f

∂r,

1r2∂f

∂θ,∂f

∂z

),

where f : R3→R is a function of r, θ and z.

. If X = (P ,Q,R) is a vector field on R3, where P ,Q,R are functions of r, θ and z,

then

divX =1rP +

∂P∂r

+∂Q∂θ

+∂R∂z.

. If X is a vector field in cylindrical coordinates, then

curlX =(

1r∂R∂θ− r ∂Q

∂z

)∂∂r

+1r

(∂P∂z− ∂R∂r

)∂∂θ

+(2Q+ r

∂Q∂r− 1r∂P∂θ

)∂∂z.


Spherical coordinates:For standard spherical coordinates on R

3,

x = ρ sinϕ cosθ, ρ2 = x2 + y2 + z2,

y = ρ sinϕ sinθ, ϕ = tan−1

√x2 + y2

z

,

z = ρcosϕ, θ = tan−1(yx

),


. dVM = ρ2 sinϕdρ∧ dϕ ∧ dθ.

gradf =(∂f

∂ρ,

1ρ2∂f

∂ϕ,

1

ρ2 sin2ϕ

∂f

∂θ

),

where f : R3→R is a function of ρ, ϕ and θ.

. If X = (P ,Q,R) is a vector field on R3, where P ,Q,R are functions of ρ, ϕ and

θ, then

divX =2ρP +

∂P∂ρ

+Qcotϕ +∂Q∂ϕ

+∂R∂θ.

. If X is a vector field in spherical coordinates, then

curlX =(2cosϕR+ sinϕ

∂R∂ϕ− 1

sinϕ∂Q∂θ

)∂∂ρ

+(

1ρ2 sinϕ

∂P∂θ− 2sinϕ

ρR− sinϕ

∂R∂ρ

)∂∂ϕ

+(

2ρ sinϕ

Q+1

sinϕ∂Q∂ρ− 1ρ2 sinϕ

∂P∂ϕ

)∂∂θ.


Appendix B

Some applications of differential formsand vector calculus

§ Extreme values

Analogous to single-variable calculus, one of the key ideas for determining extrema of isthe rate of change. For a function f : Rn→R, recall that the gradient of f at a point p,gradp f , is pointed in the direction of steepest ascent. If f obtains an extreme value at apoint p, then the gradient should be 0 at p, since a particle on the graph of f could notascend higher (or descend lower). Indeed, this is the case, hence determining the pointsp where gradp f is 0 is the starting place for extreme value theory. Since we now knowthat the vector field gradf corresponds to the covector field df , we will use differentialforms to develop a brief survey of extreme value theory and optimization.

Definition B.. Suppose f : Rn→R. A local minimum occurs at x ∈Rn if f (x) ≤ f (u)for all u in some open set U ⊂R

n containing x. A local maximum occurs at x ∈Rn iff (x) ≥ f (u) for all u in some open set U ⊂R

n containing x. A point x ∈Rn is a criticalpoint for f if either f is not differentiable at x or gradx f = 0. A critcal point a at which fdoes not attain a local extreme value is called a saddle point.

Theorem B.. Suppose f : Rn→R is differentiable on an open set U ⊂Rn containing x. If f

achieves a local extremum at x, then gradx f = 0.

This theorem states that any extrema of f occur at critical points. For, if f achieves alocal extremum at a point x where it is not differentiable, then x is still a critical point.

Example B. Let f : R2→R be defined by f (x,y) = e−(x2+y2). The differential of f is

df = −2xe−(x2+y2)dx − 2ye−(x2+y2)dy.


The covector field df equals 0 if and only if each coefficient is 0. Hence, to determineany critical points, we start by solving

−2xe−(x2+y2) = 0 and − 2ye−(x2+y2) = 0,

which yields x = 0 and y = 0, since e−(x2+y2) , 0. The function f is differentiableeverywhere, so (x,y) = (0,0) is the only critical point. Indeed, f achieves a local (in fact,global) maximum at (0,0), where f (0,0) = 1. ®

Definition B.. If f : Rn→R is defined on a set U ⊂Rn containing a, the second

derivative matrix (or Hessian matrix) of f at a is

Hf (a) =

∂2f

∂x21(a) ∂2f

∂x1∂x2(a) · · · ∂2f

∂x1∂xn(a)

∂2f∂x2∂x1

(a) ∂2f

∂x22(a) · · · ∂2f

∂x2∂xn(a)

.... . . · · · ...

∂2f∂xn∂x1

(a) ∂2f∂xn∂x2

(a) · · · ∂2f

∂x2n(a)

. (B.)

If f is C2 on U , then Hf is a symmetric matrix at all points of U .

Theorem B.. Suppose f : R2→R is a C2 function defined on a set U ⊂R2 containing a,

where grada f = 0.

a) If∂2f

∂x2 (a) > 0 and det(Hf (a)) > 0, then f attains a local minimum at a.

b) If∂2f

∂x2 (a) < 0 and det(Hf (a)) > 0, then f attains a local maximum at a.

c) If det(Hf (a)) < 0, then f has a saddle point at a.

Example B. Consider the function f : R2→R defined by f (x,y) = x siny. Then

df = siny dx+ xcosy dy.

Any critical point (x,y) of f must satisfy

siny = 0 xcosy = 0.

Since siny and cosy cannot simulataneously equal 0, we must have x = 0 and siny = 0,hence the critical points are (0,nπ) for any integer n. Now, the Hessian matrix of f is

(0 cosy

cosy −x siny,

)


which has determinant −cos2 y < 0, hence f has only saddle points. ®

This last example gives a little insight into saddle points for a function of two variables.If f : R2→R is C2, then the mixed partial derivatives are equal. Furthermore, if∂2f

∂x2 (a) = 0, then

det(Hf (a)) = det

0 ∂2f∂x∂y

∂2f∂y∂x

∂2f∂y2

= −

(∂2f

∂x∂y

)2

< 0.

In this particular scenario, any critical point must be a saddle point.

Recall that, if a continuous function f : R→R is defined on a closed interval [a,b] then fattains global extreme values (the single-variable Extreme Value Theorem). The globalextrema occur either at a critical point in the open interval (a,b) or at one of the twoenpoints. Similar to the closed interval method of single-variable calculus, determiningglobal extrema for a multivariable function f , using first derivative information only,involves determining critical points and comparison of function values. If f : Rn→R isdefined on a set U ⊂R

n, we compare the value of f at critical points and compare withthe value of f at points on the boundary, ∂U . There is a theorem, comparable to theExtreme Value Theorem of single-variable calculus, that ensures the existence of extremevalues, given assumptions on f and the set U . However, finding these extreme valuescan be another issue altogether. When n ≥ 2, the boundary of U consists of an infinitenumber of points, in general, so techniques beyond a simple comparison of critcal valuesand boundary values of f are often needed. Consider the following example.

Example B. Determine extrema of f (x,y) = 8x+ 3y defined on the disc x2 + y2 ≤ 1.First, note that f does not have any critical points, since df = 8dx+ 3dy , 0. IfU = {(x,y) |x2 + y2 ≤ 1} is the disc, the boundary is the unit circle ∂U = {(x,y) |x2 + y2 = 1}.The boundary ∂U can be parametrized by γ(t) = (cos t,sin t), where 0 ≤ t ≤ 2π. What weneed to do is determine the maximum and minimum values of f on the boundary. If werestrict f to points on ∂U only, we obtain a new function called the restriction of f to ∂U ,denoted f

∣∣∣∂U

. A formula for f∣∣∣∂U

can be obtained by evaluating f ◦γ , the compositionof f with γ . We have f (γ(t)) = 8cos t + 3sin t, which has critical points where thedifferential is 0 (or undefined). The differential of f ◦γ is d(f ◦γ) = (−8sin t + 3cos t)dt,which is 0 when tan t = 3/8. Since x = cos t, y = sin t, this yields

x = cos(arctan

38

)= ± 8√

73

y = sin(arctan

38

)= ± 3√

73.

Since the value of tan t is positive, x and y have the same sign. The maximum value of foccurs for the positive x,y values, while the minimum occurs for the negative x,y values.


Now, several comments are in order. First, note that d(f ◦γ) is precisely the pullback ofdf by γ . Indeed, df = 8dx+ 3dy, hence

γ∗(df ) = 8d(cos t) + 3d(sin t) = (−8sin t + 3cos t)dt.

Second, the equation defining the boundary, x2 +y2 = 1, can be considered as a constraint.We could arrive at the same answer as follows. Let F(x,y) = x2 + y2. It can be shown thatan extreme value may occur where the wedge product, df ∧ dF, equals 0. Here, we have

df ∧ dF = (8dx+ 3dy)∧ (2xdx+ 2y dy)= (16y − 6x)dx∧ dy,

which is 0 when y =38x. Plugging this in to the boundary equation yields

x2 +(38x)2

= 1,

which, upon solving for x and then y, gives the same solutions as above. As a final note,

if we had not already ruled out any critical points for f , we could have plugged y =38x

into f to determine where extreme values may occur in the interior of U . ®

§.. Constrained Extrema

Suppose we want to determine extrema of a function f : Rn→R subject to a constraint.For instance, it may be desired to maximize the volume of a container, while keeping thesurface area of the container fixed, or within certain limits. If f is defined on a setU ⊂R

n, determining the maximum and minimum values on the boundary ∂U is a typeof constrained extreme value problem. In practical applications, there may be multipleconstraints to consider. In some cases, the method of the previous section can besuccesfully applied. For example, if we want to determine extreme values of f subject tothe constraint F(x) = c, where F : Rn→R and c is a constant, we can attempt solvingdf ∧ dF = 0. The case of more constraints can be handled similarly. We will assume thatall constraints can be represented as some function equaling a constant. Suppose wewant to minimize f (x) subject to

F1(x) = c1

F2(x) = c2

...

Fk(x) = ck ,


where the constraints correspond to the functions Fi : Rn→R and constants ci ,i = 1, . . . , k. Using the above method, we would seek solutions to

df ∧ dF1 ∧ dF2 ∧ · · · ∧ dFk = 0.

Equivalently, if F : Rn→Rk is a vector-valued function with component functions

F1, · · · ,Fk, we can seek solutions to df ∧ dF = 0. However, the method of Lagrangemultipliers is more suitable for constrained optimization, particularly when there aremultiple constraints.

Theorem B. (Method of Lagrange multipliers). Consider the problem of minimizing (ormaximizing) f : Rn→R subject to constraints Fi = ci , where Fi : Rn→R and ci is constant,for i = 1, . . . , k, where n > k. If p = (x1, . . . ,xn) is a local solution, then there exist scalarsλ1, . . . ,λk such that

df = λ1dF1 + . . .+λk dFk , (B.)

evaluated at p.

In Theorem B., the scalars λ1, . . . ,λk are called Lagrange multipliers.

Example B. Determine the extrema of f (x,y,z) = x2 + y2 + z2 − x+ y on the set

U = {(x,y,z) |x2 + y2 + z2 ≤ 1}.The set U is the unit ball and the boundary

∂U {(x,y,z) |x2 + y2 + z2 = 1}is the unit sphere. Determining any extreme values in the interior of U can be done bysolving df = 0. We have

df = (2x − 1)dx+ (2y + 1)dy + 2zdz,

hence the only critical point is (1/2,−1/2,0), at which f (1/2,−1/2,0) = −1/2. We will determinethe extrema on the boundary in two ways.

Method ) Considering the boundary as a constraint, define

F(x,y,z) = x2 + y2 + z2.

The equation F(x,y,z) = 1 represents the boundary constraint, which is the onlyconstraint in this example. By Theorem B., there exists a scalar λ such thatdf = λdF, hence we solve

(2x − 1)dx+ (2y + 1)dy + 2zdz = λ(2xdx+ 2y dy + 2zdz).


Equivalently, by combining components, we have

(2x − 1−λ2x)dx+ (2y + 1−λ2y)dy + (2z −λ2z)dz = 0,

which reduces to the three equations

2x − 1 = λ2x, 2y + 1 = λ2y, 2z = λ2z.

The last equation holds if λ = 1 or z = 0, but if λ = 1 the first two equations have nosolution, hence we assume z = 0 (and λ , 1). Solving the first two equations for λ,we have

λ =2x − 1

2xand λ =

2y + 12y

.

Setting these equations equal yields y = −x, which, when substituted into theboundary equation along with z = 0, yields

x = ±√

22.

So, the two points (±√

22,∓√

22,0

)

on the boundary may correspond to extreme values. We will check aftercompleting the second method.

Method ) One might call this the direct method, as we will not be using Lagrange multipliers.With f and F as defined above, we compute

df ∧ dF = ((2x − 1)dx+ (2y + 1)dy + 2zdz)∧ ((2xdx+ 2y dy + 2zdz))= 2zdy ∧ dz+ 2zdz∧ dx − (2x+ 2y)dx∧ dy.

Thus, if df ∧ dF = 0, we must have z = 0 and y = −x. Substituting z = 0 and y = −xinto the boundary equation x2 + y2 + z2 = 1, we obtain 2x2 = 1, hence

x = ±√

22,

as before.

No matter which method we prefer, there are three points to consider, namely theinterior point (1

2,−1

2,0

)


and the boundary points (√2

2,−√

22,0

),

(−√

22,

√2

2,0

).

The corresponding function values are

f(12,−1

2,0

)= −1

2(minimum)

f

(√2

2,−√

22,0

)= 1−

√2

f

(−√

22,

√2

2,0

)= 1 +

√2 (maximum)

®

§ Maxwell’s equations

Let E = (E1,E2,E3) and B = (B1,B2,B3) be vector fields on R3. Here, the components

Ei : R4→R and Bi : R4→R, i = 1,2,3, are functions of t (time) and x,y,z (Cartesiancoordinates for R3). The vector fields E and B represent the (time-variable) electric andmagnetic field strength, respectively. Let ρ : R4→R be a real-valued function (the chargedensity) and let j = (j1, j2, j3) be a vector field on R

3 (the current density). Note that thecomponents of j are independent of time, t. Maxwell’s equations of electromagnetismare given by

divE = ρ (Coulomb’s Law)

curlE = −∂B∂t

(Faraday’s Law)

divB = 0 (absence of magnetic monopoles)

c2 curlB = j +∂E∂t

(Ampere’s Law)

(B.)

where div,curl are defined in terms of the spatial variables (x,y,z). Suppose M is a2-surface in R

3 with boundary ∂M. Let F(u,v) = (x(u,v), y(u,v), z(u,v)) parametrize Mand let γ : [a,b]→R

3 parametrize ∂M. The voltage around ∂M is given by the lineintegral ∫

γE · ds =

∫ b

a〈E(γ),γ ′〉dt.


The magnetic flux across M is given by the flux integral�M

⟨B(F),

∂F∂u× ∂F∂v

⟩dudv.

Documents

Vector Calculus - Mathematicscrhumber/ma213/vec_book.pdf · Vector Calculus Math 213 Course Notes Cary Ross Humber November 28, 2016