Embed Size (px)
Citation preview
SE102 Multivariable Calculus
Hyosang Kang ([email protected])Assistant Professor, School of Undergraduate Studies
Daegu-Gyeongbuk Institute of Science and Technology (DGIST)http://klein.dgist.ac.kr
(last modified: June 15, 2017)
Contents
1 Vectors
2 Matrices
3 Continuity
4 Partial derivatives
5 Differentiability
6 Chain rule
7 Taylor series
8 Lagrange multiplier
9 Vector spaces
10 Eigenvalues and eigenvectors
11 Solving system of homogeneous DE
12 Methods of solving systems of nonho-mogeneous DEs
13 Double integrals
14 Integration by substitution
15 Triple integrals
16 Line integrals
17 Curl and divergence
18 Surface integrals
19 Stokes theorem
20 Divergence theorem
21 Conservation fields
1 Vectors
Definition 1.1. A (n-dimensional) vector is a n-tuple of real numbers
⇀a = (a1, a2, · · · , an)
with the following operations.
• (vector sum)
⇀a +
⇀
b = (a1 + b1, a2 + b2, · · · , an + bn)
• (scalar multiplication)
k⇀a = (ka1, ka2, · · · , kan)
The set of all (n-dimensional) vectors is called the(n-dimensional) vector space, and denoted by Rn.
Example 1.2.
1. (Variables of functions) Recall that a functionis an assignment of a object to another object.Such objects can either be numbers, points, orcollections of numbers. For example, let us con-sider a function f sending every point in R2 to apoint in R3.
f(a1, a2) = (b1, b2, b3)1 (1.2.1)
Here the coordinates b1, b2, b3 are dependents,and a1, a2 are independents. Such functions are
1The graph (cf. Example 3.3) of such f is called a surfacein R3.
1
called multivariable functions (cf. Definition3.1). From Definition 1.1, a vector is just an or-dered collection of real numbers. Thus (1.2.1)can be written as
f(⇀a ) =
⇀
b
where⇀a = (a1, a2) and
⇀
b = (b1, b2, b3). Thatis, a multivariable function can be treated as asingle ‘variable’ function from a vector space toanother vector space2.
2. (Position vectors) Let O = (0, 0, 0) be the originand P = (a1, a2, a3) a point in 3-dimensional
space. Then the arrow−−→OP can be represented
by the vector
−−→OP = (a1, a2, a3)
The scalar multiplication is a dilation, andthe vector sum is a superposition. Let Q =(b1, b2, b3) be another point, and R = (a1 +
b1, a2 + b2, a3 + b3). Then the arrow−→PR is rep-
resented by the same vector as−−→OQ. Thus
−−→OP +
−−→OQ =
−−→OP +
−→PR =
−−→OR
3. (Velocity vectors) Let c : (−ε, ε) → R2 be aparametrization of a curve on a plane. Supposethat
c(t) = (x(t), y(t))
and x(t), y(t) are differentiable on (−ε, ε). Let usdefine c′(t) as follows.
c′(t) = (x′(t), y′(t))
We can consider c′(t) as a vector, which repre-sents the velocity at c(t).
Definition 1.3. For a (n-dimensional) vector⇀a =
(a1, a2, · · · , an),
||⇀a || =√a2
1 + a22 + · · ·+ a2
n
is called the norm of⇀a . A vector with norm 1 is
called a unit vector. The norm ||⇀a || is 0 if and only
if⇀a = (0, · · · , 0). Such vector is called the zero
vector and denoted by⇀0 . For a nonzero vector
⇀a ,
⇀u =
⇀a/||⇀a ||
is called the normalization of⇀a .
2Thus the Multivariable Calculus is often called the VectorCalculus.
Example 1.4. For each i = 1, · · · , n, the vector
⇀e i = (0, · · · , 1, · · · , 0)
with 1 at ith place is called a (unit) basis vector.Especially, 3-dimensional basis vectors are denotedby
⇀i = (1, 0, 0),
⇀j = (0, 1, 0),
⇀
k = (0, 0, 1)
A vector⇀a = (a1, a2, · · · , an) can be decomposed as
a linear sum of basis vectors.
⇀a = a1
⇀e 1 + a2
⇀e 2 + · · ·+ an
⇀e n
Thus the set{⇀e 1,
⇀e 2, · · · ,
⇀e n}
is called the basis, and generates all n-dimensionalvectors.
Definition 1.5. The inner product of two vectors⇀a ,
⇀
b is an operation defined by
⇀a ·
⇀
b = a1b1 + a2b2 + · · ·+ anbn
Proposition 1.6. The inner product satisfies the fol-lowing.
1.⇀a · (
⇀
b +⇀c ) =
⇀a ·
⇀
b +⇀a ·⇀c
2.⇀a · (k
⇀
b ) = k(⇀a ·
⇀
b )
3.⇀a ·⇀a = ||⇀a ||2
Theorem 1. Let⇀a ,
⇀
b be nonzero 2-dimensional vec-
tors. If θ is the angle between⇀a and
⇀
b , then
⇀a ·
⇀
b = ||⇀a || · ||⇀
b || · cos θ
Example 1.7. As norm measures the ‘size’ of a vec-tor, the inner product measure the direction of a vec-tor. The direction of a 2-dimensional vector
⇀a is
determined by 0 ≤ θ1, θ2 ≤ π satisfying
⇀a ·⇀e 1 = ||⇀a || cos θ1
⇀a ·⇀e 2 = ||⇀a || cos θ2
In order to determine the direction of a 3-dimensional vector
⇀a , we need three angles 0 ≤
α, β, γ ≤ π satisfying
cosα =
⇀a ·
⇀i
||⇀a ||, cosβ =
⇀a ·
⇀j
||⇀a ||, cos γ =
⇀a ·
⇀
k
||⇀a ||
Such quantities are called the direction cosines.
2
Exercise 1.8. Show that the direction cosine α, β, γsatisfy
cos2 α+ cos2 β + cos2 γ = 1
Definition 1.9. For nonzero vectors⇀a and
⇀
b withthe same dimension, the vector defined by
proj⇀b
⇀a =
⇀a ·
⇀
b
||⇀
b ||2
⇀
b
is called the projection of⇀a onto
⇀
b .
Example 1.10. Find the proj⇀b
⇀a for
⇀a = (1, 2, 1),
⇀
b = (2, 1, 2)
2 Matrices
Definition 2.1. A n×m (n-by-m) matrix is a col-lection of nm numbers (or functions) arranged in thefollowing way.
A = (aij)n×m =
a11 a12 · · · a1m
a21 a22 · · · a2m
......
...an1 an2 · · · anm
The indices i, j of an entry aij represents the row andcolumn indices respectively.
Example 2.2.
1. A n×m matrix is called square matrix if n = m.
2. If A is a square matrix and aij = 0 for all i 6= j,then A is called diagonal.
A =
a11 0. . .
0 ann
3. If the diagonal entries of a diagonal matrix are
all 1, then it is called the identity matrix, anddenoted by In.
In =
1 0. . .
0 1
4. If A is a square matrix and aij = 0 for all i <j (i < j, respectively), then A is called lowertriangle (upper triangle, respectively).∗ 0
. . .
∗ ∗
,
∗ ∗. . .
0 ∗
5. If every entry is 0, then it is called the zero
matrix, and denoted by 0.
Definition 2.3. Let A = (aij) and B = (bij) ben×m matrix. Then
A+B = (aij + bij)
k ·A = (kaij)
Thus the set of all n×m matrices
M(n,m; R) = {A = (aij) | 1 ≤ i ≤ n, 1 ≤ j ≤ m}
is a vector space Rnm. Moreover, if A is a n×m andB a m× l matrices, then A ·B is a n× l matrix whoseentries are
A ·B = (ai1b1j + ai2b2j + · · ·+ aimbmj)1≤i≤n,1≤j≤l
=
a11 a12 · · · a1m...
......
ai1 ai2 · · · aim...
......
an1 an2 · · · anm
b11 · · · b1j · · · b1lb21 · · · b2j · · · b2l...
......
bm1 · · · bmj · · · bml
Proposition 2.4. Let A,B,C be matrices. When-ever the operations are valid, the following holds.
1. (A ·B) · C = A · (B · C)
2. A · (B + C) = A ·B +A · C
3. (B + C) ·A = B ·A+ C ·A
4. k · (A ·B) = (k ·A) ·B = A · (k ·B)
Moreover, the transpose of a matrix A = (aij) de-fined by
AT = (aji)
satisfies the following.
5.(AT)T
= A
6. k ·AT = (k ·A)T
7. (A+B)T = AT +BT
3
8. (AB)T = BTAT
Definition 2.5. The determinant of 2 × 2 matrixA is defined as follows.
detA =
∣∣∣∣ a11 a12
a21 a22
∣∣∣∣ = a11a22 − a12a21
The determinant of 3 × 3 matrix B is defined asfollows.
detB =
∣∣∣∣∣∣b11 b12 b13
b21 b22 b23
b31 b32 b33
∣∣∣∣∣∣= b11b22b33 + b12b23b31 + b13b21b32
− b13b22b31 − b11b23b32 − b12b21b33
Remark 2.6. Remember the following rule:
+ + +− − −b11 b12 b13 b11 b12
b21 b22 b23 b21 b22
b31 b32 b33 b31 b32
Proposition 2.7. Let us denote |A| by the determi-nant of a matrix A. Then the following holds.
1. If A has a row (or a column) whose entries areall zero, then |A| = 0.
2. Let B be the matrix obtained by interchangingtwo rows (or columns) of A. Then |B| = −|A|.
3. Let B be a matrix obtained by multiply c on arow (or column) followed by adding it to anotherrow (or column). Then |B| = |A|.
Exercise 2.8. Show that if a row (or column) is aconstant multiple of another row (or column), thenthe determinant is 0.
Definition 2.9. Let A be a n× n matrix. A matrixB satisfying
A ·B = B ·A = In
is called the inverse of A, denoted by B = A−1. Ifan inverse matrix A−1 exists, then A is said to benon-singular. Otherwise, it is called singular.
Theorem 2.10. A matrix A is singular if and onlyif |A| = 0.
Remark 2.11. If A is a 2× 2 matrix, then A−1 is
A−1 =1
detA
(a22 −a21
−a12 a11
)For a 3× 3-matrix B, the inverse is given by
B−1 =1
detB
c11 −c21 c31
−c12 c22 −c32
c13 −c23 c33
(2.11.1)
where each cij3 is the determinant of 2 × 2-matrix
obtained by deleting ith row and jth column, for ex-ample,
c21 =
∣∣∣∣∣∣b11 b12 b13
b21 b22 b23
b31 b32 b33
∣∣∣∣∣∣ =
∣∣∣∣ b12 b13
b32 b33
∣∣∣∣Notice that the row and column indices are switchedin (2.11.1). In general, the determinant of n× n ma-trix is defined as a function
det : M(n; R) −→ R
satisfying the three properties of Proposition 2.7.One can prove that such function is unique, and thecomputation is inductively defined4 Deteminants willbe used in cross-product (cf. Definition 2.12), Jaco-bian (cf. Definition ??), orientation (cf. Definition18.6), etc.
Definition 2.12. Let⇀a = (a1, a2, a3),
⇀
b =(b1, b2, b3) be two 3-dimensional vectors. The cross-
product of⇀a ,
⇀
b is a vector defined by
⇀a ×
⇀
b = (a2b3 − a3b2, a3b1 − a1b3, a1b2 − a2b1)
Remark 2.13. Remember the formula:
⇀a ×
⇀
b =
∣∣∣∣∣∣∣⇀i
⇀j
⇀
ka1 a2 a3
b1 b2 b3
∣∣∣∣∣∣∣Proposition 2.14. Let
⇀a ,
⇀
b ,⇀c be 3-dimensional
vectors and k a constant. Then
1.⇀a ×
⇀0 =
⇀0 ×⇀
a =⇀0
3called cofactor4Let B be a nonsingular n× n-matrix. Then
B−1 =1
detB((−1)i+jcij)T
where cij is the cofactor defined similar to Remark 2.11.
4
2.⇀a ×⇀
a =⇀0
3.⇀a ×
⇀
b = −⇀
b ×⇀a
4.⇀a × (k
⇀
b ) = (k⇀a )×
⇀
b = k(⇀a ×
⇀
b )
5.⇀a × (
⇀
b +⇀c ) =
⇀a ×
⇀
b +⇀a ×⇀
c
6.⇀a · (⇀a ×
⇀
b ) =⇀
b · (⇀a ×⇀
b ) = 0
7. (⇀a ×
⇀
b )×⇀c = (
⇀a ·⇀c )
⇀
b − (⇀
b ·⇀c )⇀a
Remark 2.15. The property 6 in Proposition 2.14
shows that the cross product⇀a ×
⇀
b is the vector
perpendicular to the plane spanned by⇀a and
⇀
b .
Example 2.16. Find the equation of a plane con-taining P = (6, 3, 2) and perpendicular to
⇀n =
(−2, 1, 5).
Theorem 2.17. Let⇀a ,
⇀
b be two 3-dimensional vec-tors. Then
||⇀a ×⇀
b || = ||⇀a || · ||⇀
b || · | sin θ|
where θ is the angle between⇀a and
⇀
b .
Exercise 2.18. Show that |⇀a · (⇀
b ×⇀
b )| is the area
of parallelepiped bounded by⇀a ,
⇀
b ,⇀c .
3 Continuity
Definition 3.1. A function is called multivariableif there are at least two independent (or dependent)variables. Let us write a multivariable function fwith n independent variables and m dependent vari-ables as follows.
f(x1, x2, · · · , xn) = (y1, y2, · · · , ym) (3.1.1)
Remark 3.2. For each j = 1, · · · ,m, the variableyj is a real-valued function with n independent vari-ables.
yj = yj(x1, x2, · · · , xn)
If we view⇀x = (x1, x2, · · · , xn),
⇀y = (y1, y2, · · · , ym)
as vectors, then (3.1.1) can be written as
f(⇀x) =
⇀y = y1
⇀e 1 + · · ·+ ym
⇀em
and f ‘decomposes’ into a linear combination of mreal-valued yj ’s. Thus, we often write (3.1.1) as
f = (f1, f2, · · · , fm)
where each fj(x1, x2, · · · , xn) = yj for j = 1, · · · ,m.
Example 3.3. Let f(x, y) = z be a function definedon a set D ⊂ R2. The graph of f is the set
G(f) = {(x, y, f(x, y)) | (x, y) ∈ D}
1. The graph of z = x2 − y2 is called saddle.
2. The graph of z = x(x2 − y2) is called Monkey’ssaddle.
The graph of a function f(x, y, z) = w cannot bedrawn in 3-dimensional space5. In such case, we usethe level set
Lc(f) = {(x, y, z) ∈ R3 | f(x, y, z) = c} (3.3.1)
3. The level sets of f(x, y, z) = x2 + y2 − z2 withc = −1, 0, 1.
5We need 4-dimensional space!
5
If there are only one independent variable, namely
f(t) = (x1(t), x2(t), · · · , xm(t)),
then the trajectory is given as a curve.
4. The graph of f(t) = (t2, t3)
5. The graph of f(t) = (cos t, t, sin t)
Exercise 3.4.
1. Graph the function f(x, y) = 1 + 2x2 + 2y2
2. Find the level set of f(x, y, z) = x2 − y2 + z2 −2x+ 2y + 4z with c = 2.
Definition 3.5. Let f(x, y) be a two-variable func-tion. (For simplicity, let us assume f is defined onthe entire plane R2.) A constant L is called the limitof f at (x0, y0) if for any (arbitrary small) ε > 0,there exists a (small) δ > 0 such that whenever thedistance between (x, y) and (x0, y0) is less than δ, theinequality
|f(x, y)− L| < ε
holds. In such case, we simply write
lim(x,y)→(x0,y0)
f(x, y) = L
The function f(x, y) is said to be continuous at(x0, y0) if
lim(x,y)→(x0,y0)
f(x, y) = f(x0, y0)
Example 3.6. Let f(x, y) be a function defined as
f(x, y) =
xy√x2 + y2
(x, y) 6= (0, 0)
0 (x, y) = (0, 0)
Let us prove that f(x, y) is continuous at (0, 0). For
any ε > 0, let δ = ε. Then whenever√x2 + y2 < δ,
|x|, |y| < δ holds. Thus∣∣∣∣∣ xy√x2 + y2
− 0
∣∣∣∣∣ =|x| · |y|√x2 + y2
≤ |y| < δ = ε
Remark 3.7. The concept of continuity of a multi-variable function is the same as that of single-variablefunctions. Intuitively, we should say that a function iscontinuous if its graph ‘looks’ continuous. The graphis continuous at the point (x0, y0) if the value of func-tion f(x, y) is not far away from f(x0, y0) when aneighbor point (x, y) is sufficiently close to (x0, y0).Moreover, if (x, y) approaches to (x0, y0), the valuef(x, y) must converges to f(x0, y0). In the single-variable case, there are only two ways of approach,namely the right and left limit. However, even inthe most simple case of multivariable function, suchas f(x, y) = z, there are infinitely many direction ofapproach. A worse news is that not only the direc-tions but also the paths matters! The following is ageometric definition of continuity.
Definition 3.8. A function f(x, y) is continuous at(x0, y0) if and only if the value f(x(t), y(t)) con-verges to f(x0, y0) for any parametrized curve c(t) =(x(t), y(t)) with c(0) = (x0, y0) as t→ 0.
In general, it is impossible to check the convergencefor all parametrizations. Thus, it is easier to use Def-inition 3.5 to prove the continuity.
However, to prove the discontinuity, it is harderto use Definition 3.5. Instead, Definition 3.8 tells usthat a function is not continuous if and only if thereis at least one path to (x0, y0) along which the valueof function does not converge to f(x0, y0). Therefore,
6
• use Definition 3.5 to show continuity, and
• use Definition 3.8 to show discontinuity
Example 3.9. Let
f(x, y) =
x4 − 4y2
x2 + 2y2(x, y) 6= (0, 0)
0 (x, y) = (0, 0)
Let us prove that f(x, y) is discontinuous at (0, 0).Suppose that we approach (0, 0) along c(t) = (0, t).Then
f(0, t) =−4t2
2t2= −2→ −2
Meanwhile, if we approach along c(t) = (t, 0), then
f(t, 0) =t4
t2= t2 → 0
The limits do not coincide.
Exercise 3.10. Let
f(x, y) =
y2
|x− y|x 6= y
0 x = y
Show that f(x, y) is discontinuous at (0, 0).
4 Partial derivatives
Definition 4.1. Let f(x, y) be a function defined
near (x0, y0) and⇀u = (a, b) a unit vector. The limit
D⇀uf(x0, y0) = lim
t→0
f(x0 + at, y0 + bt)− f(x0, y0)
t
is called the⇀u-directional derivative of f at
(x0, y0).
Example 4.2. Let f(x, y) = x2 + y2. The (1, 0)-directional derivative of f at ( 1
2 , 0) is
D(1,0)f(
12 , 0)
= limt→0
f( 12 + t, 0)− f( 1
2 , 0)
t
= limt→0
(12 + t
)2 − ( 12 )2
t= 1
Remark 4.3. The directional derivative is the mostnatural way to extend the concept of derivatives ofsingle-variable functions. For y = f(x), the deriva-tive f ′(x) represents rate of change, slope of tangentline, etc. Such quantities have geometric interpre-tation using the graph of y = f(x). However, inthe two-variable case, the concept does not gener-alize naturally. For example, the rate of change off(x, y) is ambiguous, since there are infinitely many
directions. The easiest way is to fix a direction⇀u ,
and then compute the rate of change of f(x, y) along
the direction of⇀u . This is exactly what Definition
4.1 means.For example, the directional derivative in Example
4.2 can be computed in the following way. Let c(t)
be a curve passing ( 12 , 0) with direction
⇀u
c(t) = (12 , 0) + t
⇀u = ( 1
2 + t, 0)
Then
D(1,0)f(
12 , 0)
=d
dt
∣∣∣∣t=0
f(c(t))
=d
dt
∣∣∣∣t=0
(1
2+ t
)2
= 1
Example 4.4. As we can take the derivative asmany times as we want, we can do so with direc-tional derivatives. Let f(x, y) = x3 + 5x2y + y3 and⇀u = ( 3
5 ,45 ). Then the directional derivative at (x, y)
is
D⇀uf(x, y) = lim
t→0
f(x+ 3
5 t, y + 45 t)− f(x, y)
t
=29
5x2 + 6xy +
12
5y2 = g(x, y)
7
We can take the second directional derivative too.
D⇀uD⇀uf(x, y) = D⇀
ug(x, y)
(Note that we may take the second directional deriva-tive with different direction.)
Definition 4.5. The partial derivatives of f(x, y)
at (x0, y0) are directional derivatives along⇀e 1 =
(1, 0) and⇀e 2 = (0, 1).
Dxf(x0, y0) = D⇀e 1f(x0, y0)
Dyf(x0, y0) = D⇀e 2f(x0, y0)
We also denote them by ∂f∂x (x0, y0), ∂f
∂y (x0, y0), or
simply fx(x0, y0), fy(x0, y0).
Remark 4.6. As we observed in Remark 4.3, thepartial derivatives are rate of change of f along x,y-directions. Thus, in most cases, fx(x0, y0) (fy(x0, y0)respectively) be computed by taking the ‘usual’derivative with x (y respectively) while treating y (xrespectively) as a constant.
Definition 4.7. The second partial derivatives aredenoted as follows.
∂2f
∂x2= fxx = Dxxf = Dx(Dxf)
∂2f
∂x∂y= fyx = Dxyf = Dx(Dyf)
∂2f
∂y∂x= fxy = Dyxf = Dy(Dxf)
∂2f
∂y2= fyy = Dyyf = Dy(Dyf)
Example 4.8. Let
f(x, y) =
xyx2 − y2
x2 + y2(x, y) 6= (0, 0)
0 (x, y) = (0, 0)
Let us find Dyxf(0, 0). First, we compute Dxf(x, y).For (x, y) 6= (0, 0), we take the ‘usual’ partial deriva-tive (cf. Remark 4.6).
Dxf(x, y) =(x4 + 4x2y2 − y4)y
(x2 + y2)2(4.8.1)
The formula (4.8.1) is not defined at (0, 0). We haveto compute Dxf(0, 0) by definition.
Dxf(0, 0) = limt→0
f(t, 0)− f(0, 0)
t= 0
Thus
Dxf(x, y) =
(x4 + 4x2y2 − y4)y
(x2 + y2)2(x, y) 6= (0, 0)
0 (x, y) = (0, 0)
(4.8.2)Then
Dyxf(0, 0) = limt→0
Dxf(0, t)−Dxf(0, 0)
t= −1
Now try computing Dxyf(0, 0). Is it same asDyxf(0, 0)? In general,
fxy 6= fyx
Theorem 2 (Clairaut). If the partial derivatives fxy,fyx are continuous at (x0, y0), then
fxy(x0, y0) = fyx(x0, y0)
Exercise 4.9. Show that (4.8.2) is not continuous at(0, 0). (This explains why fxy(0, 0) 6= fyx(0, 0).)
5 Differentiability
Definition 5.1. Let f(x, y) be a real-valued functionsuch that fx, fy are defined at (x0, y0). Then
L(x, y) = f(x0, y0)+fx(x0, y0)(x−x0)+fy(x0, y0)(y−y0)
is called the linear approximation of f at (x0, y0).
Example 5.2. Find the linear approximation of z =√|xy| at (1, 1). What about at (0, 0)? Can we call
the equation z = L(x, y) a tangent plane?
Definition 5.3. Let L(x, y) be the linear approxi-mation of f(x, y) at (x0, y0). The function f(x, y) issaid to be differentiable at (x0, y0) if
lim(x,y)→(x0,y0)
f(x, y)− L(x, y)√(x− x0)2 + (y − y0)2
= 0 (5.3.1)
Remark 5.4.
1. As Definition 5.3 says, the differentiability de-pends on the existence of the partial derivativesfx, fy. However, they are just special kinds ofdirectional derivatives. So we may wonder if wecould define the differentiability by using direc-tional derivatives along two linearly independentvectors
⇀u1,
⇀u2. But then, we may worry that the
differentiability changes. Later, we will see thatthis does not happens.
8
2. If f(x, y) is differentiable, then z = L(x, y) is thetangent plane. However, the existence of L(x, y)does not mean that f(x, y) is differentiable.
Example 5.5. Show that
f(x, y) =
xy√x2 + y2
(x, y) 6= (0, 0)
0 (x, y) = (0, 0)
is not differentiable at (0, 0). (Note that it is contin-uous at (0, 0) as shown in Example 3.6).
Theorem 5.6. Suppose that ε1 = ε1(x, y), ε2 =ε2(x, y) satisfy
f(x, y)− f(x0, y0) = fx(x0, y0)(x− x0)
+ fy(x0, y0)(y − y0) + ε1(x− x0) + ε2(y − y0)
If ε1, ε2 → 0 as (x, y) → (x0, y0), then f(x, y) is dif-ferentiable at (x0, y0)6.
Remark 5.7. Theorem 5.6 gives a geometric mean-ing of differentiability. The function f(x, y) is differ-entiable at (x0, y0) if the linear approximation L(x, y)is a ‘good’ approximation, meaning that the value ofL(x, y) gets close to f(x0, y0) ‘faster’ then the point(x, y) gets close to (x0, y0).
Definition 5.8. Let P = (x0, y0, z0) be a point onthe graph of z = f(x, y). A plane containing P iscalled a tangent plane if it contains
(x0, y0, z0) + t · c′(t)
for all t ∈ R where c : (−ε, ε) → R3 is any curve on
the the graph of f satisfying c′(0) 6=⇀0 (cf. Example
1.2).
Example 5.9. Find the equation of a tangent planeat (1, 1, 1) of z =
√|xy|. What about at (0, 0, 0)?
6The textbook says the condition is also a necessary condi-tion. The proof rely on the chain rule and the integrability ofpartial derivatives. Not only the integrability is guaranteed inthe general situation, but also the book uses this theorem toprove the chain rule! This is a circular reasoning. So we onlystate that the condition is a sufficient condition, and will givea proof of chain rule without using this theorem.
Exercise 5.10. Let P be a point on the level set√x +√y +√z =√c. Prove that the sum of x,y,z-
intercepts of tangent plane at P is a constant (inde-pendent to P ).
Proposition 5.11. If f(x, y) is differentiable at(x0, y0), then z = L(x, y) is the tangent plane at(x0, y0, f(x0, y0)).
Example 5.12. The differentiabiliy does not implythe continuity of partial derivatives. For example, thefunction
f(x, y) =
(x2 + y2) sin
(1√
x2 + y2
)(x, y) 6= (0, 0)
0 (x, y) = (0, 0)
is differentiable at (0, 0), but the partial derivativesfx, fy are not continuous at (0, 0). This means, thatthe differentiability does not always imply that the‘slope’ of tangent plane changes smoothly.
6 Chain rule
Theorem 6.1 (The chain rule). Let x(t), y(t) besingle-variable functions which are differentiable att0. The composite function F (t) = f(x(t), y(t)) isalso a single-variable function. If f(x, y) is differen-tiable at (x0, y0) = (x(t0), y(t0)), then F (t) is differ-entiable at t0 and satisfies
d
dt
∣∣∣∣t=t0
f(x(t), y(t)) =∂f
∂x(x0, y0)
dx
dt(t0)
+∂f
∂y(x0, y0)
dy
dt(t0) (6.1.1)
Remark 6.2. The chain rule shows a true meaningof differentiability of a multivariable function. So far,we defined the differentiability, but not the deriva-tives. What is the derivatives of multivariable func-tions? To investigate this, first start with a single-variable function y = f(x). The derivative f ′(x) isdefined by the following limit.
f ′(x) = limh→0
f(x+ h)− f(x)
h
In other words, f ′(x) is a value (or a function) satis-fying
limh→0
|f(x+ h)− f(x)− f ′(x)h||h|
= 0
9
Now, replace x, y by vectors⇀x ,
⇀y , and the absolute
value by the norm of vectors.
lim⇀
h→⇀0
||f(⇀x +
⇀
h)− f(⇀x)− f ′(
⇀x) ·
⇀
h ||
||⇀
h ||= 0 (6.2.1)
This equation makes sense if f ′(x)·⇀
h has the same di-
mension as⇀y does. Let us compare it with Definition
5.3. By subsituting⇀x = (x, y) and
⇀
h =⇀x − (x0, y0),
(5.3.1) can be written as follows.
lim⇀
h→⇀0
||f(⇀x +
⇀
h)− f(⇀x)− (fx, fy) ·
⇀
h ||
||⇀
h ||= 0
(6.2.2)Comparing (6.2.1) and (6.2.2), we notice that f ′(x)must be a vector and the multiplication must be re-placed by the inner product
f ′(⇀x) = (fx, fy)
Thus, the derivative of a multivariable functionf(x, y) is a vector, which we often call the gradi-ent of f , and denoted by ∇f . Let c(t) = (x(t), y(t).The chain rule 6.1.1 can be written as follows.
(f ◦ c)′(t0) = ∇f(c(t0)) · c′(t0)
Note that (f ◦ c)′(t0) an c′(t0) are both (2-dimensional) vectors. The point c(t0) does not de-pend on the parametrization of c. Thus we conclude:
∇f is a map sending a velocity vector atc(t0) to another velocity vector at (f ◦c)(t0).
We can apply this idea to any real-valued functiony = f(x1, x2, · · · , xn). The gradient, or the deriva-tive, of f is
∇f = (fx1, fx2
, · · · , fxn)
We can consider the derivative of f as a matrix. In
that case, the vector⇀
h should also be considered asa column matrix.
f ′(⇀x) ·
⇀
h =(fx fy
)·(x− x0
y − y0
)Definition 6.3. Let f(
⇀x) =
⇀y be a multivari-
able function where⇀x = (x1, x2, · · · , xn) and
⇀y =
(y1, y2, · · · , ym). If there exists a m × n-matrix Msatisfying
lim⇀
h→0
||f(⇀x +
⇀
h)− f(⇀x)−M ·
⇀
h ||
||⇀
h ||= 0
then f is said to be differentiable at⇀x . Here,
all (m-dimensional) vectors in the numerators aretaken as column matrix. Such matrix M is calledthe derivative (or gradient) of f .
Remark 6.4. If f(⇀x) =
⇀y is differentiable, then M
is uniquely determined by
M =
∂y1∂x1
∂y1∂x2
· · · ∂y1∂xn
∂y2∂x1
∂y2∂x2
· · · ∂y2∂xn
......
...
∂ym∂x1
∂ym∂x2
· · · ∂ym∂xn
Moreover, f(
⇀x) =
⇀y is differentiable if and only if
each yj = yj(x1, x2, · · · , xn) (j = 1, · · · ,m) is differ-entiable.
Theorem 6.5. Suppose that s(x, y), t(x, y) are dif-ferentiable at (x0, y0). and f(x, y) is differentiableat (s0, t0) = (s(x0, y0), t(x0, y0)). Then the partialderivatives of F (x, y) = f(s(x, y), t(x, y)) satisfy thefollowing.
∂F
∂x=∂f
∂s
∂s
∂x+∂f
∂t
∂t
∂x,
∂F
∂y=∂f
∂s
∂s
∂y+∂f
∂t
∂t
∂y(6.5.1)
Remark 6.6. The matrix form of the chain rule(6.5.1) is the following.
(∂F∂x
∂F∂y
)=(∂f∂s
∂f∂t
)·
∂s∂x
∂s∂y
∂t∂x
∂t∂y
So the most general form of the the chain rule is thefollowing. Let f(x1, x2, · · · , xn) = (y1, y2, · · · , ym)and g(y1, y2, · · · , ym) = (z1, z2, · · · , zl) be differen-tiable functions. Then F = g ◦ f is also differentiableand
∇F = ∇g · ∇f
Example 6.7. Let w = xy + yz + xz, x = r cos θ,y = r sin θ, z = rθ. Find
∂w
∂r
∣∣∣∣(r,θ)=(2,π2 )
,∂w
∂θ
∣∣∣∣(r,θ)=(1,π4 )
Proposition 6.8. Let f(x, y) be a real-valued func-tion. The implications of the following propertieshold.
10
3
1 2
4
1. The partial derivatives fx, fy are continuous at(x0, y0).
2. The function f is differentiable at (x0, y0).
3. The directional derivative D⇀uf(x0, y0) exists for
every⇀u .
4. The function f is continuous at (x0, y0).
Exercise 6.9. Find the counter-examples for all im-plications not shown in Proposition 6.8.
7 Taylor series
Definition 7.1. Suppose that f(x, y) has continu-ous second partial derivatives at (x0, y0). Then thepolynomial
Q(x, y) = f(x0, y0) + fx(x0, y0)(x− x0) (7.1.1)
+ fy(x0, y0)(y − y0) +1
2fxx(x0, y0)(x− x0)2
+ fxy(x0, y0)(x− x0)(y − y0)
+1
2fyy(x0, y0)(y − y0)2
is called the Taylor polynomial of f of second de-gree at (x0, y0).
Remark 7.2. Let ∆⇀
h = (∆x,∆y) and
∇ =
(∂
∂x,∂
∂y
)Then (7.1.1) simplifies as
Q(x, y) =
2∑n=0
1
n!(⇀
h · ∇)nf(x0, y0)
Note that fxy(x0, y0) = fyx(x0, y0) by the assumption(cf. Theorem 2).
Theorem 7.3 (Taylor). Suppose that all third par-tial derivatives fxxx, fxyx, · · · , fyyy of a functionf(x, y) are continuous on a rectangular region D ={(x, y) | |x − x0| ≤ ε1, |y − y0| ≤ ε2}. Then for each(x, y) ∈ D, there exists a constant 0 ≤ c ≤ 1 satisfy-ing
f(x, y) = Q(x, y) +R2(x, y)
where
R2(x, y) =1
3!(⇀
h ·∇)3f(x0 + c(x−x0), y0 + c(y− y0))
Example 7.4. Find the Taylor polynomial of seconddegree at (1, 1) of f(x, y) =
√xy, and find the error
bound for |x− 1|, |y − 1| ≤ 0.1.
Definition 7.5. Let f(x, y) be a function f(x, y) de-fined on a region D.
• A point (x0, y0) is said to be local maximal(minimal, respectively) at (x0, y0) if there existsa (sufficiently small) ε > 0 such that for all (x, y)satisfying
√(x− x0)2 + (y − y0)2 < ε,
f(x0, y0) ≥ f(x, y) ( f(x0, y0) ≤ f(x, y), respectively)
holds. Such points are called extremal.
• A point (x0, y0) is called a critical point if itsatisfies one of the following.
1. fx(x0, y0) = fy(x0, y0) = 0
2. fx or fy does not exist at (x0, y0)
3. f is discontinuous at (x0, y0).
A critical point which is not an extremal pointis called a saddle point.
Definition 7.6. Suppose that all second partialderivatives of a function f(x, y) are continuous on aregion R containing (x0, y0). Then
∆ = fxx(x0, y0)fyy(x0, y0)− fxy(x0, y0)2
is called the discriminant of f .
Theorem 7.7. Let f be a function satisfying condi-tions in Definition 7.6 and (x0, y0) a critical point off .
• If ∆ > 0 and fxx(x0, y0) > 0, then f(x0, y0) is alocal minimum.
• If ∆ > 0 and fxx(x0, y0) < 0, then f(x0, y0) is alocal maximum.
11
• If ∆ < 0, then f(x0, y0) is a saddle point.
• If ∆ = 0, then we do not know local extremity off(x0, y0).
Example 7.8. Find the point on the graph of y2 =9 + xz which is the closest to the origin.
8 Lagrange multiplier
Proposition 8.1. Let c = f(x0, y0) and Lc(f) be thelevel set c = f(x, y) on the xy-plane. (cf. (3.3.1)).Then ∇f(x0, y0) is perpendicular to the tangent lineat (x0, y0) of the curve Lc(f).
Remark 8.2.
Theorem 8.3 (Lagrange multiplier). Let g(x, y),f(x, y) be differentiable functions. Let (x0, y0) be apoint on the level set g(x, y) = c where f(x, y) is lo-
cally extremal. If ∇g(x0, y0) 6=⇀0 , then there exists λ
such that
∇f(x0, y0) = λ∇g(x0, y0)
Example 8.4. Find the point on the circle x2 +y2 =10 at which the function f(x, y) = 3x+ y is maximalor minimal.
Corollary 8.5. The gradient ∇f(x0, y0) is the di-rection where the value of function f(x, y) changesfastest from (x0, y0).
Exercise 8.6. Let
f(x, y) = 100 +100
x2 + 2y2 + 9
be a function whose graph represent the contour of amountain. Suppose that a water flow from the point(1, 0, 110) down the valley. Find the trajectory of thewater path.
Exercise 8.7. Find the point on the graph xy2z3 = 2which is the closest to the origin.
Exercise 8.8. Find the minimal and maximal valueof f(x, y, z) = x3+y3+z3 on the sphere x2+y2+z2 =1 on the first octant. (Figure below shows a level setof f .)
9 Vector spaces
Definition 9.1. A vector space V is a set of ele-ments, called vectors, satsifying the following proper-ties.
1. For any two vectors⇀v ,
⇀w, the sum
⇀v +
⇀w is also
a vector and satisfy⇀v +
⇀w =
⇀w +
⇀v
(⇀v +
⇀w) +
⇀r =
⇀v + (
⇀w +
⇀r )
2. For any scalar k, such as real numbers, k · ⇀v isalso a vector and satisfy
k(⇀v +
⇀w) = k
⇀v + k
⇀w
3. There exists a unique zero vector⇀0 satisfying
⇀0 +
⇀v =
⇀v +
⇀0 =
⇀v
for every vector⇀v in V .
Definition 9.2. Let⇀v 1,
⇀v 2, · · · ,
⇀v n be vectors in Rn
and c1, c2, · · · , cn real numbers. Then the vector
c1⇀v 1 + c2
⇀v 2 + · · ·+ cn
⇀v n
is called the linear combination (or linear sum) of⇀v i’s. A set of vectors {⇀v 1,
⇀v 2, · · · ,
⇀v n} is said to be
linearly independent if the constants c1, c2, · · · , cnsatisfy
c1⇀v 1 + c2
⇀v 2 + · · ·+ cn
⇀v n =
⇀0
if and only if c1 = c2 = · · · = cn = 0.
12
Definition 9.3. Let S = {⇀v 1,⇀v 2, · · · ,
⇀v n} be a set
of (n-dimensional) vectors. A vector space spanned
by S is the set of all linear combinations of⇀v i’s.
V = span〈S〉 = {c1⇀v 1 + c2
⇀v 2 + · · ·+ cn
⇀v n | ci ∈ R}
Note that V is a vector space by Definition 9.1. IfS is linearly independent, then it is called the basisfor V , and the elements
⇀v 1,
⇀v 2, · · · ,
⇀v n are called
basis vectors The size of S, namely n, is called thedimension of V , and denoted by
n = dimR V
Example 9.4. Show that two sets of vectors
S1 = {(1, 1, 1), (1, 0,−1)}, S2 = {(0, 1, 2), (2, 1, 0)}
spans the same vector space in R3 of dimension 2.
Remark 9.5. A finite dimensional vector space canbe identified as Rn. However, a vector space is calledinfinite dimensional if no finite collection of vectorscan span the vector space. Here are some examplesof infinite dimensional vector spaces.
1. The set of all continuous function on [0, 1].
C([0, 1]) = {f : [0, 1]→ R |f is continuous on [0, 1]}
2. The set of infinite-tuples.
R∞ = {(x1, x2, x3, · · · ) |xi ∈ R, xi = 0 except for finitely many i’s}
Definition 9.6. Let A be a n× n square matrix.
A =
a11 a12 · · · a1m
a21 a22 · · · a2m
......
...an1 an2 · · · anm
(9.6.1)
Let⇀c 1,
⇀c 2, · · · ,
⇀c n be the n column vectors from A.
⇀c 1 =
a11
a21
...an1
,⇀c 2 =
a12
a22
...an2
, · · · ,⇀c n =
a1n
a2n
...ann
The rank of A is the maximal number of linearlyindependent
⇀c i’s.
Example 9.7. Find the rank of
A =
−2 1 −1−1 1 22 −1 −1
Remark 9.8. Two questions arise naturally.
Q1 Why does rank of a matrix always defineduniquely?
Q2 What if we use row vectors, instead of columnvectors, to define the rank?
To answer the first question, we observe the followingidentity.
rank A = dimR span〈S〉
where S = {⇀c 1,⇀c 2, · · · ,
⇀c n}. Thus the uniqueness
of rank A is equivalent to the uniqueness of dimen-sion. Suppose that two sets S1, S2 of (n-dimensional)vectors
S1 = {v1, · · · , vk}S2 = {w1, · · · , wl}, l 6= k
spans the same vector space V . If k > l, then S1
cannot be linearly independent. Thus the dimensionis uniquely determined.
Let R = {⇀r 1,⇀r 2, · · · ,
⇀r n} be the set of row vectors
of A in (9.6.1).
⇀r 1 =
(a11 a12 · · · a1n
)⇀r 2 =
(a21 a22 · · · a2n
)...
⇀r n =
(an1 an2 · · · ann
)Then Q2 is equivalent to
dimR span〈S〉 ?= dimR span〈R〉
13
This identity is always true. This follow from thesimple algorithm to compute the rank of a matrixusing an echelon form. An echelon form
0 0 · · · ∗...
... . .. ...
0 ∗ · · · ∗0 ∗ · · · ∗
can be obtained by the following column operations:
1. replace a column with itself by multiplying a con-stant,
2. interchaning two columns, and
3. multiply a column by a scalar, and then subtractfrom another column.
The number of nonzero entries, ∗, on the diagonal isthe rank. One can obtain an echelon form
∗ · · · ∗ ∗...
. . ....
...0 · · · ∗ ∗0 · · · 0 0
by taking a row operations, which is similarly definedas above. The crucial fact is that either way, the num-ber of nonzero diagonal entries are the same. Thatis,
dimR span〈S〉 = rank A = dimR span〈R〉
Theorem 9.9. A n×n matrix A nonsingular if andonly if rank A = n.
10 Eigenvalues and eigenvec-tors
Definition 10.1. Let A be a n × n-matrix. If anonzero vector
⇀v and a contant λ satisfy
A ·⇀v = λ⇀v ,
then⇀v is called the eigenvector, and λ the eigen-
value of A. Here we consider vector⇀v as a n × 1-
matrix.
Example 10.2. Show that the vectors
⇀v 1 =
(01
),
(11
)
are eigenvectors of
A =
(2 01 1
)What are the eigenvalues?
Theorem 10.3. A constant λ is a eigenvalue of Aif and only if it is a root of
det(A− λIn) = 0 (10.3.1)
The polynomial (10.3.1) is called the characteristicpolynomial of A.
Example 10.4. The fundamental theorem of alge-bra states that every polynomial has at least one root.Thus there are always n eigenvalues (real and com-plex) counting multiplicity. Let us see how to get theeigenvectors for each eigenvalue.
1. Let
A =
(−1 01 3
)The characteristic polynomial is (−1 − λ)(3 −λ) = 0. Thus λ = −1, 3 are two eigenvalues. If⇀v is an eigenvector for −1, then
A ·⇀v = −⇀v =⇒ (A+ I) ·⇀v = 0
That is, (0 01 4
)(v1
v2
)=
(00
)Thus v1 + 4v2 = 0 and
⇀v = (−4, 1) is an eigen-
vector. Similarly, we get (0, 1) as an eigenvectorfor 3.
2. Let
A =
(1 1−1 3
)Since characteristic polynomial is (λ − 2)2 = 0,there is only one eigenvalue λ = 2. The corre-sponding eigenvector
⇀v satisfies(
−1 1−1 1
)·(v1
v2
)=
(00
)so
⇀v = (1, 1) is an eigenvector. Note that such
eigenvector is unique up to a scalar multiplica-tion.
14
3. Let
A =
2 1 11 2 11 1 2
The characteristic polynomial is (λ−1)2(λ−4) =
0. For λ = 1, the eigenvector⇀v must satisfy1 1 1
1 1 11 1 1
v1
v2
v3
=
000
There are two linearly independent eigenvectors.For example,
⇀v 1 = (−1, 1, 0),
⇀v 2 = (−1, 0, 1)
are linearly independent eigenvectors for λ = 1.
4. The eigenvalue can be a complex number. Let
A =
(1 −11 1
)Then the characteristic polynomial is λ2 − 2λ+2 = 0, whose roots are 1 ± i7. The eigenvectorcorresponding a complex eigenvalue may havecomplex entries. For λ = 1 + i, the eigenvec-tor
⇀v satisfies(
−i −11 −i
)(v1
v2
)=
(00
)Then
⇀v = (i, 1) is an eigenvector for λ = 1 + i.
We can check that the conjuagte⇀v = (−i, 1) is
an eigenvector for λ = 1− i.
Proposition 10.5. Let λ be a complex eigenvaluefor A and
⇀v a corresponding eigenvector. Then λ is
also an eigenvalue for A and⇀v is a corresponding
eigenvector.
Definition 10.6. Let⇀x = (x1, x2, · · · , xn) be a vec-
tor whose entries xi’s are functions of t. Let us write
dx1
dt
dx2
dt
...
dxndt
=d
dt
x1
x2
...
xn
7Note that complex roots always come as pair.
The system of linear DE
dx1
dt= a11x1 + a12x2 + · · ·+ a1nxn + f1(t)
dx2
dt= a21x1 + a22x2 + · · ·+ a2nxn + f2(t)
...
dxndt
= an1x1 + ab2x2 + · · ·+ abnxn + fn(t)
can be written as
d⇀x
dt= A ·⇀x +
⇀
f (10.6.1)
where
A =
a11 a12 · · · a1m
a21 a22 · · · a2m
......
...an1 an2 · · · anm
and
⇀
f = (f1, f2, · · · , fn). The system of DE (10.6.1)
is called homogeneous if⇀
f =⇀0 , or nonhomoge-
neous otherwise.
11 Solving system of homoge-neous DE
Definition 11.1. Let
d⇀x
dt= A ·⇀x (11.1.1)
be a system of homogeneous DE. Let λ1, · · · , λnbe the eigenvalues of A (counting multiplicity), and⇀v 1,
⇀v 2, · · · ,
⇀v n corresponding eigenvectors. Then
the vectors
⇀x1 = eλ1t⇀v 1,
⇀x2 = eλ2t⇀v 2, · · · , ⇀
xn = eλnt⇀v n
are called the solution vectors.
Proposition 11.2. Let⇀x1,
⇀x2, · · · ,
⇀xn be solu-
tion vectors for (11.1.1). Then for any constantsc1, c2, · · · , cn,
c1⇀x1 + c2
⇀x2 + · · ·+ cn
⇀xn
is also a solution.
15
Definition 11.3. Let⇀x1,
⇀x2, · · · ,
⇀xn be solution
vectors. If
c1⇀x1 + c2
⇀x2 + · · ·+ cn
⇀xn = 0
hold if and only if c1 = c2 = · · · = cn = 0, then⇀x1,
⇀x2, · · · ,
⇀xn are called linearly independent.
Otherwise, it is called linearly dependent.
Proposition 11.4. Let
⇀x1 =
x11
x21
...xn1
,⇀x2 =
x12
x22
...xn2
, · · · , ⇀xn =
x1n
x2n
...xnn
be solution vectors where each entry xij is an ana-
lytic8 function. Then⇀x1,
⇀x2, · · · ,
⇀xn is linearly in-
dependent if and only if the Wronskian
W (⇀x1,
⇀x2, · · · ,
⇀xn) =
∣∣∣∣∣∣∣∣∣x11 x12 · · · x1n
x21 x22 · · · x2n
......
...xn1 xn2 · · · xnn
∣∣∣∣∣∣∣∣∣is nonzero.
Theorem 11.5. Let λ1, λ2, · · · , λn be eigenvalues forA counting multiplicity. Suppose that correspondingeigenvectors
⇀v 1,
⇀v 2, · · · ,
⇀v n are linearly independent.
Then
⇀x = c1
⇀x1 + c2
⇀x2 + · · ·+ cn
⇀xn
= c1eλ1t⇀v 1 + c2e
λ2t⇀v 2 + · · ·+ cneλnt⇀v n
is the general solution of (11.1.1).
Example 11.6. Find the general solution of
dx
dt= 2x
dy
dt= x+ 3y − 2z
dz
dt= −x+ z
Theorem 11.7. Suppose that λ1 be an eigenvaluefor A with multiplicity m9 (m < n). Furthermore,
8A function x(t) is analytic if its Taylor expansion alwaysconverges to x(t).
9That is, counting multiplicity, the spectrum of eigenvaluesis
λ1 = λ2 = · · · = λm, λm+1, · · · , λn
suppose that⇀v 1 is the only eigenvector corresponding
to λ1. Then for⇀v 2,
⇀v 3, · · · ,
⇀vm satisfying
(A− λ1I)⇀v 1 =
⇀0
(A− λ1I)⇀v 2 =
⇀v 1
...
(A− λ1I)⇀v n =
⇀v n
The vectors
⇀x1 = eλ1t⇀v 1
⇀x2 = eλ1t⇀v 2 + teλ1t⇀v 1
...
⇀xm = eλ1t⇀vm + teλ1t⇀vm−1 + · · ·+ tm
(m− 1)!eλ1t⇀v 1
are solution vectors for (11.1.1).
Example 11.8. Find the general solution of
dx
dt= x+ y
dy
dt= −x+ 3y
Theorem 11.9. Suppose that λ1 be a complex eigen-values of A, and
⇀v 1 a corresponding (complex) eigen-
vectors. Then
⇀x1 = eλ1t⇀v 1,
⇀x2 =
⇀x1 = eλ1t⇀v 2
are (complex) solution vectors for (11.1.1). In terms
of real functions, if λ = α+ iβ and⇀v 1 =
⇀p + i
⇀q ,
⇀t 1 = eαt
(⇀p cosβt+
⇀q sinβt
)⇀t 2 = eαt
(⇀p cosβt−⇀
q sinβt)
are solution vectors.
Example 11.10. Find the general solution of
dx
dt= 3x− y
dy
dt= 2x+ y
12 Methods of solving systemsof nonhomogeneous DEs
Theorem 12.1. Let
d⇀x
dt= A ·⇀x +
⇀
f (12.1.1)
16
be a system of nonhomogeneous DE. A solution⇀xu
satisfying
d⇀x
dt= A ·⇀x
is called a complementary solution of (12.1.1).
A solution⇀xp without any parameters satisfying
(12.1.1) is called a particular solution. If⇀xu is
the general complementary solution, then the generalsolution of (12.1.1) is given by
⇀x =
⇀xu +
⇀xp
Example 12.2 (Method of determined coefficients).Find the general solution of
dx
dt= x+ 3y − 3
dy
dt= 3x+ y + 2
Theorem 12.3. Let
⇀xu = c1
⇀x1 + c2
⇀x2 + · · ·+ cn
⇀xn
be the complementary solution of (12.1.1). Let usdenote
Φ =
x11 x12 · · · x1n
x21 x22 · · · x2n
......
...xn1 xn2 · · · xnn
Then
⇀xp = Φ ·
∫Φ−1 ·
⇀
f dt
is a particular solution of (12.1.1).
Example 12.4. Find the general solution of
dx
dt= 3x+ 2y − e−t
dy
dt= x+ 2y − 2
13 Double integrals
Definition 13.1. Let D = [a, b] × [c, d] be a rect-angular region on R2. Let us subdivide the intervalsa = x0 < x1 < · · · < xn = b, c = y0 < y1 < · · · <yn = d such that for each i = 1, · · · , n
∆x = xi − xi−1 =b− an
, ∆y = yi − yi−1 =d− cn
Let f(x, y) be a function defined on D. If for anychoice of xi−1 ≤ x∗i ≤ xi and yi−1 ≤ y∗i ≤ yi, thelimit
limn→∞
n∑i,j=1
f(x∗i , y∗i )∆x∆y
converges to a unique value R, then f is said to beintegrable on D. The value R is called the definiteintegral of f , and denoted by
R =
∫∫D
f(x, y)dxdy
Theorem 13.2. If f is continuous on the closed re-gion D = [a, b]× [c, d], then f is integrable on D.
Remark 13.3. The function f(x, y) on [0, 1]× [0, 1]defined by
f(x, y) =
{1 x or y is rational
0 otherwise
is not integrable on [0, 1]× [0, 1].
Theorem 13.4 (Fubini I). Let f(x, y) be a continu-ous function on D = [a, b]× [c, d]. Then∫∫
D
f(x, y)dxdy =
∫ b
a
(∫ d
c
f(x, y)dy
)dx
=
∫ d
c
(∫ b
a
f(x, y)dx
)dy
Example 13.5. Compute the followings.
1. The volume under the graph z = 10 + x3 + 3y2
on [0, 1]× [0, 2].
2. The volume of paraboloidx2
4+y2
9+ z = 1 on
[−1, 1]× [−2, 2].
3. The volume bounded by z = x sec2 y, z = 0,x = 0, x = 2, y = 0, and y = π/4.
4.
∫ 1
0
∫ 1
0
y
1 + xydxdy.
5.
∫∫[−1,1]×[0,π]
xy cos ydA.
6.
∫∫[0,1]×[0,2]
xy3
x2 + 1dA.
7.
∫∫[1,2]×[1,2]
1
xydA.
17
8. The volume under the plane z = 2 − x − y on[0, 1]× [0, 1].
9. The volume under the graph z = 2 sinx cos y on[0, π/2]× [0, π/4].
Exercise 13.6. Let
f(x, y) =
x2 − y2
(x2 + y2)2(x, y) 6= (0, 0)
0 (x, y) = (0, 0)
Show that∫ 1
0
∫ 1
0
f(x, y)dydx 6=∫ 1
0
∫ 1
0
f(x, y)dxdy.
Explain why Fubini’s theorem 13.4 does not apply.
Definition 13.7. Let D be a bounded region in R2,on which the function f(x, y) is defined. We say f isintegrable on D if for some (sufficiently large) rect-angular region [a, b]×[c, d] containing D, the functiondefined by
F (x, y) =
{f(x, y) (x, y) ∈ D0 (x, y) /∈ D
is integrable on [a, b]× [c, d] (cf. Definition 13.1), and∫∫D
f(x, y)dxdy =
∫∫[a,b]×[c,d]
F (x, y)dxdy
is called the definite integral of f .
Theorem 13.8 (Fubini II). Let D = {(x, y) | a ≤x ≤ b, g1(x) ≤ y ≤ g2(x)} be a region in R2 andf(x, y) a continuous function on D. Then∫∫
D
f(x, y)dxdy =
∫ b
a
(∫ g2(x)
g1(x)
f(x, y)dy
)dx
Similarly, if D = {(x, y) | c ≤ y ≤ c, h1(x) ≤ y ≤h2(x)}, then∫∫
D
f(x, y)dxdy =
∫ d
c
(∫ h2(x)
h1(x)
f(x, y)dx
)dy
Example 13.9. Compute the following integral.
1.
∫ 1
0
∫ s2
0
cos(s3)dtds.
2.
∫∫D
y2dA where D is the triangle with vertices
(0, 1), (2, 2), (4, 1).
3.
∫∫D
(2x− y)dA where D is the disk with radius
2 and center (0, 0).
4.
∫ 1
0
∫ y3
0
3y4exydxdy
5.
∫∫D
e−y2
dA where D = {(x, y) | 0 ≤ x ≤ 1, 0 ≤
y ≤ x}.
Definition 13.10. Let D be a region in R2 whosedensity at (x, y) is δ(x, y). Then the point (x, y)defind by
x =
∫D
xδ(x, y)dxdy∫D
δ(x, y)dxdy
, y =
∫D
yδ(x, y)dxdy∫D
δ(x, y)dxdy
is called the center of mass.
Example 13.11. Find the center of mass of D ={(x, y) |x2 + y2 ≤ 4} where the density is δ(x, y) =2x− y.
14 Integration by substitution
Definition 14.1. A function
T (x, y) = (u(x, y), v(x, y))
is called a transformation if it is one-to-one. TheJacobian of T is defined by
JT =
∣∣∣∣∂(u, v)
∂(x, y)
∣∣∣∣ = det
[ux uyvx vy
]Example 14.2. Find the Jacobian of T (x, y) = (x−y, x + y). Let D = [0, 1] × [0, 1]. Let D′ = T (D).What is the region D′? What is its area? Verify that
area D′ = |JT | · area D
Find the area of codomain of the transformationT (x, y) = (ax+ by, cx+ dy) with the domain [0, 1]×[0, 1].
Theorem 14.3. Let T : D → D′ be a transforma-tion, and f(x, y) be a continuous function defined onD′. Then∫∫
D′f(u, v)dudv =
∫∫D
f ◦ T (x, y)|JT |dxdy
18
Example 14.4. The map
T (r, θ) = (r cos θ, r sin θ)
is called the polar transformation. Compute thefollowing.
1. The area outside r = 1 and inside r = 1 + cos θ.
2. The area inside r2 = 4 cos2 θ.
Exercise 14.5. Compute the following.
1.
∫ 1
0
∫ 1−x
0
ey/(x+y)dydx
2.
∫D
x2+y2dxdy, where D is bounded by x2−y2 =
1, x2 − y2 = 9, xy = 2, and xy = 4.
3.
∫ 2
1
∫ x
√x
sin
(πx
2y
)dydx+
∫ 4
2
∫ 2
√x
sin
(πx
2y
)dydx
4.
∫D
e−(x2+xy+y2)dxdy where D = {(x, y) |x2 +
xy + y2 ≤ 1}
Remark 14.6. The notation JT =∂(u, v)
∂(x, y)is useful
to remember the subsitution rule:∫∫D
f(u, v)dudv =
∫∫T (D)
f(u, v)
∣∣∣∣∂(u, v)
����∂(x, y)
∣∣∣∣���dxdy
For the inverse transformation T−1(u, v) =(x(u, v), y(u, v)), the following holds.
JT−1 =1
JT,
∂(x, y)
∂(u, v)=
1
∂(u, v)
∂(x, y)
Definition 14.7. The integral
∫∫D
f(x, y)dxdy is
called improper if it satisfies one of the following.
• the region D is unbounded, or
• the function diverges at some point in D.
Exercise 14.8. Compute the following integrals.
1.
∫∫R2
e−x2−y2dxdy
2.
∫ ∞0
e−x2
dx (Hint: use Fubini’s theorem)
15 Triple integrals
Definition 15.1. Let f(x, y, z) be a function definedon the box [a, b]× [c, d]× [e, f ]. The triple integral off(x, y, z) is denoted by∫∫∫[a,b]×[c,d]×[e,f ]
f(x, y, z)dxdydz or
∫∫∫[a,b]×[c,d]×[e,f ]
f(x, y, z)dV
Theorem 15.2 (Fubini III). Let D = {(x, y, z) | a ≤x ≤ b, g1(x) ≤ y ≤ g2(x), h1(x, y) ≤ z ≤ h2(x, y)}be a region in R3 on which the function f(x, y, z) iscontinuous. Then∫∫∫
D
f(x, y, z)dxdydz
=
∫ b
a
(∫ g2(x)
g1(x)
(∫ h2(x,y)
h1(x,y)
f(x, y, z)dz
)dy
)dx
Exercise 15.3. Find the integral∫ 3
0
∫ 4
0
∫ y/2+1
y/2
2x− y2
+z
3dxdydz
Definition 15.4. The map
T (x, y, z) = (u(x, y, z), v(x, y, z), w(x, y, z))
defined on a region D ⊂ R3 is called a transforma-tion if it is one-to-one. The Jacobian of T is
JT =∂(u, v, w)
∂(x, y, z)=
∣∣∣∣∣∣ux uy uzvx vy vzwx wy wz
∣∣∣∣∣∣Theorem 15.5. Let T be a transformation from D ⊂R3 to D′ ⊂ R3 and f(x, y, z) be a continuous functionon D. Then∫∫∫
D′f(u, v, w)dudvdw
=
∫∫∫D
(f ◦ T )(x, y, z)|JT |dxdydz
Remark 15.6. As we have seen in Remark 14.6,the Jacobian of the inverse transformation is the re-ciprocal of original Jacobian. This also works forT (x, y, z) = (u, v, w).
JT =1
JT−1
,∂(u, v, w)
∂(x, y, z)=
1∂(x,y,z)∂(u,v,w)
Example 15.7. The map
T (r, θ, z) = (r cos θ, r sin θ, z)
is called the cylinderical transformation whoseJacobian is JT = r. Compute the following.
19
1. The volume between z = 9 − x2 − y2 andz = 3x2 + 3y2 − 16.
2. The area of the regionbounded by two cylindersx2 + y2 ≤ 1, y2 + z2 ≤ 1.
3. The area of the regionbounded by three cylindersx2 + y2 ≤ 1, y2 + z2 ≤ 1,and x2 + z2 ≤ 1.
4. Find the common area between x2 + y2 ≤ 1,x2 + z2 ≤ 1, and z ≥ 1/2.
Example 15.8. The map
T (r, θ, φ) = (r sinφ cos θ, r sinφ sin θ, r cosφ)
is called the spherical transformation whose Ja-cobian is JT = r2 sinφ. Compute the following.
1.
∫∫∫D
√1− x2 − y2 − z2dxdydz where D =
{(x, y, z) |x2 + y2 + z2 < 1}.
2.
∫ ∞0
∫ ∞0
∫ ∞0
e−x2−y2−z2dxdydz
Definition 15.9. Let D be a region in R3 andδ(x, y, z) the density at the point (x, y, z) ∈ D. Thepoint (x, y, z) where
x =
∫∫∫D
xδ(x, y, z)dV∫∫∫D
δ(x, y, z)dV
y =
∫∫∫D
yδ(x, y, z)dV∫∫∫D
δ(x, y, z)dV
z =
∫∫∫D
zδ(x, y, z)dV∫∫∫D
δ(x, y, z)dV
is called the center of mass.
16 Line integrals
Definition 16.1. Let I be an interval in R. A func-tion c : I → Rn
c(t) = (x1(t), x2(t), · · · , xn(t))
is said to be piecewise Cn if each component xi(t)(i = 1, 2, · · · , n) is Cn10 on I except for finitely manypoints.
Definition 16.2. Let C be a curve on R2
parametrized by a piecewise C1 function c : [a, b] →R2. The line integral of f along C is∫
C
fds =
∫ b
a
f(c(t)) · ||c′(t)||dt
where c′(t) is the velocity vector at c(t).
Remark 16.3. We can define a line integral alonga curve on R3 in similar way. Note that we need aparametrization to compute a line integral, and theorientation of parametrization matters. However, ifthe orientations are the same, any two parametriza-tions of the same curve give the same line integral.
Example 16.4. Compute the line integral
∫C
2 +
x2yds along C where C is the unit circle at the origin.
Definition 16.5. A (2-dimensional) vector field onD ⊂ R2 is an assignment of a (2-dimensional) vector⇀
F (x, y) to each point (x, y) ∈ D. Similarly, we candefine a 3-dimensional vector field.
Definition 16.6. Let⇀
F (x, y) be a vector field onD, and C a curve in D. Let c : [a, b] → R2 be a
parametrization of C. The line integral of⇀
F alongC is ∫
C
⇀
F · d⇀s =
∫ b
a
⇀
F (x, y) · c′(t)dt
Example 16.7. Find the line integral of⇀
F (x, y) =12 (−y, x) along c(t) = (r cos t, r sin t) for 0 ≤ t ≤ 2π.
Theorem 16.8. Let⇀
F be a vector field on D ⊂ R2
and c : [a, b]→ D a piecewise C1-curve. For any one-to-one, C1-function f : [c, d]→ [a, b], the compositionc = c ◦ f gives a reparametrization of C = c([a, b]),and the following identity holds.∫C
⇀
F · d⇀s =
∫ b
a
F ◦ c(t) · c′(t)dt =
∫ b
a
F ◦ c(t) · c′(t)dt
Example 16.9. Let⇀a be a vector field defined by
⇀a =
(− y
x2 + y2,
x
x2 + y2
)(16.9.1)
10A function x(t) is called Cn if the n-th derivative x(n) iscontinuous.
20
Note that⇀a is defined on R2 \ {0}. The line integral
of⇀a along a curve C gives an angular displacement
of start point and end point. Compute∫C
⇀a · d⇀s
with the parametrization c(t) = (cosnt, sinnt), 0 ≤t ≤ 2π.
Definition 16.10. Let⇀
F be a vector field. A func-tion ϕ(x, y) satisfying
∇ϕ =⇀
F
is called the potential function of⇀
F .
Theorem 16.11. Suppose that ϕ(x, y) is C1 on aregion D ⊂ R2. For any curve C from p0 to p1 lyingin D ∫
C
∇ϕ · d⇀s = ϕ(p1)− ϕ(p0)
Corollary 16.12. If a vector field⇀
F admits a poten-tial function on D, then∮
C
⇀
F · d⇀s = 0
for any closed curve C in D.
Example 16.13. Show that the function
θ(x, y) = arctan(y/x)
is a potential function of⇀a in (16.9.1) on the first
quadrant x, y > 0. Prove that⇀a does not admit a
potential function on R2 \ {(0, 0)}.
Definition 16.14. A vector field⇀
F (x, y) =(f(x, y), g(x, y)) is said to be closed if
fy = gx
holds.
Theorem 16.15. A vector field is closed if it admitsa potential function.
Example 16.16.
1. Show that⇀
F = (x2+y2, x2−y2) has no potentialfunction.
2. On R2 \ {(0, 0)}, the vector field⇀a in (16.9.1) is
closed, but does not admit any potential func-tion.
17 Curl and divergence
Definition 17.1. Let⇀
F (x, y) = (f(x, y), g(x, y)) bea C1-vector field. The function
div⇀
F =∂f
∂x+∂g
∂y
is called the divergence of⇀
F .
Example 17.2. A vector field can be viewed as a‘flow’. The divergence measures the rate of change ofarea which flows along the vector field. For example,the divergence of the position vector field
⇀r (x, y) = (x, y)
is 2. Suppose that the area
D = {(r, θ) | a ≤ r ≤ b, 0 ≤ θ ≤ φ}
flows along⇀r . The area at t = 0 is
A(0) =1
2(b− a)2φ
After a short period of time ∆t, the area changes to
A(∆t) =1
2(b− a)2(1 + ∆t)2φ
Thus the rate of change is
lim∆t→0
A(∆t)− 12 (b− a)2φ
∆t= div
⇀r A(0)
Definition 17.3. If div⇀
F = 0, then⇀
F is called in-compressible.
Example 17.4. The vector field⇀
F = (−y, x) is in-compressible.
Definition 17.5. Let⇀
F (x, y) = (f(x, y), g(x, y)) bea C1-vector field. The function
rot⇀
F =∂g
∂x− ∂f
∂y
is called the rotational vector field. If
rot⇀
F =⇀0
then⇀
F is said to be irrotational.
Theorem 17.6. A vector field is irrotational if itadmits a potential function.
21
Proposition 17.7. Let⇀
F be a vector field defined onD ⊂ R2 and Cr be a circle at p ∈ D of radius r.
rot⇀
F (p) = limr→0
1
πr2
∮Cr
⇀
F · d⇀s
Theorem 17.8 (Poincare lemma). Let D be a convex
region in R2 and⇀
F a vector field defined on D. If⇀
F is
closed, then⇀
F admits a potential function. Further-more, if D is connected, then the potential functionis unique up to constant.
Theorem 17.9 (Green). Let D be a connected regionin R2 and C = ∂D be the boundary curve. For any
C1-vector field⇀
F ,∮C
⇀
F · d⇀s =
∫∫D
rot⇀
FdA
where the curve in the line integral is parametrizedcounter-clockwise orientation.
Remark 17.10. For⇀
F = (f(x, y), g(x, y)) and acurve C,∫
C
f(x, y)dx+ g(x, y)dy =
∫C
⇀
F · d⇀s
is alternative expression for the line integral.
Example 17.11. Find the line integrals usingGreen’s theorem. (All closed curves are orientedcounter-clockwise.)
1.
∮C
(x− y)dx+ (x+ y)dy where C is the circle at
the origin with radius 2.
2. Let C be the boundary of the region bounded byy = x2 and x = y2.∮
C
(y + e√x)dx+ (2x+ cos2 y)dy
3. Let C be the circle (x−2)2+(y−3)2 = 1 and⇀
F =
(y− log(x2 + y2), 2 tan−1(y/x)). Find
∫C
⇀
F · d⇀s .
18 Surface integrals
Definition 18.1. Let D be a region in R2 and S asuface in R3. A one-to-one function X : D → S
X(u, v) = (x(u, v), y(u, v), z(u, v))
is called the parametrization of S if every partialderivative of x, y, z is continuous on D.
Example 18.2. Find the parametrization of the fol-lowing surfaces.
1. x2 + y2 − z2 = 1
2. x2 − y2 − z2 = 1
Definition 18.3. Let f(x, y, z) be a function definedon a surface S, parametrized by X : D → S. Thesurface integral of f on S is defined by∫∫
S
f(x, y, z)dS =
∫∫D
(f ◦X)(u, v)||⇀
Xu×⇀
Xv||dudv
(18.3.1)
Example 18.4. Find the surface integral off(x, y, z) = z on the surface z =
√x2 + y2 over the
disk x2 + y2 ≤ 1.
Remark 18.5. If f(x, y, z) is a density function, thesurface integral (18.3.1) is the mass of the surface.
1. Find the center of mass of z =√
1− x2 − y2
2. Show that∫∫S
dS√(x− 1)2 + (y − 2)2 + (z − 3)2
= 16π
where S is x2 + y2 + z2 = 16.
Definition 18.6. Let X : D → S be a parametriza-tion of S. If the vector field
⇀n =
⇀
Xu ×⇀
Xv
||⇀
Xu ×⇀
Xv||(18.6.1)
is continuous, then we say S has an orientationand
⇀n is called an orientation.
Example 18.7. The vector field
⇀
A =(x, y, z)√x2 + y2 + z2
is an orientation of the sphere x2 + y2 + z2 = 1.
However, the Mobius band does not have an orienta-tion, since there are two orientation at each point onthe surface.
22
Definition 18.8. Let S be an oriented surface, and⇀
F a vector field defined on S. Then the surface
integral of⇀
F is defined by∫∫S
⇀
F · d⇀
S =
∫∫S
⇀
F ·⇀ndS
and called the flux of⇀
F on S.
Example 18.9. Let S be the plane 2x+ 4y + z = 8bounded by x = 0, y = 0, z = 0. Find the flux of the
vector field⇀
F = (x,−y, 0) on the surface S.
Remark 18.10. Let X : D → S be the parametriza-tion of S. For
⇀n in (18.6.1),∫∫
S
⇀
F · d⇀
S =
∫∫D
(⇀
F ◦X)(u, v) · (⇀
Xu ×⇀
Xv)dudv
Exercise 18.11. Let the orientation⇀n of the surface
z = 1−x2− y2, z ≥ 0 is upward, i.e.⇀n ·
⇀
k ≥ 0. Findthe surface integral of
⇀
F =(x, y, z)
x2 + y2 + z2
on the surface.
19 Stokes theorem
Definition 19.1. Let⇀
F = (P,Q,R) be a (3-dimensional) vector field. The vector field
∇×⇀
F =
∣∣∣∣∣∣∣⇀i
⇀j
⇀
k∂∂x
∂∂y
∂∂z
P Q R
∣∣∣∣∣∣∣= (Ry −Qz, Pz −Rx, Qx − Py)
is called the curl of⇀
F .
Theorem 19.2 (Stokes). Let S be an oriented sur-
face with a piecewise continuous boundary C. For⇀
Fbe a continuous vector field defined on S. Then∮
C
⇀
F · d⇀s =
∫∫S
∇×⇀
F · d⇀
S
where C and S are positively oriented11.
Example 19.3. Using the same assumption in Ex-
ample 18.9, find the line integral
∮C
⇀
F · d⇀s where C
is the boundary curve of S oriented positively.
Exercise 19.4. Let S be the surface bounded byz = 1 − x2 − y2, z ≥ 0 with upward orientation (
⇀n ·
⇀
k ≥ 0). Confirm that Stokes’ theorem holds for⇀
F =(y,−x, 0).
Remark 19.5. Stokes theorem transforms a line in-
tegral into a surface integral. Suppose that⇀
F isdefined on a simply connected region in R3. Then
∇ ·⇀
F = 0 if and only if⇀
F = ∇ ×⇀
G for some⇀
G. In
this case, the surface integral of⇀
F can be transformedinto a line integral.∫∫
S
⇀
F · d⇀
S =
∮C
⇀
G · d⇀s
Exercise 19.6.
1. Compute Example 18.9 again by using Stokes
theorem. (Note that it only works for ∇·⇀
F = 0)
2. Let C be the intersection of z = 1 − 2(x2 + y2)and z = x2−y2 oriented counter-clockwise. Find∮C
⇀
F · d⇀s where
⇀
F = (y cos(x)− yz, sinx, ez)
20 Divergence theorem
Definition 20.1. Let⇀
F = (P,Q,R) be a (3-dimensional) vector field. The function
∇ ·⇀
F = Px +Qy +Rz
is called the divergence of⇀
F .
Theorem 20.2. Let V be a region in R3 whoseboundary S = ∂V is a closed surface.12 For a vector
field⇀
F defined on V ,∫∫S
⇀
F · d⇀
S =
∫∫∫V
∇ ·⇀
FdV
where the orientation of S is outward.11The boundary is postively oriented if the direction is
counter-clockwise with⇀n being upward.
12A surface is called closed if it has no boundary curve.
23
Example 20.3. Find the surface integral
∫∫S
⇀
F ·d⇀
S
where
⇀
F = (z2x,1
3y3 + tan z, x2z + y2)
and S is the closed surface x2 + y2 + z2 = 1.
Remark 20.4. The divergence theorem can be used
to compute
∫∫S
⇀
F ·d⇀
S for a surface S with boundary.
For example, let S be a parabola x2+y2+z = 2 abovethe plane z = 1. To find the flux of
⇀
F = (z tan−1(y2), z3 ln(x2 + 1), z)
to the upward direction, use the divergence theoremon volume 1 ≤ z ≤ 2− x2 − y2, x2 + y2 ≤ 1.
Remark 20.5. The assumption that⇀
F must be de-fined on V is crucial in the divergence theorem. Forexample, let
⇀
F =(x, y, z)
(x2 + y2 + z2)3/2
Since ∇ · F = 0, in order to compute the flux of⇀
Fon z = 4− x2 − y2, z ≥ 0, one should use the surfacethe hemisphere x2 + y2 + z2 = 4, z ≥ 0, not the diskx2 + y2 ≤ 1 on xy-plane.
Exercise 20.6.
1. Let S = ∂V be a closed surface. Prove the fol-lowing.
(a) For any constant vector field⇀
C,∫∫S
⇀
C · d⇀
S = 0
(b) For any vector field⇀
F ,∫∫S
∇×⇀
F · d⇀
S = 0
2. Find the flux of⇀
F = (x, y, z) on the surface be-low. (It is a surface of a cube with a cornerremoved.)
21 Conservation fields
Theorem 21.1. The following identities hold.
1. ∇× (∇f) = 0
2. ∇ · (∇×⇀
F ) = 0
3. ∇ · (⇀
F +⇀
G) = ∇ ·⇀
F +∇ ·⇀
G
4. ∇× (⇀
F +⇀
G) = ∇×⇀
F +∇×⇀
G
5. ∇ · (f⇀
F ) = f∇ ·⇀
F +⇀
F · ∇f
6. ∇× (f⇀
F ) = f∇×⇀
F +∇f ×⇀
F
7. ∇ · (⇀
F +⇀
G) =⇀
G · (∇×⇀
F )−⇀
F · (∇×⇀
G)
8. ∇ · (∇f ×∇g) = 0
9. Denote ∇2 = ∇ · ∇. Then
∇× (∇×⇀
F ) = ∇(∇ ·⇀
F )−∇2 ·⇀
F
Definition 21.2. Let⇀
F be a 3-dimensional vectorfield.
• If the line integral
∫C
⇀
F ·d⇀s only depends on the
start and end point, i.e.∮C
⇀
F · d⇀s = 0
for any closed curve C, then⇀
F is called conser-vative.
• If a function ϕ(x, y, z) satisfies
∇ϕ =⇀
F ,
then ϕ is called a potential function of⇀
F .
24
• If ∇×⇀
F =⇀0 , then
⇀
F is called closed.
Theorem 21.3. Let D be a simply connected region
and⇀
F defined on D. Then the following hold.∮C
⇀
F · d⇀s ⇐⇒⇀
F = ∇ϕ⇐⇒ ∇×⇀
F = 0
Theorem 21.4 (2-dimensional divergence theorem).Let D be a connected region in R2 and C = ∂D aboundary curve. Let
⇀n be a 2-dimensional vector field
along C which is perpendicular (i.e. normal) to thecurve. Then ∮
C
⇀
F ·⇀nds =
∫∫D
∇ ·⇀
FdA
Exercise 21.5.
1. Does⇀
F = (2xy, x2 + 2yz, y2) admit a potentialfunction? If so, find one.
2. Does
⇀
F = (log x+sec2(x+y), sec2(x+y)+y2
y2 + z2,
z
y2 + z2)
admits a potential function?
3. Is there a vector field⇀
G satifsying ∇ ×⇀
G =(x sin y, cos y, z − xy)? If so, find one.
4. Is⇀
F = (2x− 3,−z, cos z) a conservative field?
25
