Math 20E - Vector CalculusPrepared by Jason Gaddis

7.2 Line Integrals

Remark 1. Let’s back up a little to the last lecture on path integrals. In thiscase, we had a multivariable function f : R3 → R and a path c : [a, b]→ R3.We want to integrate the function f over the path c, that is, to find the areaof the curtain or fence with length ||c(t)|| and height at each point t givenby f(c(t)).

To do this, we partition the interval I = [a, b] as

a = t0 < t1 < t2 < · · · < tN = b.

The arc length of each subpath is

∆si =

∫ ti+1

ti

||c′(t)|| dt.

When N is large, ∆si is small and so we approximate the area as

SN =N−1∑i=0

f(xi, yi, zi)∆si,

where (xi, yi, zi) = c(t) for some t ∈ [ti, ti+1]. By the MVT for integrals, weknow

∆si = ||c′(t∗i )||∆tifor some t∗i ∈ [ti, ti+1] and ∆ti = ti+1 − ti. Then

limN→∞

SN = limN→∞

N−1∑i=0

f(xi, yi, zi)||c′(t∗i )||∆ti

=

∫I

f(x(t), y(t), z(t))||c′(t)||∆t =

∫c

f(x, y, z).

1

Remark 2. Now we discuss a related topic but with vector fields. Supposea particle moves along a path in a force field, say a gravitational force fieldin space. We want to measure the work done by the field on the particle.

The simplest case is when c is a straight line displacement given by the vectord and F is a constant force. Then the work done by F in moving the particlealong the path is the dot product

F · d = (magnitude of force) × (displacement in direction of force).

What if the path is curved? In this case, we divide the path into several smallintervals and approximate on each interval using the above formula. For asmall ∆t, the vector displacement is approximated using the MVT,

∆s = c(t+ ∆t)− c(t) ≈ c′(t)∆t.

(Note, in the previous section, vector displacement was replaced by arclength). Therefore, the work done in going from c(t) to c(t+ ∆t) is given by

F(c(t)) ·∆s ≈ F(c(t)) · c′(t)∆t.

Summing over the intervals of t gives us an approximation, taking a limitgives us the formua for a line integral.

Definition 1. Let F be a vector field on R3 that is continuous on the C1

path c : [a, b]→ R3. We define∫c F · ds, the line integral of F along c, by

the formula ∫c

F · ds =

∫ b

a

F(c(t)) · c′(t) dt.

Example 1. Let F = xi + yj + zk. Evaluate the integral of F along the pathc(t) = (t2, 3t, 2t3), −1 ≤ t ≤ 2.

Remark 3. Let c(t) be a path, a ≤ t ≤ b. Let T(t) = c′(t)/||c′(t)|| denotethe unit tangent vector of c. Then we have∫

F · ds =

∫ b

a

F(c(t)) · c′(t) dt =

∫ b

a

[F(c(t)) ·T(t)] ||c′(t)|| dt

It is also common to write line integrals in differential form,∫c

F · ds =

∫c

F1 dx + F2 dy + F3 dz =

∫ b

a

F1

(dx

dt+ F2

dy

dt+ F3

dz

dt

)dt

2

Example 2. Evaluate

∫c

y dx+(3y3−x) dy+z dz for the path c(t) = (t, t2, 0),

0 ≤ t ≤ 1.

Note that dx/ dt = 1, dy, dt = 2t and dz/dt = 0. Then∫c

y dx + (3y3 − x) dy + z dz

=

∫ 1

0

y(t)dx

dt+ (3y(t)3 − x(t))

dy

dt+ z(t)

dz

dtdt

=

∫ 1

0

(t2(1) + (2t6 − t)(t) + 0

)dt

Remark 4. In general,

∫c1

F · ds 6=∫c2

F · ds. This is true in the case that

c1 is a reparametrization of c2, that is, they describe the same curve.

Definition 2. let h : I → I1 be a C1 real-valued function that is a 1-1 mapof an interval I = [a, b] onto another interval I1 = [a1, b1]. Let c : I1 → R3 bea piecewise C1 path. Then we call the composition

p = c ◦ h : I → R3

a reparametrization of c.

Theorem 1. Let F be a vector field continuous on the C1 path c : [a1, b1]→R3, and let p : [a, b] → R3 be a reparametrization of c. If p is orientation-preserving (carries endpoints to endpoints) then∫

p

F · ds =

∫c

F · ds

and if p is orientation-reversing (reverses endpoints) then∫p

F · ds = −∫c

F · ds.

Theorem 2. Suppose f : R3 → R is of class C1 and that c : [a, b]→ R3 is apiecewise C1. Then ∫

c

∇f · ds = f(c(b))− f(c(a)).

This should remind you of the FTC.

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Definition 3. A simple curve is one that does not intersect itself, formally,we define a simple curve C to be the image of a piecewise C1 map c : I → R3

that is 1-1 on an interval I (parametrization of C). An oriented simplecurve is a curve along with a sense of direction. A closed curve is one suchthat the endpoints are the same. A simple closed curve c on [a, b] is 1-1on [a, b) with the property that c(a) = c(b).

Remark 5. If C be an oriented simple curve or a simple closed curve and cany orientation preserving parametrization of C. Then∫

C

F · ds =

∫c

F · ds.

Example 3. Evaluate the integral of F = (z3 + 2xy)i + x2j + 3xz2k aroundthe circumference of a square with vertices (±1,±1).

In this case, we divide c into four curves winding around the square.

c1(t) = (1, t− 1) t ∈ [0, 2]

c2(t) = (3− t, 1) t ∈ [2, 4]

c3(t) = (−1, 5− t) t ∈ [4, 6]

c4(t) = (t− 7,−1) t ∈ [6, 8].

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7.3 Parametrized surfaces

Remark 6. We now move on to discussing integrals over surfaces, as opposedto the paths in 7.1 and 7.2. We have previously defined surfaces as the graphsof functions of the form z = f(x, y). This is too restrictive. Just as manypaths cannot be written as y = f(x), surfaces such as the torus cannot bewritten as the graph of z = f(x, y).

Definition 4. A parametrization of a surface is a function Φ : D ⊂ R2 →R3. The surface S corresponding the the function Φ is its image: S = Φ(D).We can write

Φ(u, v) = (x(u, v), y(u, v), z(u, v)).

If Φ is differentiable or is of class C1, we call S a differentiable or C1 surface.

Example 4. Let P be a plane which is parallel to the vectors α and β andthat passes through the tip of another vector γ. Then the set of all points onP can be described as the set of vectors that are γ plus a linear combinationof α and β. That is, P is the set of vectors

Φ(u, v) = αu+ βv + γ.

Remark 7. Let Φ be a parametrization of a surface. If we fix u or v at u0 orv0, respectively, then the functions t 7→ (t, v0) and t 7→ (u0, t) describe curveson the surface. We can therefore look at the vectors tangent to these curvesat the point (u0, v0). We denote these

Tu =∂Φ

∂u=∂x

∂u(u0, v0)i +

∂y

∂u(u0, v0)j +

∂z

∂u(u0, v0)k,

Tv =∂Φ

∂v=∂x

∂v(u0, v0)i +

∂y

∂v(u0, v0)j +

∂z

∂v(u0, v0)k.

We say that the surface is regular (or smooth) at Φ(u0, v0) if Tu×Tv 6= 0,that is, the normal plane given by the cross product will give us a tangentplane to the surface at that point. The surface is regular if the above is truefor all choices of points.

Definition 5. Suppose the parametrized surface Φ : D ⊂ R2 → R3 is regularat Φ(u0, v0). We define the tangent plane of the surface at Φ(u0, v0) tobe the plane determined by the vectors Tu and Tv. Thus, n = Tu × Tv is

5

a normal vector, and an equation of the tangnet plane at (x0, y0, z0) on thesurface is given by

(x− x0, y − y0, z − z0) · n = 0,

where n is evaluated at (u0, v0). In other words, the tangent plane is theset of (x, y, z) satisfying the above equation. If n = (n1, n2, n3), the formulabecomes

n1(x− x0) + n2(y − y0) + n3(z − z0) = 0.

Example 5. Find the equation of the plane tangent to the surface x = u2,y = v2, z = u2 + v2 at the point u = 1, v = 1.

We compute the tangent vectors,

Tu =∂Φ

∂u= 2ui + 0j + 2uk = 2i + 2k,

Tv =∂Φ

∂v= 0i + 2vj + 2vk = 2j + 2k.

Next we calculate the normal vector,

n = Tu ×Tv =

∣∣∣∣∣∣i j k2 0 20 2 2

∣∣∣∣∣∣ = −4i− 4j + 4k.

(Note that, we could just as well take n = i + j− k.) Note that, at (u, v) =(1, 1) we have (x0, y0, z0) = (1, 1, 2). Thus, the equation of the plane is

−4(x− 1)− 4(y − 1) + 4(z − 2) = 0,

or equivalently,(x− 1) + (y − 1)− (z − 2) = 0,

or equivalently,x+ y = z.

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7.4 Area of a surface

Remark 8. Given a parametrized surface, we would like to compute its area.This will be the first step in defining the surface integral. We will not do thisfor an arbitrary surface, but only for the ‘nice’ ones. By this we mean Sis the image of Φ : D → R3 where D is piecewise regular, each Di is anelementary region in the plane, Φi is of class C1 and is 1-1, except possiblyon the boundary of D, and Si − Φi(D) is regular, except possibly at a finitenumber of points.

Definition 6. We define the surface area A(S) of a parametrized surfaceby

A(S) =

∫∫D

||Tu ×Tv|| du dv,

where ||Tu ×Tv|| is the norm of Tu ×Tv. If S is a union of surfaces Si, itsarea is the sum of the areas of the Si.

Remark 9. One checks that

||Tu ×Tv|| =

√[∂(x, y)

∂(u, v)

]2

+

[∂(y, z)

∂(u, v)

]2

+

[∂(x, z)

∂(u, v)

]2

where∂(x, y)

∂(u, v)=

∣∣∣∣∂x∂u ∂x∂v

∂y∂u

∂y∂v

∣∣∣∣ .Remark 10. One can justify this formula using Riemann integration. Sup-pose we take a small, regular rectangle in D with lower left-hand corner(ui, vj), height ∆v and width ∆u. The image in S = Φ(D) of this rectanglecan be approximated as a rectangle Pij with width ∆uTui and eight ∆vTvj .The area of this rectangle is then

A(Pij) = ||∆uTui ×∆vTvj || = ||Tui ×Tvj ||∆u∆v.

Summing over all such rectangles and taking a limit gives us the result.

7

Example 6. Find the area of the surface defined by Φ(u, v) 7→ (x, y, z),where x = h(u, v) = u + v, y = g(u, v) = u, and z = f(u, v) = v; 0 ≤ u ≤ 1,0 ≤ v ≤ 1.

We can do this in two ways. One is to compute each tangent vector and thentake the cross product, and then its norm, or to compute each Jacobian anduse the formula. We will do the latter.

∂(x, y)

∂(u, v)=

∣∣∣∣∂x∂u ∂x∂v

∂y∂u

∂y∂v

∣∣∣∣ =

∣∣∣∣1 11 0

∣∣∣∣ = −1

∂(y, z)

∂(u, v)=

∣∣∣∣∂y∂u ∂y∂v

∂z∂u

∂z∂v

∣∣∣∣ =

∣∣∣∣1 00 1

∣∣∣∣ = 1

∂(x, y)

∂(u, v)=

∣∣∣∣∂x∂u ∂x∂v

∂z∂u

∂z∂v

∣∣∣∣ =

∣∣∣∣1 10 1

∣∣∣∣ = 1.

Thus, the area is given as

A(S) =

∫∫D

||Tu ×Tv|| du dv

=

∫ 1

0

∫ 1

0

√[∂(x, y)

∂(u, v)

]2

+

[∂(y, z)

∂(u, v)

]2

+

[∂(x, z)

∂(u, v)

]2

du dv

=

∫ 1

0

∫ 1

0

√3 du dv =

√3.

8

Example 7. Find the area of the surface defined by z = xy and x2 + y2 ≤ 2.

We reparametrize to cylindrical coordinates. The conditions then becomez = (r cos θ)(r sin θ) = r2 cos θ sin θ and r2 ≤ 2. Our parametrization is then

Φ(r, θ) = (r cos θ, r sin θ, r2 cos θ sin θ).

We compute the tangent vectors

Tr = cos θi + sin θj + 2r cos θ sin θk

Tθ = −r sin θi + r cos θj + r2(cos2 θ − sin2 θ)k.

The cross product is then

Tr × Tθ =

∣∣∣∣∣∣i j k

cos θ sin θ 2r cos θ sin θ−r sin θ r cos θ r2(cos2 θ − sin2 θ)

∣∣∣∣∣∣ = −r2 sin θi− r2 cos θj + rk.

We compute the magnitude of the normal vector

||Tr × Tθ|| =√r4 sin2 θ + r4 cos2 θ + r2 = r

√r2 + 1.

Finally, we compute the integral,

A(S) =

∫∫D

||Tr ×Tθ|| dr dθ

=

∫ √2

0

∫ 2π

0

r√r2 + 1 dr dθ

= 2π

∫ √2

0

r√r2 + 1 dr

= 2π

[1

3(r2 + 1)3/2

]√2

0

=2π

3

(√27− 1

)

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7.5 Integrals of scalar functions over surfaces

Remark 11. Let Φ : D ⊂ R2 → R3 where D is an elementary region whichwe write as

Φ(u, v) = (x(u, v), y(u, v), z(u, v)).

Definition 7. If f(x, y, z) is a real-valued continuous function defined on aparametrized surface S, we define the integral of f over S to be∫∫

S

f(x, y, z) dS =

∫∫S

f dS =

∫∫D

f(Φ(u, v))||Tu ×Tv|| du dv.

Remark 12. We will not discuss at length the derivation of this formula,only a few brief remarks. We can take some small (rectangular) subset of thedomain D, call it Rij and let Sij = Φ(Rij). The area A(Sij) is given by theformula for area of a surface which we derived previously. Thus, the integralover Sij is approximated by f(Φ(ui, vj)A(Sij), where (ui, vj) ∈ Rij. Takingsums gives us an approximation of the entire integral, and taking limits ofthe sums gives us the actual value.

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Example 8. Evaluate the integral of f(x, y, z) = z + 6 over the surface Sgiven by Φ(u, v) = (u, v/3, v), u ∈ [0, 2], v ∈ [0, 3].

The real work in this case is to determine ||Tu ×Tv||. We will do this usingthe Jacobian method.

∂(x, y)

∂(u, v)=

∣∣∣∣∂x∂u ∂x∂v

∂y∂u

∂y∂v

∣∣∣∣ =

∣∣∣∣1 00 1/3

∣∣∣∣ = 1/3

∂(y, z)

∂(u, v)=

∣∣∣∣∂y∂u ∂y∂v

∂z∂u

∂z∂v

∣∣∣∣ =

∣∣∣∣0 1/30 1

∣∣∣∣ = 0

∂(x, y)

∂(u, v)=

∣∣∣∣∂x∂u ∂x∂v

∂z∂u

∂z∂v

∣∣∣∣ =

∣∣∣∣1 00 1

∣∣∣∣ = 1.

Then

||Tu ×Tv|| =

√[∂(x, y)

∂(u, v)

]2

+

[∂(y, z)

∂(u, v)

]2

+

[∂(x, z)

∂(u, v)

]2

=

√1

9+ 1 =

√10

3.

Now the integral is∫∫S

f dS =

∫∫D

f(Φ(u, v))||Tu ×Tv|| du dv

=

∫ 3

0

∫ 2

0

(v + 6)

√10

3du dv

=2√

10

3

∫ 3

0

v + 6 dv

=2√

10

3

[1

2v2 + 6v

]3

0

=2√

10

3

(45

2

)= 15

√10

11

Example 9. Evaluate∫∫

S z dS, where S is the upper hemisphere of radius

a, that is, the set of (x, y, z) with z =√a2 − x2 − y2.

We make a change to cylindrical coordinates. The integral then becomes,∫∫S

z dS =

∫∫D

zr dr dθ.

The surface D is then parametrized by (r, θ, g(r, θ)) where z = g(r, θ) =√a2 − r2. Now ∂g

∂r = −r(a2 − r2)−1/2 and ∂g∂θ = 0. Then√(

∂g

∂r

)2

+

(∂g

∂θ

)2

+ 1 =

√(−r(a2 − r2)−1/2

)2+ 1

=

√r2

a2 − r2+ 1 =

a√a2 − r2

Now∫∫D

zr dr dθ =

∫∫D

√a2 − r2 · a√

a2 − r2r dr dθ =

∫ a

0

∫ 2π

0

ar dθ dr

= 2aπ

∫ a

0

r dr = 2aπ

[r2

2

]a0

= a3π.

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7.6 Surface integrals of vector fields

Remark 13. Previously, in studying line integrals, we considered the physicalnotion of work. That is, the amount of force exerted times a given displace-ment. We will consider a generalization of that notion, known as flux. Thebook gives an example of a vector field determined by water flowing down ariver. One might want to know what volume of water flows across a stretch ofthe river over some given time. Other examples of flux come from consideringan electric or magnetic field. We are working toward a mathematical gener-alization of principles from these scientific notions. Specifically, the relationbetween flux of a vector field and current in a bounding loop.

Definition 8. Let F be a vector field defined on S, the image of Φ, a

parametrized surface. The surface integral of F over Φ, denoted

∫∫Φ

F· dS,

is defined by ∫∫Φ

F · dS =

∫∫D

F · (Tu ×Tv) du dv.

Remark 14. We will not derive this formula, at least not today. However,this is the formula one would expect when thinking of surface integrals as ageneralization of line integrals.

Example 10. Consider the vector field F = xyi+yzj+xzk. Find the surfaceintegral of F over the parametrized surface Φ(u, v) = (u − v, u + v, 1), with0 ≤ u ≤ 3 and 0 ≤ v ≤ 2.

In this case, Tu ×Tv = 2k. Then∫∫Φ

F · dS =

∫∫D

F · (Tu ×Tv) du dv

=

∫ 2

0

∫ 3

0

[(u2 − v2)i + (u+ v)j + (u− v)k

]· 2k du dv

=

∫ 2

0

∫ 3

0

2(u− v) du dv = 2

∫ 3

0

[u2

2− uv

]2

0

dv

= 2

∫ 3

0

2− 2v dv = 2[2v − v2

]30

= −6.

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Definition 9. An oriented surface S is a two-sided surface with one sidespecified as the outside (positive) side and the other the inside (nega-tive) side. At each point (x, y, z) ∈ S there are two unit normal vectors n1

and n2 with n1 = −n2. Each can be associated with one side of the surface.To specify a side of S, at each point we choose a unit normal vector thatpoints away from the positive side of S.

Remark 15. A Mobius strip is an example of a non-oriented surface.

Remark 16. Let Φ : D → R3 be a parametrization of the oriented surfaceS and suppose S is regular at Φ(u0, v0) so that (Tu0 × Tv0)/||Tu0 × Tv0|| isdefined. If n(Φ(u0, v0)) denotes the unit normal to S at Φ(u0, v0), then

(Tu0 ×Tv0)/||Tu0 ×Tv0|| = ±n(Φ(u0, v0)).

We say Φ is orientation-preserving in the positive case and in the negativecase we say it is orientation-reversing.

Any 1-1 parametrized surface for which Tu ×Tv never vanishes can be con-sidered as an oriented surface with a positive side determined by the directionof Tu ×Tv.

Example 11. The sphere is given in cartesian coordinates as x2+y2+z1 = 1.We select the unit normal vector n(x, y, z) = xi + yj + zk, which points tothe outside of the sphere.

On the other hand, we can parametrize S with spherical coordinates,

Φ(θ, φ) = (r cos θ sinφ, r sin θ sinφ, r cosφ)

Then,Tθ × Tφ = (− sin2 φ cos θ,− sin2 φ sin θ,− sin θ cosφ).

That is,Tθ × Tφ = − sinφ(sinφ cos θ, sinφ sin θ, cosφ).

Because sinφ ≥ 0 for all 0 ≤ φ ≤ π, then this vector is negative and sopoints inward on the sphere. Thus, switching Tφ and Tθ would parametrizethe surface such that it points outwards.

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Theorem 3. Let S be an oriented surface and let Φ1 and Φ2 be two regu-lar orientation-preserving parametrizations, with F a continuous vector fielddefined on S. Then ∫∫

Φ1

F · dS =

∫∫Φ2

F · dS.

If Φ1 is orientation-preserving and Φ2 is orientation-reversing, then∫∫Φ1

F · dS = −∫∫

Φ2

F · dS.

If f is a real-valued continuous function defined on S, and if Φ1 and Φ2 areparametrizations of S, then∫∫

Φ1

f · dS =

∫∫Φ2

f · dS.

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Theorem 4. The surface integral of F over S is equal to the integral of thenormal component of F over the surface. That is,∫∫

S

F · dS =

∫∫S

F · n dS.

Proof. See page 406.

Example 12. Evaluate∫∫

S F · dS where F(x, y, z) = 2xi− 2yj + z2k and Sis the cylinder x2 + y2 = 4 with z ∈ [0, 1].

We parametrize S in cylindrical coordinates, so Φ(θ, z) = (2 cos θ, 2 sin θ, z).The tangent vectors are Tθ = (−2 sin θ, 2 cos θ, 0) and Tz = (0, 0, 1). Thenthe cross product is Tθ × Tz = 2 cos θi + 2 sin θj. Now,∫∫

S

F · dS =

∫∫D

F · (Tθ × Tz) dS

=

∫ 2π

0

∫ 1

0

(4 cos θ)(2 cos θ)− (4 sin θ)(2 sin θ) dz dθ

= 8

∫ 2π

0

cos(2θ) dθ = 8

[1

2sin(2θ)

]2π

0

= 0.

Remark 17. Recall that if x, y, and z are vectors in R3, then x · (y × z) isthe volume of the parallelpiped determined by the three vectors. Thus, wecan interpret the surface integral as the volume of the object determined bythe surface and the vector field F. Formally, one can write this out usingRiemann sums.

Consider F as the velocity field of a fluid pointing in the direction in whichfluid is moving across the surface near each point. The expression

|F · (T∆u×Tv∆v)|

approximations the amount of fluid that passing through the tangent paral-lelogram per unit of time. Thus, the surface integral is the net quantity offluid to flow across the surface per unit of time. This is called the flux of Facross the surface.

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Example 13. Compute the heat flux across the unit sphere S if the temper-ature at point T (x, y, z) = x.

The temperature gradient is ∇T and heat flows with the vector field −k∇T =F where k is a positive constant. Thus

∫∫S F · dS is the total rate of heat

flow or flux across the surface S.

We parametrize the unit sphere using spherical coordinates

Φ(θ, φ) = (r cos θ sinφ, r sin θ sinφ, r cosφ)

and in this case r = 1. Moreover, 0 ≤ θ ≤ 2π and 0 ≤ φ ≤ π. Then

Tφ × Tθ = (sin2 φ cos θ, sin2 φ sin θ, sin θ cosφ).

Moreover, ∇T = i so F = −k∇T = −ki. It turns out that it won’t matterwhat k is. ∫∫

S

F · dS =

∫ π

0

∫ 2π

0

−k sin2 φ cos θ dθ dφ

= −k∫ π

0

sin2 φ [sin θ]2π0 dφ = 0

Flow in = Flow out.

Remark 18 (Gauss’ Law). Let E be an electric field and S a closed surface.Then the flux of the field over S is equal to the net charge Q enclosed by thesurface. That is ∫∫

S

E · dS = Q.

17

Graphs

Remark 19. Suppose a surface S can be given in the form z = g(x, y). Howdoes this fit into the theory of parametrization that we have developed overthe last two lectures? This graph can be parametrized by setting

x = u, y = u, z = g(u, v).

When g is of class C1 and the parametrization is smooth. This gives,

Tu = i +∂g

∂uk, Tv = j +

∂g

∂vk.

Then

Tu ×Tv = −∂g∂u

i− ∂g

∂vj + k = −∂g

∂xi− ∂g

∂yj + k.

Thus, the surface area formula reduces to

A(S) =

∫∫D

√(∂g

∂x

)2

+

(∂g

∂y

)2

+ 1 dA.

Now suppose we want to integrate a scalar function over the surface describedabove. Then the integral formula reduces to∫∫

D

f(x, y, z) dS =

∫∫D

f(x, y, g(x, y))

√(∂g

∂x

)2

+

(∂g

∂y

)2

+ 1 dx dy.

Remark 20. Now let θ be the angle of the normal with the unit vectork at the point (x, y, g(x, y)). The surface is then described by φ(x, y, z) =z − g(x, y) = 0 so N = ∇θ is normal to the surface and

N = −∂g∂x

i− ∂g

∂yj + k.

Then

cos θ =N · k√

(∂g/∂x)2 + (∂g/∂y)2 + 1.

Let n = N/||N|| so cos θ = n · k and

dS =dx dy

n · k.

18

Remark 21. As before, consider the surface S described by z = g(x, y),(x, y) ∈ D, where S is oriented with the upward-pointing unit normal

n =−∂g∂xi−

∂g∂y j + k√

(∂g/∂x)2 + (∂g/∂y)2 + 1.

Now

Tx = i +∂g

∂xk, Ty = j +

∂g

∂yk.

Tx ×Ty = −∂g∂x

i− ∂g

∂yj + k.

If F = F1i + F2j + F3k is a continuous vector field, then∫∫S

F · dS =

∫∫D

F · (Tx ×Ty) dx dy

=

∫∫D

[F1

(−∂g∂x

)+ F2

(−∂g∂y

)+ F3

]dx dy.

19

8.1 Green’s Theorem

Remark 22. The Fundamental Theorem of Calculus ties together differentialcalculus with integral calculus. We have already seen one generalization ofthis idea. We will expand on this idea and generalize it even further.

Theorem 5. Suppose f : R3 → R is of class C1 and that c : [a, b]→ R3 is apiecewise C1. Then ∫

c

∇f · ds = f(c(b))− f(c(a)).

Remark 23. A simple closed curve C that is the boundary of an elemen-tary region D has two orientations: counterclockwise (positive) and clockwise(negative). We denote the former by C+ and the latter by C−. A good wayto remember it: If you walk along C with the counterclockwise orientationthen the region D is on the left.

We can decompose C+ into multiple paths. The first case we will considerwill be when the region is y-simple.

Lemma 1. Let D be a y-simple region amd let C be its boundary. SupposeP : D → R is of class C1. Then∫

C+

P dx = −∫∫

D

∂P

∂ydx dy.

Proof. Suppose D is described as a ≤ x ≤ b and φ1(x) ≤ yφ2(x). Wedecompose C+ as

C+ = C+1 +B+

2 + C−2 B−1 .

By the FTC,∫∫D

∂P

∂y(x, y) dx dy =

∫ b

a

∫ φ2(x)

φ1(x)

∂P

∂y(x, y) dx dy

=

∫ b

a

[P (x, φ2(x))− P (x, φ1(x))] dx.

We parametrize C+1 as x 7→ (x, φ1(x)), a ≤ x ≤ b, and C+

2 as x 7→ (x, φ2(x)),

20

a ≤ x ≤ b. Then,∫ b

a

P (x, φ1(x)) dx =

∫C+

1

P (x, y) dx∫ b

a

P (x, φ2(x)) dx =

∫C+

2

P (x, y) dx = −∫C−

2

P (x, y) dx.

On the other hand, since x is constant on B+2 and B−1 we have∫

B+2

P dx = 0 =

∫B−

1

P dx.

Now ∫C+P dx =

∫C+

1

P dx +

∫B+

2

P dx +

∫C−

2

P dx +

∫B−

1

P dx

=

∫C+

1

P dx +

∫C−

2

P dx

= −∫ b

a

[P (x, φ2(x))− P (x, φ1(x))] dx

= −∫∫

D

∂P

∂ydx dy

Remark 24. The next lemma is similar and so we omit the proof.

Lemma 2. Let D be a x-simple region amd let C be its boundary. SupposeQ : D → R is of class C1. Then∫

C+

Q dy =

∫∫D

∂Q

∂xdx dy.

Remark 25. Our main theorem is now seen to be an easy consequence ofthe previous two lemmas.

Theorem 6 (Green’s Theorem). Let D be a simple region and let C be itsboundary. Suppose P : D → R and Q : D → R are of class C1. Then∫

C+

P dx +Q dy =

∫∫D

(∂Q

∂x− ∂P

∂y

)dx dy.

21

Example 14. Let D = [−1, 1] × [−1, 1], P (x, y) = −y and Q(x, y) = x.Verify Green’s Theorem.

We can parametrize each curve seperately,

C1 : c1 = (t− 1,−1), t ∈ [0, 2]

C2 : c2 = (1, t− 1), t ∈ [0, 2]

C3 : c3 = (1− t, 1), t ∈ [0, 2]

C4 : c4 = (−1, 1− t), t ∈ [0, 2].

Then ∫Ci

−y dx + x dy =

∫ 2

0

1 dt = 2.

Thus,∫C+ P dx +Q dy = 8.

First we compute,∫∫D

(∂Q

∂x− ∂P

∂y

)dx dy =

∫ 1

−1

∫ 1

−1

(1− (−1)) dx dy = 8

Remark 26. Green’s theorem applies to any region which can be piecewisedivided into simple regions with the proper orientation.

Example 15. Let D be the triangle in the xy plane with vertices at (−1, 1),(0, 0), and (3, 5), and let C = ∂D. Evaluate∫

C

(x2 − xy) dx + (y2 + 3xy) dy.

By Green’s Theorem,∫C

(x2 − xy) dx + (y2 + 3xy) dy

=

∫∫D

(∂

∂x(y2 + 3xy)− ∂

∂y(x2 − xy)

)dx dy

=

∫∫D

(3y + x) dx dy

=

∫ 0

−1

∫ x+2

−x3y + x dx dy +

∫ 3

0

∫ 53x

−x3y + x dx dy

22

Theorem 7. If C is a simple closed curve that bounds a region to whichGreen’s Theorem applies, then the area of the region D bounded by C = ∂Dis

A =1

2

∫∂D

x dy − y dx.

Proof. Let P (x, y) = −y and Q(x, y) = x. By Green’s Theorem,

1

2

∫∂D

x dy − y dx =1

2

∫∫D

[∂x

∂x− ∂(−y)

∂y

]dx dy

=1

2

∫∫D

[1 + 1] dx dy = A

Example 16. Find the area of the disc of radius R using Green’s Theorem.

We parametrize the boundary of the disk via c(θ) = (R cos θ, R sin θ), 0 ≤θ ≤ 2π. Note that this is simple and has the proper orientation (counter-clockwise). Thus, by Green’s Theorem,

A =1

2

∫∂D

x dy − y dx

=1

2

∫ 2π

0

(R cos θ)(R cos θ)− (R sin θ)(−R sin θ) dθ

=1

2

∫ 2π

0

R2 dθ = πR2.

23

Review of Divergence and Curl

Remark 27. Divergence measures the rate of expansion per unit of volume(or area in the case of graphs), given the flow of whatever. Divergence iscomputed in the following way. We define the del operator to be

∇ = i∂

∂x+ j

∂

∂y+ k

∂

∂z.

This can be restricted to two-space, and extended to n-space. This shouldbe recognizeable from work previous on the gradient vector. However, whenwe write ∇f , we mean the operator applied to f . It will be advantageousto think of the operator as its own entitity, without reference to a specificfunction.

Definition 10. If F = F1i + F2j + F3k, the divergence of F is the scalarfield

divF = ∇ · F =∂F1

∂x+∂F2

∂y+∂F3

∂z.

Example 17. Find the divergence of the vector field

V(x, y, z) = exyi− exyj + eyzk.

Remark 28. We will be discussing curl more in the coming section. Fornow, we focus on its computational aspect.

Definition 11. If F = F1i + F2j + F3k, the curl of F is the vector field

curlF = ∇× F =

∣∣∣∣∣∣i j k∂∂x

∂∂y

∂∂z

F1 F2 F3

∣∣∣∣∣∣ .Example 18. Compute the curl in the preceeding example.

24

8.1 (cont) Green’s Theorem

Remark 29. We are now ready to state the vector form of Green’s Theoremusing curl. Let F = P i +Qj. Then

(∇× F) · k =∂Q

∂x− ∂P

∂y.

We verify this now,

curlF = ∇× F =

∣∣∣∣∣∣i j k∂∂x

∂∂y

∂∂z

P Q 0

∣∣∣∣∣∣ .Since we will dot this with k, we only care about the k component of the

cross product, which is(∂Q∂x −

∂P∂y

)k. The result is now clear.

This leads to the following version of Green’s Theorem.

Theorem 8. Let D ⊂ R2 be a region to which Green’s Theorem applies. Let∂D be its boundary (oriented positively), and let F = P i+Qj be a C1 vectorfield on D. Then∫

∂D

F · ds =

∫∫D

(curlF) · k dA =

∫∫D

(∇× F) · k dA.

Theorem 9 (Divergence Theorem). Let D ⊂ R2 be a region to which Green’sTheorem applies and let ∂D be its boundary. Let n denote the outward unitnormal to ∂D. If

c : [a, b]→ R2

t 7→ (x(t), y(t)).

is a positively oriented parametrization of ∂D, n is given by

n =(y′(t),−x′(t))√[x′(t)]2 + [y′(t)]2

.

Let F = P i +Qj be a C1 vector field on D. Then∫∂D

F · n =

∫∫D

divF dA.

25

Example 19. Verify the divergence theorem with F = yi− xj and D is theunit disc.

We parametrize the unit disc with c(t) = (cos t, sin t), 0 ≤ t ≤ 2π. Then

n = cos ti + sin tj,

and so ∫∂D

F · n =

∫∂D

(sin t)(cos t)− (sin t)(cos t) = 0.

On the other hand,

divF =∂

∂x(y) +

∂

∂y(x) = 0.

Thus, ∫∫D

divF dA = 0.

26

8.2 Stoke’s Theorem

Remark 30. We now push Green’s Theorem one step further, relating lineintegrals to integrals over a surface.

We consider a surface S that is the graph of a function f(x, y). Then Smay be parametrized by Φ(u, v) = (u, v, f(u, v)), for (u, v) in some domainD ⊂ R2. We will assume D is a region whose boundary is a simple closedcurve to which Green’s Theorem applies. We must choose an orientation forthis curve. We will choose the one which validates Green’s Theorem and callthis orientation positive.

Suppose c : [a, b] → R2, c(t) = (x(t), y(t)) is a parametrization of ∂D. Wedefine the boundary curve of S to be the oriented simple closed curve that isthe image of the mapping p : t 7→ (x(t), y(t), f(x(t), y(t)).

The way to remember whether your orientation is correct. If you are walkingalong the boundary of the surface with the positive unit normal as yourupright direction, the surface should be on your left.

Theorem 10 (Stokes’ Theorem for Graphs). Let S be the oriented surfacedefined by a C2 function z = f(x, y), where (x, y) ∈ D, a region to whichGreen’s Theorem applies, and let F be a C1 vector field on S. Then if ∂Sdenotes the oriented boundary curfce of S as just defined, we have∫∫

S

curlF · dS =

∫∫S

(∇× F ) · dS =

∫∂S

F · dS.

Remark 31. Recall that if F = F1i + F2j + F3k and z = f(x, y), then∫∫S

F · dS =

∫∫D

[F1

(−∂z∂x

)+ F2

(−∂z∂y

)+ F3

]dx dy.

Thus, ∫∫S

curlF · dS

=

∫∫D

[(∂F3

∂y− ∂F2

∂z

)(−∂z∂x

)+

(∂F1

∂z− ∂F3

∂x

)(−∂z∂y

)+

(∂F2

∂x− ∂F1

∂y

)]dx dy.

27

Example 20. Verify Stoke’s Theorem for

S = {(x, y, z) : z = 1− x2 − y2, z ≥ 0},

(oriented as a graph) ∂S = {(x, y) : x2 +y2 = 1}, F = zi+xj+(2zx+2xy)k.

We have,curlF = (2x, 1− 2z − 2y, 1).

Hence,∫∫S

curlF · dS =

∫∫D

(2x)(2x) + (1− 2z − 2y)(2y) + 1 dx dy

=

∫∫D

4(x2 − y2)− 4zy + 2y + 1 dx dy

=

∫∫D

4(x2 − y2)− 4(1− x2 − y2)y + 2y + 1 dx dy

=

∫∫D

4(x2 − y2) + 4(x2 + y2)y − 2y + 1 dx dy

=

∫ 1

0

∫ 2π

0

4r2 cos(2θ) + 4r2 sin θ − 2 sin θ + 1(r dθ dr)

=

∫ 1

0

[2r2 sin(2θ)− 4r2 cos θ + 2 cos θ + θ

]2π0

(r dr)

=

∫ 1

0

2rπ dr =[πr2]1

0= π.

On the other hand, we may parametrize ∂S via

c(t) = (cos t, sin t, 0).

Then ∫∂S

F · ds =

∫ 2π

0

〈0, cos t, 2 cos t sin t〉 · 〈− sin t, cos t, 0〉 dt = π.

28

Example 21. Let C be the closed, piecewise smooth curve formed by trav-eling in a straight lines between the points (0, 0, 0), (2, 0, 4), (3, 2, 6), (1, 2, 2)and back to the origin, in that order. (So S is the surface lying in the interiorof the plane z = 2x.) Use Stokes’ theorem to evaluate the integral,∫

C

(z cosx) dx + (x2yz) dy + (yz)dz.

Let F be the given vector field, we have∫C

F · ds =

∫∫S

curlF · dS

=

∫∫D

[(z − x2y)(−2) + (2xyz − 0)

]dx dy

=

∫ 2

0

∫ 3

0

−4x+ 6x2y dx dy

=

∫ 2

0

[−2x2 + 2x3y

]30

dy

=

∫ 2

0

−18 + 54y dy

=[−18y + 27y2

]20

= 72.

Remark 32. We now generalize Stokes’ Theorem for parametrized surfaces.It may happen in this case (i.e. the sphere) where the boundary curve ∂S isnot well-defined. In this case, the line integral

∫∂S F · ds will be defined to

be zero. We generalize this more when we discuss conservative vector fields.

Theorem 11 (Stokes’ Theorem for parametrized surfaces). Let S be anoriented surface defined by a 1-1 parametrization Φ : D ⊂ R2 → S, whereD is a region to which Green’s Theorem applies. Let ∂S denote the orintedboundary of S and let F be a C1 vector field on S. Then∫∫

S

(∇× F ) · dS =

∫∂S

F · ds.

29

Example 22. Calculate∫∫

S(∇ × F ) · dS, where S is the hemisphere x2 +y2 + z2 = 1, x ≥ 0, and F = x3i− y3j.

Then ∂S is the circle y2 + z2 = 1 which we can parametrize by c(t) =(0, cos(t), sin(t)), 0 ≤ t ≤ 2π. Then∫

∂S

F · dS =

∫ 2π

0

(cos3 t,− sin3 t, 0) · (0,− sin t, cos t) dt

=

∫ 2π

0

− sin4 t dt

=

∫ 2π

0

−(

1− cos 2t

2

)2

dt

= −1

4

∫ 2π

0

(1− 2 cos 2t+ cos2(2t)) dt

= −1

4

∫ 2π

0

(1− 2 cos 2t+1 + cos 2t

2) dt

= −1

4

[t− sin 2t+

1

2

(t+

1

2sin 2t

)]2π

0

dt

= −3π

4.

30

Theorem 12 (MVT for surface integrals). If F is a continuous vector field,then ∫∫

S

F · n dS = [F(Q) · n(Q)]A(S)

for some point Q ∈ S, where A(S) is the area of S.

Remark 33. We are now ready to give a physical interpretation of curl interms of circulation.

Let V be the velocity field of a fluid. Let P be a point and a unit vector n.Let Sρ denote the disc of radius ρ and center P , which is perpendicular to n.By Stokes’ Theorem,∫∫

Sρ

curlV · dS =

∫∫Sρ

curlV · n dS =

∫∂Sρ

V · ds,

where ∂Sρ has the orientation induced by n. By the MVT for integrals, thereis a point Q in Sρ such that∫∫

Sρ

curlV · n dS = [curlV (Q) · n]A(Sρ),

where A(Sρ) = πρ2. Thus,

limρ→0

1

A(Sρ)

∫∂Sρ

V · ds = limρ→0

1

A(Sρ)

∫∫Sρ

curlV · dS

= limρ→0

curlV (Q) · n = curlV (P ) · n.

Thus,

curlV (P ) · n = limρ→0

1

A(Sρ)

∫∂Sρ

V · ds.

Suppose that V points in the dirction tangent to the oriented curve C. Ingeneral,

∫C V · ds represents the net amount of turning of the fluid in a

counterclockwise direction around C. We refer to∫C V · ds as the circulation

of V around C.

Remark 34 (Circulation and Curl). The dot product of curlV (P ) with unitvector n equals the circulation of V per unit area at P on a surface perpen-dicular to n.

31

8.4 Gauss’ Theorem

Remark 35. You should take some time to review Section 4.4 if you havenot already. We have discussed curl and divergence, but also you shouldunderstand the vector identities listed on page 255.

Also recall that flux is defined as the integral∫∫

S F · dS or the net quantityof fluid (or whatever) to flow across the surface S per unit of time (rate offluid flow).

Remark 36. Gauss’ Theorem equates the flux of a vector field out of a closedsurface to the integral of the divergence of that vector field over the volumeenclosed by the surface.

The type of regions we will consider are elementary regions in space. Inparticular, the boundary of an elementary region is a surface made up of afinite number of surfaces (2 ≤ n ≤ 6), each of which can be described asgraphs of functions from R2 to R. Examples include a cube and a sphere (oneither extreme). There are two normals, one pointing into the object and onepointing out.

Suppose S is such a closed surface and F a vector field on S. Then∫∫S

F · ds =∑i

∫∫Si

F · dS.

Given S the outward (resp. inward) orientation, the integral∫∫

S F · dsmeasures the amount of fluid leaving (resp. entering) the region bounded byS per unit of time.

Let n(x, y, z) be the unit normal vector at each point of S (with the properorientation), then we have∫∫

S

F · ds =

∫∫S

(F · n) dS.

32

Example 23 (Page 462, the cube). We can decompose the following integralwith signs coming from the six normal vectors on the faces of the cube W .Let F = F1i + F2j + F3k.∫∫

∂W

F · dS =

∫∫S

F · n dS

= −∫∫

S1

F3 dS +

∫∫S2

F3 dS−∫∫

S3

F1 dS

+

∫∫S4

F1 dS−∫∫

S5

F2 dS +

∫∫S6

F2 dS

Example 24. Let W = [0, 1] × [0, 1] × [0, 1]. Compute the flux out of thevector field F = zyi + xzj + xyk.

For example, we have ∫∫S1

F3 =

∫ 1

0

∫ 1

0

xy dx dy =1

4.

On the other hand, ∫∫S2

F3 =

∫ 1

0

∫ 1

0

xy dx dy =1

4.

Hence,

−∫∫

S1

F3 dS +

∫∫S2

F3 dS = 0.

The other casese are similar.

The vector field in this case is known as conservative. We will see more ofthis in the next section.

33

Theorem 13 (Gauss’ Divergence Theorem). Let W be a symmetric el-emetary region in space. Denote by ∂W the oriented closed surface thatbounds W . Let F be a smooth vector field defined on W . Then∫∫∫

W

(∇ · F) dV =

∫∫∂W

F · dS,

or equivalentely, ∫∫∫W

(divF) dV =

∫∫∂W

(F · n) dS,

Remark 37. Here is what this theorem says. If W is a region in R3, thenthe flux of a vector field F outward across the closed surface ∂W is equal tothe integral of divF over W .

Example 25. Compute the last example using the divergence theorem.

We need only note that divF = 0.

Example 26. Evaluate∫∫

F · dS, where F = 3xy2i + 3x2yj + z3k and S isthe surface of the unit sphere.

By the Divergence Theorem,∫∫

F · dS =∫∫∫

W (divF) dV. We compute thedivergence of F,

divF = 3y2 + 3x2 + 3z2.

Thus, ∫∫F · dS =

∫∫∫W

(3y2 + 3x2 + 3z2) dx dy dz,

where W is the region enclosed by the unit sphere. We make a change ofspherical coordinates, so this integral becomes,∫∫

F · dS =

∫ 2π

0

∫ π

0

∫ 1

0

3r2(r2 sinφ) dr dφ dθ

= 2π [− cosφ]π0

[3

5r5

]1

0

=12π

5.

34

Remark 38 (Divergence as the flux per unit of volume). For a point P , F(P )is the rate of net outward flux at P per unit volume. By Gauss’ theorem andthe MVT for integrals, if Wρ is a ball in R3 or radius ρ centered at P , thenthere is a point Q ∈ Wρ such that∫∫

∂Wρ

F · n dS =

∫∫∫Wρ

divF dV = divF(Q) · vol(Wρ).

Thus,

divF(Q) = limρ→0

divF(Q) = limρ→0

1

V (Wρ)

∫∫∂Wρ

F · n dS.

A C1 vector field F on R3 is said to be divergence-free if divF = 0. In thiscase,

∫∫S F · dS for all closed surfaces S. The converse also holds.

Remark 39. The following is a more general version of Gauss’s Theorem.

Theorem 14 (Gauss’ Law). Let M be a symmetric elementary region in R3.Then if (0, 0, 0) /∈ ∂M , we have∫∫

∂M

r · nr3

dS =

{4π (0, 0, 0) ∈M0 (0, 0, 0) /∈M

where r(x, y, z) = xi + yj + zk and r(x, y, z) = ||r(x, y, z)|| =√x2 + y2 + z2.

35

8.3 Conservative Fields

Remark 40. Recall that if F = ∇F , then∫c F · ds = F(c(b)) − F(c(a))

where a and b are the endpoints of the curve c. Hence, the line integraldepends only on the endpoints of the path and not on the path itself. Weclarify this idea in this section.

Our first result is actually a corollary of a subsequent result, but it seems amore natural place to start.

Corollary 1. F is a C1 vector field on R2 of the form P i +Qj that satisfies∂P/∂y = ∂Q/∂x, then F = ∇f for some f on R2. When this holds, we sayF is a conservative vector field.

Example 27. Show that the vector field F = 2xy2+1i−

2y(x2+1)(y2+1)2 j. is conservative.

Calculate∫C F· ds where C is parametrized by x = t3−1, y = t6−t, 0 ≤ t ≤ 1.

Theorem 15 (Conservative Fields). Let F be a C1 vector field defined onR3, except possibly for a finite number of points. The following conditionson F are all equivalent:

(i) For any oriented simple closed curve C,∫C F · ds = 0.

(ii) For any two oriented simple curves C1 and C2 that have the same end-points, ∫

C1

F · ds =

∫C2

F · ds.

(iii) F is the gradient of some function f ; that is, F = ∇f (and if F hasone or more exceptional points where it fails to be defined, f is alsoundefined there).

(iv) ∇× F = 0.

Remark 41. Any vector field satisfying one (and hence all) of the conditionsabove is called a conservative vector field.

Remark 42. As a consequence of the previous theorem is that a field F hasno circulation if and only if curlF = 0. Such a vector is called irrotational.Hence, a vector field is irrotational if and only if it is the gradient field forsome function, F = ∇f . The function f is called the potential for F.

36

Example 28. Show that the vector field F(x, y, z) = (ex cos y,−ex sin y, π)is conservative and find the potential f .

Theorem 16. If F is a C1 vector field on all of R3 with divF = 0, then thereexists a C1 vector field G with F = curlG.

Example 29. Let F = xzi− yzj + yk. Verify that ∇·F = 0. Find a G suchthat F = ∇×G.

37

Review

Remark 43. A simple closed curve C that is the boundary of an elemen-tary region D has two orientations: counterclockwise (positive) and clockwise(negative). We denote the former by C+ and the latter by C−. A good wayto remember it: If you walk along C with the counterclockwise orientationthen the region D is on the left.

Theorem 17 (Green’s Theorem). Let D be a simple region and let C be itsboundary. Suppose P : D → R and Q : D → R are of class C1. Then∫

C+

P dx +Q dy =

∫∫D

(∂Q

∂x− ∂P

∂y

)dx dy.

Example 30. Verify Green’s Theorem for the line integral,∫C

x3 dy − y3 dx,

when C is the unit circle.

The boundary C can be parametrized as c(t) = (cos t, sin t), t ∈ [0, 2π]. Thus,∫C

x3 dy − y3 dx =

∫ 2π

0

cos3(t)(cos t)− sin3(t)(− sin t) dt

=

∫ 2π

0

cos4(t) + sin4(t) dt

=

∫ 2π

0

(1 + cos(2t)

2

)2

+

(1− cos(2t)

2

)2

dt

=1

4

∫ 2π

0

(1 + 2 cos(2t) + cos2(2t)

)+(1− 2 cos(2t) + cos2(2t)

)dt

=1

4

∫ 2π

0

2 + 2 cos2(2t) dt =1

4

∫ 2π

0

2 + 2

(1 + cos(4t)

2

)dt

=1

4

∫ 2π

0

3 + cos(4t) dt =3π

2.

38

On the other hand, by Green’s Theorem,∫C

x3 dy − y3 dx =

∫C

−y3 dx + x3 dy

=

∫∫D

((3x2)− (−3y2)

)dx dy = 3

∫ 2π

0

∫ 1

0

r2(r dr dθ)

= 3

∫ 2π

0

∫ 1

0

r3 dr dθ =3π

2

Theorem 18 (Stokes’ Theorem for Graphs). Let S be the oriented surfacedefined by a C2 function z = f(x, y), where (x, y) ∈ D, a region to whichGreen’s Theorem applies, and let F be a C1 vector field on S. Then if ∂Sdenotes the oriented boundary curfce of S as just defined, we have∫∫

S

curlF · dS =

∫∫S

(∇× F ) · dS =

∫∂S

F · dS.

Theorem 19 (Stokes’ Theorem for parametrized surfaces). Let S be anoriented surface defined by a 1-1 parametrization Φ : D ⊂ R2 → S, whereD is a region to which Green’s Theorem applies. Let ∂S denote the orintedboundary of S and let F be a C1 vector field on S. Then∫∫

S

(∇× F ) · dS =

∫∂S

F · ds.

39

Example 31. Calculate the line integral∫C F · ds where F = 3zi−7xj+5yk

and C is the intersection z = x+ 4 with x2 + y2 = 4.

The curve C is parametrized by c(t) = (2 cos t, 2 sin t, 2 cos t+ 4), 0 ≤ t ≤ 2π.Then∫C

F · ds =

∫ 2π

0

F (c(t)) · c′(t) dt

=

∫ 2π

0

(6 cos t+ 12,−14 cos t, 10 sin t) · (−2 sin t, 2 cos t,−2 sin t) dt

=

∫ 2π

0

−12 sin t cos t− 24 sin t− 28 cos2 t− 10 sin2 t dt

Nevermind, let’s use Stokes theorem:∫C

F · ds =

∫∫S

(∇× F ) · ndA.

The line integral is equal to the surface integral of the graph z = x + 4 overthe surface of the cylinder. We have curlF = 5i + 3j− 7k and n = (−i + k).Then∫

C

F · ds =

∫∫S

(∇× F ) · ndA =

∫∫x2+y2

−5− 7 dx dy = −48π.

Theorem 20 (Gauss’ Divergence Theorem). Let W be a symmetric el-emetary region in space. Denote by ∂W the oriented closed surface thatbounds W . Let F be a smooth vector field defined on W . Then∫∫∫

W

(∇ · F) dV =

∫∫∂W

F · dS,

or equivalentely, ∫∫∫W

(divF) dV =

∫∫∂W

(F · n) dS,

Example 32. Evaluate∫∫

S F · dS where F = xi + yj + 3k and S is thesurface of the unit sphere.

We apply the divergence theorem. Note that divF = 2. Then∫∫∂W

F · dS =

∫∫∫W

(divF) dV =

∫∫∫W

2 dV = 2vol(W ) =8π

3.

40