Calculus III

Chapter 16 - Vector Calculus

1. Vector Fields

Previously we have studied vector valued functions. Recall that we defined functions f :

R → V2 (resp. V3) and these defined space curves in R2 (resp. R3). Now we consider

generalizations of this concept: vector fields. Vector fields can be used to model a wide

assortment of physical phenomenon. In particular, energy and heat flux, gravitational force

field, and forces acting on a charge at a certain position.

Let D be a set in R2. A vector field on R2 is a function F : D → V2. Similarly, if E is a

subset of R3, then a vector field on R3 is a function F : E → V3.

Suppose F is a vector field on R2 with domain D. Then for every (x, y) ∈ D,

F(x, y) = P (x, y)i +Q(x, y)j = 〈P (x, y), Q(x, y)〉.

We will sometimes simply write P = P (x, y) and Q = Q(x, y). We call P and Q the

component functions of F. When P and Q are scalars, F my be called a scalar field. As with

our discussion in Chapter 13, F is continuous if and only if P and Q are continuous. All of

this extends to vector fields in R3 where we write

F(x, y, z) = P (x, y, z)i +Q(x, y, z)j +R(x, y, z)k = 〈P (x, y), Q(x, y)〉.

Example 1. Let F : R2 → R2 be the vector field defined by F(x, y) = yi− 2xj. We sketch

some vectors for F below.

The vectors seem to be moving in an elliptic path.

Let x = 2xi + yj be a position vector, then

x · F(x, y) = 2xy − 2yx = 0.

Hence, F(x, y) is perpendicular to the position vec-

tor 〈2x, y〉 and so it is tangent to an ellipse with

center the origin and radius |x| =√

4x2 + y2.

Moreover,

|F(x, y)| =√y2 + (−2x)2 = |x|,

so the magnitude of F(x, y) is equal to the distance

from the origin to that point on the ellipse.1

One example of a vector field we have already seen is the gradient vector field ∇.

Example 2. Define f(x, y) = 12(x2 − y2). Then the gradient of f is ∇f = 〈x,−y〉.

The gradient vectors are perpendicular to the level

curves of f (as discussed in Chapter 14). The

length of the vectors actually depends on how close

together the level curves are. This follows because

the length of the gradient vector is the value of the

directional derivative of f and closely spaced level

curves indicate a steep graph.

Given a vector field F, one can ask whether there exists a scalar function f , called a potential

function for F, such that F = ∇f . If such a f exists, we say that F is a conservative vector

field. It is not clear yet why we would care about such vector fields.

Example 3. Newton’s Law of Gravitation states that the magnitude of the gravitational

force between two objects with masses m and M is

|F| = mMG

r2

where r is the distance between the objects and G is the gravitational constant. Assume the

object with mass M is located at the origin in R3. Let the position vector of the object with

mass m be x = 〈x, y, z〉. Then r = |x|, so r2 = |x|2. The gravitational force exerted on this

second object acts toward the origin, and the unit vector in this direction is

− x

|x|.

Therefore the gravitational force acting on the object at x = 〈x, y, z〉 is

F(x) = −mMG

|x|3x.

This is a vector field, called a gravitational field, and it associates a vector with every point

x in space.

2

2. Line integrals

In this section, we consider the problem of integrating over a curve C (as opposed to an

interval). Imagine having a long winding bar in space, then dropping a curtain from that.

The area of the curtain is the line integral.

Let C be a smooth curve in R2 with parametric equations r(t) = 〈x(t), y(t)〉 where a ≤ t ≤ b.

Recall that C smooth is equivalent to r′ continuous and r′(t) 6= 0. We divide the interval

[a, b] into n subintervals [ti−1, ti] of length ∆si (because s is typically used for arc length).

Choose a sample point Pi(x∗i , y∗i ) in each interval. Now let f be a function whose domain

includes C. Then the line integral of f along C is∫C

f(x, y) ds = limn→∞

n∑i=1

f(x∗i , y∗i )∆si.

Recall that the length of the curve C is

L =

∫ b

a

√(dx

dt

)2

+

(dy

dt

)2

.

Let s(t) be the length of C between r(a) and r(t), then

ds

dt=

√(dx

dt

)2

+

(dy

dt

)2

or ds =

√(dx

dt

)2

+

(dy

dt

)2

dt.

Thus, we can reparametrize the line integral to get∫C

f(x, y) ds =

∫ b

a

f(x(t), y(t))

√(dx

dt

)2

+

(dy

dt

)2

dt.

The formula for a line integral in three variables can be obtained similarly to get∫C

f(x, y, z) ds =

∫ b

a

f(x(t), y(t), z(t))

√(dx

dt

)2

+

(dy

dt

)2

+

(dz

dt

)2

dt.

In both cases, representing C by the vector function r(t), a ≤ t ≤ b, gives the generic formula∫ b

a

f(r(t))|r′(t)| dt.

Taking f(x, y, z) = 1 recovers the formula for arc length.

Example 4. Consider the curve C : x = t, y = cos 2t, z = sin 2t, 0 ≤ t ≤ 2π, and the line

integral∫C

(x2 + y2 + z2) ds. Now x(t)2 + y(t)2 + z(t)2 = t2 + 1 and furthermore,√(dx

dt

)2

+

(dy

dt

)2

+

(dz

dt

)2

=√

1 + (−2 sin 2t)2 + (2 cos 2t)2 =√

5.

3

Then ∫C

(x2 + y2 + z2) ds =

∫ 2π

0

(t2 + 1)√

5 dt =√

5[t3 + t

]2π0

=√

5(8π3 + 2π).

The definition of a line integral required that C be a smooth curve. However, this is not

strictly necessary. If C is the union of a finite number of smooth curves C1, . . . , Cn, then we

say that C is a piecewise-smooth curve and∫C

f(x, y) ds =

∫C1

f(x, y) ds + · · ·+∫Cn

f(x, y) ds.

The line integral we have seen so far is sometimes called the line integral with respect to arc

length. We can also consider the line integral with respect to the x-axis:∫C

f(x, y) dx = limn→∞

n∑i=1

f(x∗i , y∗i )∆xi

and the line integral with respect to the y-axis:∫C

f(x, y) dy = limn→∞

n∑i=1

f(x∗i , y∗i )∆yi.

In essence, we are only considering change with respect to x or y, respectively. If x = x(t)

and y = y(t) on the interval a ≤ t ≤ b, then dx = x′(t) dt and dy = y′(t) dt and so the above

formulas become,∫C

f(x, y) dx =

∫ b

a

f(x(t), y(t))x′(t) dt and

∫C

f(x, y) dy =

∫ b

a

f(x(t), y(t))y′(t) dt.

We will frequent abbreviate the combined line integral with respect to x and y as∫C

P (x, y) dx+

∫C

Q(x, y) dy =

∫C

P (x, y) dx+Q(x, y) dy.

Example 5. Let C be the arc of the circle x2 + y2 = 4 from (2, 0) to (0, 2), followed by the

line segment from (0, 2) to (4, 3). We will evaluate the line integral∫Cx2 dx+ y2 dy.

The curve C1 can be parameterized by x = 2 cos(t),

y = 2 sin(t), where 0 ≤ t ≤ π/2.

The curve C2 can be parameterized using our stan-

dard vector representation of a line segment:

r(t) = (1− t)r0 + tr1,

for 0 ≤ t ≤ 1. Here r0 = 〈0, 2〉 and r1 = 〈4, 3〉, so

r(t) = 〈x(t), y(t)〉 = (1−t)〈0, 2〉+t〈4, 3〉 = 〈4t, 2+t〉.4

Now we evaluate the two line integral separately. First we evaluate along C1,∫C1

x2 dx+ y2 dy =

∫ π/2

0

4 cos2 t(−2 sin t) dt+

∫ π/2

0

4 sin2 t(2 cos t) dt

= 8

[1

3cos3 t

]π/20

+ 8

[1

3sin3 t

]π/20

= −8

3+

8

3= 0.

Now we evaluate along C2,∫C2

x2 dx+ y2 dy =

∫ 1

0

(4t)2(4) dt+

∫ 1

0

(2 + t)2(1) dt

= 64

∫ 1

0

t2 dt+

∫ 1

0

(4 + 4t+ t2) dt

=64

3+

(4 + 2 +

1

3

)=

64 + 12 + 6 + 1

3=

83

3.

Hence, ∫C

x2 dx+ y2 dy =

∫C1

x2 dx+ y2 dy +

∫C2

x2 dx+ y2 dy = −2

3+

83

3= 27.

Suppose in the previous example we instead considered C1 as the circle from (0, 2) to (2, 0).

This reverses the orientation and we obtain∫C1x2 dx+ y2 dy = 2

3. In general, we have∫

−Cf(x, y) dx = −

∫C

f(x, y) dx and

∫−C

f(x, y) dy = −∫C

f(x, y) dy.

However, arc length does not depend on orientation and so∫−C

f(x, y) ds =

∫C

f(x, y) ds.

If f(x) is a variable force moving a particle from a to b along the x-axis, then the work done

by the force is W =∫ baf(x) dx. On the other hand, if F is a constant force moving an object

from point P to point Q in space, then the work done by the force is W = F · D, where

D =−→PQ.

Here we consider a continuous force field F = P i + Qj + Rk on R3. Suppose F moves a

particle along a smooth curve C, described by the parameter t, a ≤ t ≤ b. If we divide [a, b]

into regular subintervals, then this in turn divides C into subarcs Pi−1Pi with lengths ∆si.

Select a sample point Pi(x∗i , y∗i , z∗i ) on the ith subinterval corresponding to t∗i . Assuming ∆si

is small, then it moves from Pi−1 to Pi (approximately) in the direction of T(t∗i ). Thus, the

work done by F in moving the particle from Pi−1 to Pi is approximately,

F(x∗i , y∗i , z∗i ) · [∆siT(t∗i )] = [F(x∗i , y

∗i , z∗i ) ·T(t∗i )] ∆si.

5

It follows that the work W done by the force field F is

W = limn→∞

n∑i=1

[F(x∗i , y∗i , z∗i ) ·T(t∗i )] ∆si =

∫C

F(x, y, z) ·T(x, y, z) ds =

∫C

F ·T ds.

We call this last integral the line integral of F along C. If C is described by the vector

equation r(t), a ≤ t ≤ b, then T(t) = r′(t)/|r′(t)| and so

W =

∫ b

a

[F(r(t)) · r′(t)

|r′(t)|

]|r′(t)| dt =

∫ b

a

F(r(t)) · r′(t) dt.

Hence, we have ∫C

F · dr =

∫ b

a

F(r(t)) · r′(t) dt =

∫C

F ·T ds.

Example 6. Let F(x, y, z) = (x + y2)i + xzj + (y + z)k. and let r(t) = t2i + t3j − 2tk,

0 ≤ t ≤ 2. Then∫C

F · dr =

∫ 2

0

F(r(t)) · r′(t) dt

=

∫ 2

0

((t2 + t6)i + (−2t3)j + (t3 − 2t)

)· (2ti + 3t2j− 2k) dt

=

∫ 2

0

2t7 − 6t5 + 4t dt =

[1

4t8 − t6 + 2t2

]20

= 8.

6

3. The Fundamental Theorem for Line Integrals

Just as the integral acts as a sort of inverse operation to differentiation, so too does the line

integral act as a sort of inverse operation to taking gradients.

Theorem 7. Let C be a smooth curve given by the vector function r(t), a ≤ t ≤ b. Let f

be a differentiable function of two or three variables whose gradient vector ∇f is continuous

on C. Then ∫C

∇f · dr = f(r(b))− f(r(a)).

Proof. The proof of this theorem is striaghtforward after a change of variable. Recall from

the definition of a line integral along a curve,∫C

∇f · dr =

∫ b

a

∇f(r(t)) · r′(t) dt =

∫ b

a

(∂f

∂x

dx

dt+∂f

∂y

dy

dt+∂f

∂z

dz

dt

)=

∫ b

a

d

dtf(r(t)) (by the Chain Rule) = f(r(b))− f(r(a)).

The last step follows from the (usual) FTC. �

By the theorem, we now know that the line integral of ∇f tells us the net change of f .

Example 8. Find the work done by the gravitational field

F(x) = −mMG

|x|3x

in moving a particle with mass m from the point (1, 1, 1) to the point (2, 3, 0).

The vector field F is conservative since F = ∇f where

f =mMG√

x2 + y2 + z2.

Hence, by the FTC for line integrals,

W =

∫C

F · dr =

∫C

∇f · dr = f(1, 1, 1)− f(2, 3, 0)

=mMG√

12 + 12 + 12− mMG√

22 + 32 + 02= mMG

(1√3− 1√

13

).

Let F be a continuous vector field with domain D. We say that∫CF · dr is independent of

path (ion D) if∫C1∇F · dr =

∫C2

F · dr for any two paths C1 and C2 in D with the same

initial and terminal points. One consequence of the FTC is that line integrals of conservative7

vector fields are independent of path. That is, if C1 and C2 are two piecewise-smooth curves

from A to B, then ∫C1

∇f · dr =

∫C2

∇f · dr.

In general, there is no reason that we should expect this to hold.

A curve C is closed if its terminal point coincides with its initial point.

Theorem 9. Let F be a continuous vector field with domain D. Then∫CF · dr is indepen-

dent of path in D if and only if∫CF · dr = 0 for every closed path C in D.

Proof. Suppose∫CF · dr = 0 for every closed path C in D. Let C1 and C2 be paths in D

from point A to point B and let C be the curve C1 followed by −C2. Then

0 =

∫C

F · dr =

∫C1

F · dr +

∫−C2

F · dr =

∫C1

F · dr−∫C2

F · dr.

It follows that∫C1

F · dr =∫C2

F · dr. The other direction is similar. �

A consequence of this theorem is that if F is conservative, then∫CF · dr = 0 for every closed

path C. By the next theorem, conservative vector fields are the only vector fields that are

independent of path.

A region D ⊂ R2 is open if for every point P ∈ D, there isa. disk with center P that lies

entirely in D (i.e., D contains no boundary points). The region D is connected if any two

points in D can be joined by a path that lies in D.

Theorem 10. Suppose F is a vector field that is continuous on an open connected region

D. If∫CF · dr is independent of path in D, then F is a conservative vector field on D.

We’ve seen benefits and alternative characterizations of conservative vector field. The next

question is whether it is possible to determine whether a field is conservative.

Theorem 11. If F(x, y) = P (x, y)i +Q(x, y)j is a conservative vector field where P and Q

have continuous first-order partial derivatives on a domain D, then throughout D we have

∂P

∂y=∂Q

∂x.

Proof. Since F is conservative, there exists a function f such that ∇f = F. Then

P =∂f

∂xand Q =

∂Q

∂y.

8

Now by Clairaut’s Theorem,

∂P

∂y=

∂2f

∂y∂x=

∂2f

∂x∂y=∂Q

∂x. �

This theorem doesn’t tell us when a vector field is conservative, but it does tell us when it

is not conservative.

Example 12. The vector field F(x, y) = (xy)i+(x2)j is not conservative since ∂P∂y

= x while∂Q∂x

= 2x. (Unless D is restricted to {0}, which isn’t a particularly interesting region.)

The next theorem is a partial converse to the previous one. First, we need some additional

terminology. A curve C with equation r(t) on a ≤ t ≤ b is simple if r(t1) 6= r(t2) whenever

a < t1 < t2 < b (no intersections other than the endpoints). A region D is simply connected

if every simple closed curve in D encloses points only in D.

Theorem 13. Let F = P i + Qj be a vector field on an open simply-connected region D.

Suppose that P and Q have continuous first-order partial derivatives and

∂P

∂y=∂Q

∂x.

Then F is conservative.

Example 14. Suppose that F = (3 + 2xy2)i+ (2x2y)j. The domain of F is all of R2, which

is simply connected. Now P = 3 + 2xy2 and Q = 2x2y, and

∂P

∂y= 4xy =

∂Q

∂x.

Thus, F is conservative.

Here we explain the procedure for finding f such that ∇f = F. We know such an f exists,

then fx(x, y) = 3 + 2xy2 and fy(x, y) = 2x2y. We integrate fx(x, y) with respect to x to get

f(x, y) =

∫fx(x, y) dx =

∫3 + 2xy2 dx = 3x+ x2y2 + g(y).

Here, the constant of integration is a function only in terms of y. Differentiating f(x, y) with

respect to y gives

2x2y = fy(x, y) = 2x2y + g′(y).

It follows that g(y) = K for some constant. Thus, f(x, y) = 3x+ x2y2 +K.

Now suppose that C is the arc of the hyperbola y = 1/x from (1, 1) to (4, 14). Using the

above, we have that∫C

F · dr =

∫C

∇f · dr = f(4, 1/4)− f(1, 1) = (12 + 1)− (3 + 1) = 9.

9

4. Green’s Theorem

We now come to the second “big” theorem of this chapter. Green’s Theorem connects certain

line integrals to double integrals.

Let D be a region bounded by a simple closed curve. We choose counterclockwise to be the

positive orientation of C (this is our convention but it is important that we stick to it).

Green’s Theorem. Let C be a positively oriented, piecewise-smooth, simple closed curve

in the plane and let D be the region bounded by C. If P and Q have continuous partial

derivatives on an open region that contains D, then∫C

P dx+Q dy =

∫∫D

(∂Q

∂x− ∂P

∂y

)dA.

We will sometimes replace∫C

with∮C

to emphasize that C is closed and positively oriented.

In other cases, we may write∫δD

in place if∫C

to emphasize that we are working over the

boundary of D (again, positively oriented).

We say that D is a simple region if it is both type I and type II.

Proof of Green’s Theorem in the simple case. Assume D is a simple region. Then it is type

I and so there are continuous functions g1(x) and g2(x) such that

D = {(x, y) : a ≤ x ≤ b, g1(x) ≤ y ≤ g2(x)}.

Thus, by the FTC,∫∫D

∂P

∂ydA =

∫ b

a

∫ g2(x)

g1(x)

∂P

∂y(x, y) dy dx =

∫ b

a

P (x, g2(x))− P (x, g1(x)) dx.

Now we break δD into the union of four curves as shown below,

The curve C1 is the graph of y = g1(x). Similarly, −C3 is

the graph of y = g2(x) (because of our orientation). Hence,∫C1

P (x, y) dx =

∫ b

a

P (x, g1(x)) dx∫C3

P (x, y) dx = −∫−C3

P (x, y) dx = −∫ b

a

P (x, g2(x)) dx.

10

On the other hand, at C2 and C4, x is constant and so∫C2

P (x, y) dx = 0 =

∫C4

P (x, y) dx.

Putting this all together we get∫C

P (x, y) dx =

∫C1

P (x, y) dx+

∫C2

P (x, y) dx+

∫C33

P (x, y) dx+

∫C4

P (x, y) dx

=

∫ b

a

P (x, g1(x)) dx−∫ b

a

P (x, g2(x)) dx

=

∫ b

a

P (x, g2(x))− P (x, g1(x)) dx =

∫∫D

∂P

∂ydA.

A similar computation (but expressing D as a type II region) shows that∫C

Q dy =

∫∫D

∂Q

∂xdA.

Combining these two gives Green’s Theorem for D a simple region. �

Example 15. Let C be the ellipse x2 + 2y2 = 2. We will evaluate∫Cy4 dx+ 2xy3 dy. First

observe that, by Green’s Theorem,∫C

y4 dx+ 2xy3 dy =

∫∫D

(2y3)− (4y3) dA =

∫∫D

−2y3 dA.

To evaluate this integral, we make a change of coordinates x =√

2u and y = v. Then our

new region, D′, has boundary 2 = (√

2u)2 + 2(v)2 = 2(u2 + v2), or equivalently u2 + v2 = 1.

The Jacobian of this transformation is√

2 and so we subsequently make a change to polar

coordinates to get,∫C

y4 dx+ 2xy3 dy =

∫∫D

−2y3 dA =

∫∫D′−2v3

√2 du dv

=

∫ 1

0

∫ 2π

0

−2√

2(r cos θ)3r dθ dr

= −2√

2

∫ 1

0

∫ 2π

0

r4 cos θ(1− sin2 θ) dθ dr

= −2√

2

∫ 1

0

r4[sin θ − 1

3sin3 θ

]2π0

dr = 0.

Example 16. Let C be the triangle from (0, 0) to (1, 1) to (0, 1) to (0, 0) (remember: ori-

entation matters!). Let F = 〈√x2 + 1, arctanx〉. Note that F is not conservative. However,

we can evaluate∫CF · dr using Green’s Theorem.

11

The regionD enclosed by C is type I and can be written, D = {(x, y) : 0 ≤ x ≤ 1, x ≤ y ≤ 1}.Hence,∫

C

F · dr =

∫C

√x2 + 1 dx+ arctanx dy =

∫∫D

(∂

∂x(arctanx)− ∂

∂y(√x2 + 1)

)dA

=

∫ 1

0

∫ 1

x

(1

x2 + 1− 0

)dy dx =

∫ 1

0

[y

x2 + 1

]1x

dx =

∫ 1

0

1

x2 + 1− x

x2 + 1dx

=

[arctanx− 1

2ln |x2 + 1|

]10

=π

4− 1

2ln(2).

Green’s Theorem can easily be extended to the case where D is a finite union of simple

regions. Suppose D = D1 ∪D2 where both D1 and D2 are simple (as below).

The boundary of D1 is C1 ∪ C3 while the boundary of D2 is C2 ∪ (−C3). Hence, applying

Green’s Theorem to both regions independently gives,∫C1∪C3

P dx+Q dy =

∫∫D1

(∂Q

∂x− ∂P

∂y

)dA∫

C2∪(−C3)

P dx+Q dy =

∫∫D2

(∂Q

∂x− ∂P

∂y

)dA

Now,∫∫D

(∂Q

∂x− ∂P

∂y

)dA =

∫∫D1

(∂Q

∂x− ∂P

∂y

)dA+

∫∫D2

(∂Q

∂x− ∂P

∂y

)dA

=

∫C1∪C3

P dx+Q dy +

∫C2∪(−C3)

P dx+Q dy

=

∫C1

P dx+Q dy +

∫C2

P dx+Q dy =

∫C

P dx+Q dy.

12

Green’s Theorem can also be extended to regions with finitely many holds. Suppose D has

one hole. Let C1 be the outer boundary of D and C2 the boundary for the hole as below.

We divide into D′ ∪D′′. The curves on the boundary between D′ and D′′ cancel (as in the

above argument) and so∫∫D

(∂Q

∂x− ∂P

∂y

)dA =

∫∫D′

(∂Q

∂x− ∂P

∂y

)dA+

∫∫D′′

(∂Q

∂x− ∂P

∂y

)dA

=

∫C

P dx+Q dy.

13

5. Curl and Divergence

Suppose we have fluid moving through a pipe occupying a region E in three-dimensional

space. We let V(x, y, z) denote the velocity of the fluid at a point (x, y, z) ∈ E. Then

V(x, y, z) is a vector field. At the same time, the fluid at point (x, y, z) may rotate around

some axis, denoted curlV(x, y, z) and the length of the curl denotes how quickly the fluid

rotates around the axis. When there is no rotation, then curlV = 0.

We denote by ∇ (pronounced “del”) the operator

∇ = i∂

∂x+ j

∂

∂y+ k

∂

∂z.

(Yes yes yes, this is the same symbol for gradient.) Given a scalar function f , we have

∇f =∂f

∂xi +

∂f

∂yj +

∂f

∂zk.

Let = Pi + Qj + Rk be a vector field on R3. Suppose the partial derivatives of P , Q, and

R all exist. The curl of F is defined as

curlF = ∇× F =

∣∣∣∣∣∣∣i j k∂∂x

∂∂y

∂∂z

P Q R

∣∣∣∣∣∣∣ .Example 17. Let F = x3yz2j + y4z3k. Then

curlF = ∇× F =

∣∣∣∣∣∣∣i j k∂∂x

∂∂y

∂∂z

0 x3yz2 y4z3

∣∣∣∣∣∣∣=

∣∣∣∣∣ ∂∂y

∂∂z

x3yz2 y4z3

∣∣∣∣∣ i−∣∣∣∣∣ ∂∂x ∂

∂z

0 y4z3

∣∣∣∣∣ j +

∣∣∣∣∣ ∂∂x ∂∂y

0 x3yz2

∣∣∣∣∣k= (4y3z3 − 2x3yz)i + 0j + 3x2yz2k.

Theorem 18. If f is a function of three variables that has continuous second-order partial

derivatives, then curl(∇f) = 0. Hence, if F is conservative, then curlF = 0.

Proof. We have

curl(∇f) = ∇×∇f =

∣∣∣∣∣∣∣i j k∂∂x

∂∂y

∂∂z

∂f∂x

∂f∂y

∂f∂z

∣∣∣∣∣∣∣ =

∣∣∣∣∣ ∂∂y ∂∂z

∂f∂y

∂f∂z

∣∣∣∣∣ i−∣∣∣∣∣ ∂∂x ∂

∂z∂f∂x

∂f∂z

∣∣∣∣∣ j +

∣∣∣∣∣ ∂∂x ∂∂y

∂f∂x

∂f∂y

∣∣∣∣∣k = 0

by Clairaut’s Theorem. �14

One can use the above theorem to show that F is not conservative, but it does not guarantee

that F is conservative. The next theorem is a partial converse when F is defined on all of

R3.

Theorem 19. If F is a vector field defined on all of R3 whose component functions have

continuous partial derivatives and curlF = 0, then F is a conservative vector field.

Back to our velocity field example, V. One can measure from a point (x, y, z) the net rate

of change (with respect to time) of the mass of fluid flowing from the point (x, y, z) per unit

of volume. This is known as the divergence and is denoted divV(x, y, z).

If F = P i + Qj + Rk is a vector field on R3 and ∂P/∂x, ∂Q/∂y, ∂R/∂z all exist, then the

divergence of F is defined as

divF = ∇ · F =∂P

∂x+∂Q

∂y+∂R

∂z.

Example 20. Let F = x3yz2j + y4z3k as in Example 17. Then

divF = ∇ · F = 0 + (x3 + y2) + (3y4z2).

The next theorem is a straightforward computation. It follows for essentially the same reason

that a · (a× b) = 0.

Theorem 21. If F = P i+Qj+Rk is a vector field on R3 and P , Q, and R have continuous

second-order partial derivatives, then div curl = 0.

Suppose F = P i +Qj. Then

curlF =

∣∣∣∣∣∣∣i j k∂∂x

∂∂y

∂∂z

P Q R

∣∣∣∣∣∣∣ =

(∂Q

∂x− ∂P

∂y

)k.

It follows that we can write Green’s Theorem in vector form as∮C

F · dr =

∫∫D

(curlF) · k dA.

One can also show (see Stewart) that if n is the outward unit normal vector to C, then∮C

F · n ds =

∫∫D

divF(x, y) dA.

15

6. Parametric surfaces and their areas

Much of this section will be done via analogy with parametric curves. Suppose that

r(u, v) = x(u, v)i + y(u, v)j + z(u, v)k

is a vector-valued function defined on a region D in the uv-plane. The parametric surface

S is the set of all points (x, y, z) such that x = x(u, v), y = y(u, v), and z = z(u, v) for all

(u, v) ∈ D. The three equations defining S are called the parametric equations of S.

In some sense, we’ve seen this already with spherical coordinates.

Example 22. The sphere x2 + y2 + z2 = a2 is represented by the parametric functions

x = a sinφ cos θ y = a sinφ sin θ z = a cosφ,

with 0 ≤ φ ≤ π and 0 ≤ θ ≤ 2π.

Example 23. Suppose r(u, v) = 〈u2, u cos v, u sin v〉. Since y2 + z2 = u2 (a cylinder), then

this surface resembles a helix winding around the x-axis. This is known as a helicoid.

Now suppose that r(u, v) = x(u, v)i + y(u, v)j + z(u, v)k defines a smooth surface S. Let P0

be a point on S with position vector r(u0, v0). If we keep u constant, putting u = u0, then

r(u0, v) becomes a vector function of v and defines a curve C1 (called a grid curve). The

tangent vector to C1 at P0 is then

rv =∂x

∂v(u0, v0)i +

∂y

∂v(u0, v0)j +

∂z

∂v(u0, v0)k.

Similarly,

ru =∂x

∂u(u0, v0)i +

∂y

∂u(u0, v0)j +

∂z

∂u(u0, v0)k.

Since S is smooth, ru× rv 6= 0 and gives the normal vector for the tangent plane to S at P0.

Let R be a small rectangle of D (the domain of r(u, v)) with lower left corner (u∗i , v∗j ). Let

r∗u and r∗v be the tangent vectors at that point. Let ∆u and ∆v denote the dimension of R.

We can use the tangent plane to approximate the area of the image of R on S as

|(∆ur∗u)× (∆vr∗v)| = |r∗u × r∗v|∆u∆v.

It follows that

A(S) = limm,n→∞

m∑i=1

n∑j=1

|r∗u × r∗v|∆u∆v =

∫∫D

|ru × rv| dA.

16

Example 24. Consider the helicoid r(u, v) = u cos vi + u sin vj + vk, 0 ≤ u ≤ 1, 0 ≤ v ≤ π.

We have ru = cos vi + sin vj and rv = −u sin vi + u cos vj + k. Then

ru × rv =

∣∣∣∣∣∣∣i j k

cos v sin v 0

−u sin v u cos v 1

∣∣∣∣∣∣∣ = sin vi− cos vj + uk.

Hence,

A(S) =

∫∫D

|ru × rv| dA =

∫ 1

0

∫ π

0

√1 + u2 dv du

= π

∫ 1

0

√1 + u2 du let u = tan θ so du = sec2 θ dθ

= π

∫ π/4

0

sec3 θ dθ = π

[1

2(sec θ tan θ + ln | sec θ + tan θ|)

]π/40

=π

2

(√2 + ln(

√2 + 1)

).

If a surface is represented by the equation z = f(x, y), then we can write this in parametric

form as

x = x y = y z = f(x, y).

We have rx = i + ∂f∂xk and ry = j + ∂f

∂yk. From this we recover our definition of surface area

of a surface

A(S) =

∫∫D

√1 +

(∂z

∂x

)2

+

(∂z

∂y

)2

dA.

17

7. Surface integrals

Previously we studied line integrals in which (initially) we took the integral with respect to

arc length. We extend this to study surface integrals with respect to surface area.

Suppose that a surface S has vector equation

r(u, v) = x(u, v)i + y(u, v)j + z(u, v)k (u, v) ∈ D.

Assume thatD is a rectangle (the general case is similar). We divideD into regular rectangles

Rij of area ∆A = ∆u∆v. This divides S into patches r(Rij) = Sij. Given a function f ,

defined on S, we choose a point P ∗ij in each patch. The surface integral of f over the surface

S is defined as ∫∫S

f(x, y, z) dS = limm,n→∞

f(P ∗ij)∆Sij,

where ∆Sij is the surface area of the patch Sij. To actually evaluate this, we need to

reparameterize in terms of u and v (just as we did with t for line integrals). We approximate

each ∆Sij by the tangent plane, so

∆Sij ≈ |ru × rv|∆u∆v.

Hence, ∫∫S

f(x, y, z) dS =

∫∫D

f(r(u, v))|ru × rv| dA.

Note that, ∫∫1 dA =

∫∫D

|ru × rv| dA = A(D).

Example 25. Let S be the cone with parametric equations

x = u cos v, y = u sin v, z = u, 0 ≤ u ≤ 1, 0 ≤ v ≤ π/2.

We will evaluate the surface integral∫∫

Sxyz dS. First, we compute ru × rv,

ru × rv =

∣∣∣∣∣∣∣i j k

cos v sin v 1

−u sin v u cos v 0

∣∣∣∣∣∣∣ = −u cos vi− u sin vj + (u cos2 v + u sin2 v)k

= −u cos vi− u sin vj + uk.

Hence, |ru × rv| =√

(−u cos v)2 + (−u sin v)2 + u2 = u√

2. Now we have∫∫S

xyz dS =

∫∫D

(u cos v)(u sin v)(u)(u√

2) dA =√

2

∫ π/2

0

∫ 1

0

u4 cos v sin v du dv

=

√2

5

∫ π/2

0

cos v sin v dv =

√2

5

[1

2sin2 v

]π/20

=1

5√

2.

18

If S is a piecewise-smooth surface, so S = S1 ∪ · · ·Sn and intersections are only along the

boudary, then we can decompose the surface integral as∫∫S

f(x, y, z) dS =

∫∫S1

f(x, y, z) dS + · · ·+∫∫

Sn

f(x, y, z) dS.

In our discussion of curves, it was useful to define direction and this played a role in certain

line integrals. We want a similar notion for curves, as it will be useful in defining surface

integrals over vector fields. A classic example of an oriented surface is a sphere. We can

define a continuous function f(x, y, z)→ V3 that at each point gives the unit normal vector

to the sphere at the point (x, y, z). A surface is oriented if such a function exists, but not

all surfaces are orientable in this way. Consider the Mobius strip, which really only has one

side.

Example 26. Let z = g(x, y) be the graph of a surface S. Then S can be parameterized by

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

Hence, rx = i +(∂g∂x

)k and ry = j +

(∂g∂y

)k, so

rx × ry = −(∂g

∂x

)i−(∂g

∂y

)j + k

and so the surface integral formula becomes,∫∫S

f(x, y, z) dS =

∫∫D

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

√(∂z

∂x

)2

+

(∂z

∂y

)2

+ 1 dA.

Now we also have that a unit normal is

n =rx × ry|rx × ry|

=−(∂g∂x

)i−(∂g∂y

)j + k√(

∂g∂x

)2+(∂g∂y

)2+ 1

.

In some sense, this is the natural orientation because the k-component is positive. Thus,

this gives the upward orientation of the surface.

In general, if S is a smooth orientable surface given in parametric form by a vector function

r(u, v), we associate to it an orientation by setting

n =rx × ry|rx × ry|

.

An opposite orientation is given by −n.19

Example 27. Consider the parameterization of the sphere by spherical coordinates,

r(φ, θ) = a sinφ cos θi + a sinφ sin θj + a cosφk.

A computation shows that |rφ × rθ| = a2 sinφ (this is just the Jacobian). Hence,

n =rφ × rθ|rφ × rθ|

=1

ar(φ, θ).

This tells us that the normal vector points in the same direction as the position vector (so

outward from the sphere).

For a closed surface E (boundary is a solid region) the positive orientation is the one in which

the normal vectors point outward from E. (Like many things, this is just convention but it

is pretty standard).

Now we’re in a place to define surface integrals over vector fields, much as we did over curves.

Let S be an oriented surface with unit normal vector n and suppose fluid flows through S

with density ρ(x, y, z) and velocity v(x, y, z). The rate of (fluid) flow (mass per unit time)

per unit area if ρv. Hence, dividing S into small patches Sij we can approximate the mass

of fluid per unit time crossing Sij in the direction of n by

(ρv · n)A(Sij).

Hence, the rate of flow through S is given by∫∫S

ρv · n dS.

Hence, if F is a continuous vector field defined on an oriented surface S with unit normal

vector n, then the surface integral of F over S (or the flux of F across S) is∫∫S

F · dS =

∫∫S

F · n dS.

In the above example, F = ρv.

If S is given by r(u, v), then we can obtain n as before to get∫∫S

F · dS =

∫∫S

F · ru × rv|ru × rv|

dS

=

∫∫D

[F(r(u, v)) · ru × rv

|ru × rv|

]|ru × rv| dA

=

∫∫D

F · (ru × rv) dA.

20

Example 28. Let S be the hemisphere x2 + y2 + z2 = 4, z ≥ 0, oriented downward, and let

F = yi− xj + 2zk. We will compute∫∫

SF · dS.

We parameterize S by usng spherical coordinates

r(φ, θ) = 2 sinφ cos θi + 2 sinφ sin θj + 2 cosφk.

Hence,

F(r(φ, θ)) = 2 sinφ sin θi− 2 sinφ cos θ + 4 cosφk.

We know from before that

rφ × rθ = 4 sin2 φ cos θi + 4 sin2 φ sin θj + 4 sinφ cosφk.

Now

F(r(φ, θ)) · (rφ × rθ) = 8(sin3 φ sin θ cos θ − sin3 φ sin θ cos θ + 2 sinφ cos2 φ) = 16 sinφ cos2 φ.

The domain D is just the projection of the hemisphere into the xy-plane, so it’s a disk of

radius 2 centered at the origin. But we need to be careful. Because our sphere is oriented

downward and this is the negative orientation of the sphere, we pick up an extra negative

sign. Hence,∫∫S

F · dS = −∫∫

D

F · (ru × rv) dA = −∫ π/2

0

∫ 2π

0

16 sinφ cos2 φ dθ dr

= −32π

∫ π/2

0

sinφ cos2 φ dr = −32π

[−1

3cos3 φ

]π/20

= −32π

3.

If z = g(x, y) is the graph of a surface, then we can parameterize in the usual way and get

that

F · (rx × ry) = (P i +Qj + rk) ·(−(∂g

∂x

)i−(∂g

∂y

)j + k

)so that ∫∫

S

F · dS =

∫∫D

(−P

(∂g

∂x

)−Q

(∂g

∂y

)+R

)dA.

As an exercise, repeat the previous example by taking z =√

4− x2 − y2

21

8. Stokes’ Theorem

Stokes’ Theorem. Let S be an oriented piecewise-smooth surface that is bounded by a

simple, closed, piecewise-smooth boundary curve C with positive orientation. Let F be a

vector field whose components hav continuous partial derivatives on an open region in R3

that contains S. Then ∫C

F · dr =

∫∫S

curlF · dS.

Stokes’ Theorem connects line integrals to surface integrals, essentially what Green’s Theo-

rem does with (ordinary) double integrals. If S lies in the xy-plane with upward orientation,

then the unit normal is just k. Hence,∫CF · dr =

∫∫S

curlF · dS =∫∫

S(curlF) · k dA,

which is just Green’s Theorem. Hence, Green’s Theorem is a special case of Stokes’ Theo-

rem. However, the proof of Stokes’ Theorem that we give uses Green’s Theorem so it’s not

enough to just prove Stokes’ Theorem to get Green’s Theorem.

Proof of a special case. Suppose S has equation z = g(x, y), (x, y) ∈ D, where g has contin-

uous second-order partial derivatives and D is a simple plane region whose boundary curve

C1 corresponds to C. We assume the orientation of S is upward, in which case the positive

orientation of C corresponds to the positive orientation of C1. Suppose C1 is parameterized

by x = x(t) and y = y(t), a ≤ t ≤ b. Then C is parameterized by x = x(t), y = y(t), and

z = (g(x(t), y(t)), a ≤ t ≤ b. By the Chain Rule,∫C

F · dr =

∫ b

a

(Pdx

dt+Q

dy

dt+R

dz

dt

)dt =

∫ b

a

[Pdx

dt+Q

dy

dt+R

(∂z

∂x

dx

dt+∂z

∂y

dy

dt

)]dt

=

∫ b

a

[(P +R

∂z

∂x

)dx

dt+

(Q+R

∂z

∂y

)dy

dt

]dt

=

∫C1

(P +R

∂z

∂x

)dx+

(Q+R

∂z

∂y

)dy

=

∫∫D

[∂

∂x

(Q+R

∂z

∂y

)− ∂

∂y

(P +R

∂z

∂x

)]dA by Green’s Theorem

=

∫∫D

[(∂Q

∂x+∂Q

∂z

∂z

∂x+∂R

∂x

∂z

∂y+∂R

∂z

∂z

∂x

∂z

∂y+R

∂2z

∂x∂y

)−(∂P

∂y+∂P

∂z

∂z

∂y+∂R

∂y

∂z

∂x+∂R

∂z

∂z

∂y

∂z

∂x+R

∂2z

∂y∂x

)]dA

=

∫∫D

[−(∂R

∂y− ∂Q

∂z

)∂z

∂x−(∂P

∂z− ∂R

∂x

)∂z

∂y+

(∂Q

∂x− ∂P

∂y

)]dA

=

∫∫S

curlF · dS. �

22

Example 29. Suppose S is the cone x =√y2 + z2, 0 ≤ x ≤ 2, oriented in the direction of

the positive x-axis. Let F = arctan(x2yz2)i + x2yj + x2z2k. We will evaluate∫∫

ScurlF · dS

using Stoke’s Theorem.

The boundary curve C is y2 + z2 = 4, in the counterclockwise direction when viewed from

the front. Hence, C is parameterized by r(t) = 2i + 2 cos tj + 2 sin tk 0 ≤ t ≤ 2π. Now

F(r(t)) = arctan(32 cos t sin2 t)i + 8 cos ti + 16 sin2 t r′(t) = −2 sin tj + 2 cos tk

F(r(t)) · r′(t) = −16 cos t sin t+ 32 cos t sin2 t

Using Stokes’ Theorem,∫∫S

curlF · dS =

∮C

F · dr =

∫ 2π

0

F(r(t)) · r′(t) dt

= 16

∫ 2π

0

2 cos t sin2 t− cos t sin t dt = 16

[2

3sin3 t dt− 1

2sin2 t

]2π0

= 0

Now we’ll use Stokes’ Theorem the other way. Recall that when S is a surface given by

z = g(x, y), then we have∫∫S

F · dS =

∫∫D

(−P

(∂g

∂x

)−Q

(∂g

∂y

)+R

)dA.

Example 30. Let C be the boundary of the part of the plane 3x + 2y + z = 1 in the first

octant. Let F = i+ (x+ yz)j+ (xy−√z)k. We’ll use Stokes’ Theorem to evaluate

∫CF · dr.

A straightforward computation shows that curlF = (x − y)i − yj + k. The region S is the

portion of the plane 2x + 2y + z = 1 over D ={

(x, y) : 0 ≤ x ≤ 13, 0 ≤ y ≤ 1

2(1− 3x)

}. We

orient S upward and since S is given by z = 1− 3x− 2y, then by Stokes’ Theorem,∫C

F · dr =

∫∫S

curlF · dS =

∫∫D

[−(x− y)(−3)− (−y)(−2) + (1)] dA

=

∫ 1/3

0

∫ (1−3x)/2

0

(3x− 5y + 1) dA =

∫ 1/3

0

[(1 + 3x)y − 5

2y2](1−3x)/20

dy dx

=

∫ 1/3

0

[1

2(1 + 3x)(1− 3x)− 5

8(1− 3x)2

]dx

= −1

8

∫ 1/3

0

81x2 − 30x+ 1 dx = −1

8

[9x3 − 15x2 + x

]1/30

= −1

8· −1

3=

1

24.

Suppose S1 and S2 are two oriented surfaces with the same oriented boundary curve C, both

satisfying the hypotheses of Stokes’ Theorem. Then∫∫S1

curlF · dS =

∫C

F · dr =

∫∫S2

curlF · dS.

23

9. The Divergence Theorem

In vector form, one can rewrite Green’s Theorem as∫C

F · n ds =

∫∫D

divF(x, y) dA.

The Divergence Theorem is a higher-dimensional analog of this statement. It tells us that,

under the right conditions, the flux of a vector field F across the boundary surface of E is

equal to the triple integral of the divergence of F over E. We give the statement for simple

solid regions, i.e., those that are simultaneously type 1, 2, and 3.

The Divergence Theorem. Let E be a simple solid region and let S be the boundary

surface of E, given with positive (outward) orientation. Let F be a vector field whose

component functions have continuous partial derivatives on an open region that contains E.

Then ∫∫S

F · dS =

∫∫∫E

divF dV.

Proof. Let F = P i +Qj +Rk. Then∫∫∫E

divF dV =

∫∫∫E

∂P

∂x+∂Q

∂y+∂P

∂zdV =

∫∫∫E

∂P

∂xdV +

∫∫∫E

∂Q

∂ydV +

∫∫∫E

∂P

∂zdV

and∫∫S

F · dS =

∫∫S

(P i +Qj +Rk) · n dS =

∫∫S

P i · n dS +

∫∫S

Qj · n dS +

∫∫S

Rk · n dS.

We will prove∫∫∫

E∂Q∂y

dV =∫∫

SQj · n dS. The statement

∫∫∫E∂P∂z

dV =∫∫

SRk · n dS

is in the text, and the statement∫∫∫

E∂P∂x

dV =∫∫

SP i · n dS is part of the final reading

assignment. It is clear that the Divergence Theorem follows from these statements.

Since E is simple, it is a type 3 region:

E = {(x, y, z) : (x, z) ∈ D : u1(x, z) ≤ y ≤ u2(x, z)}

The boundary surface S consists of three pieces: a left surface S1, a right surface S2, and a

horizontal surface S3. The equation of S2 is y = u2(x, z) and the outward normal n points

in the positive y-direction. Conversely, on S1, the equation is y = u1(x, z) and the normal

points in the negative y-direction. Finally, on S3, j · n = 0 (because j is horizontal and n is24

vertical). Hence,∫∫

S3Qj · n dS = 0, while∫∫

S2

Qj · n dS =

∫∫D

Q(x, u2(x, z), z) dA,∫∫S1

Qj · n dS = −∫∫

D

Q(x, u1(x, z), z) dA.

Thus, ∫∫S

Qj · n dS =

∫∫S1

Qj · n dS +

∫∫S2

Qj · n dS +

∫∫S2

Qj · n dS

=

∫∫D

[Q(x, u2(x, z), z −

∫∫D

Q(x, u1(x, z), z) dA

]dA

=

∫∫∫E

∂Q

∂ydV by the FTC. �

Example 31. Let S be the sphere with center the origin and radius 2 and let

F = (x3 + y2)i + (y3 + z3)j + (x3 + y3)k.

We will calculate the flux of F across S. That is, we will compute∫∫

SF · dS.

Note that divF = 3(x2 + y2 + z2) and the region E is just the solid sphere of radius 2

centered at the origin. We make a change to spherical coordinates, so divF = 3ρ2. Then by

the Divergence Theorem.∫∫S

F · dS =

∫∫∫E

divF dV =

∫ 2

0

∫ π

0

∫ 2π

0

(3ρ2)(ρ2 sinφ) dθ dφ dρ

= 6π

∫ 2

0

∫ π

0

ρ4 sinφ dφ dρ = 6π

∫ 2

0

[ρ4(− cosφ)

]π0dρ

= 12π

∫ 2

0

ρ4 dρ = 12π

[1

5ρ5]20

=384π

5.

Example 32. Let S be the surface of the solid bounded by the cylinder x2 + y2 = 4 and the

planes z = y− 2 and z = 0. Let F = (xy+ 2xz)i+ (x2 + y2)j+ (xy− z2)k. We will compute∫∫SF · dS.

We have divF = (y + 2z) + (2y)− 2z = 3y. Let D be the circle of radius 2 in the xy-plane

centered at the origin. Then the region E bounded by S (in cylindrical coordinates), is

E = {(r, θ, z) : 0 ≤ r ≤ 2, 0 ≤ θ ≤ 2π, r sin θ − 2 ≤ z ≤ 0}.25

In cylindrical coordinates, divF = 3r sin θ. Now by the Divergence Theorem,∫∫S

F · dS =

∫∫∫E

divF dV =

∫ 2π

0

∫ 2

0

∫ r sin θ−2

0

(3r sin θ)(r) dz dr dθ

=

∫ 2π

0

∫ 2

0

(3r2 sin θ)(r sin θ − 2) dr dθ

=

∫ 2π

0

∫ 2

0

3r3 sin2 θ − 6r2 sin θ dr dθ

=

∫ 2π

0

[3

4r4 sin2 θ − 2r3 sin θ

]20

dθ

=

∫ 2π

0

12 sin2 θ − 16 sin θ dθ

=

∫ 2π

0

6(1− cos(2θ))− 16 sin θ dθ

=

[6

(θ − 1

2sin(2θ)

)+ 16 cos θ

]2π0

= 12π.

26

10. Summary

We’ll summarize the important methods from this chapter. As the book notes, all of our

major theorems are in fact just generalizations of the Fundamental Theorem of Calculus:∫ b

a

F ′(x) dx = F (b)− F (a).

In this section, we do not focus on the strict hypotheses of these methods, though those are

important, as one can refer back for the details. Instead, we will focus on choosing the right

method for solving a problem.

Line integrals Given a (scalar) function f(x, y, z) defined on a curve C with parameteriza-

tion r(t), a ≤ t ≤ b, the line integral of f along C is∫C

f(x, y, z) ds =

∫ b

a

f(r(t))|r′(t)| dt.

If F is a vector field defined along C, then the line integral of F along C is∫C

F · dr =

∫ b

a

F(r(t)) · r′(t) dt.

If F is conservative (F = ∇f), then the Fundamental Theorem for Line Integrals gives∫C

F · dr =

∫C

∇f · dr = f(r(b))− f(r(a)).

Example 33. Let F(x, y, z) = (y2z+ 2xz2)i+ 2xyzj+ (xy2 + 2x2z)k and let C be the curve

parameterized by r(t) =√ti + (t+ 1)j + t2k, 0 ≤ t ≤ 1.

First, let’s try to evaluate∫CF · dr using the definition. We have

F(r(t)) =[(t(t+ 1))2 + 2t3

]i + 2(

√t(t+ 1)t2)j + (

√t(t+ 1)2 + 2t3)k

We can already tell that F(r(t)) · r′(t) is going to be ridiculously complicated.

As an alternative, note that curlF = 0. Since F is defined on all of R3 and its components

have continuous partial derivatives, this means that F is conservative. So, we look for a

potential function. Since F = ∇f for some scalar function f , then

f(x, y, z) =

∫fx(x, y, z) dx =

∫y2z + 2xz2 dx = xy2z + x2z2 + g(y, z)

for some function g(y, z). Since 2xyz = fy(x, y, z) = 2xyz + g′(y, z), then g(y, z) = h(z), a

function in z. Thus, xy2 + 2x2z = fz(x, y, z) = xy2 + 2x2z + h′(z), so h(z) = K, a constant.

Thus, we take f = xy2z + x2z2. Note that r(1) = 〈1, 2, 1〉 and r0 = 〈0, 1, 0〉. Hence,∫C

F · dr =

∫C

∇f · dr = f(1, 2, 1)− f(0, 1, 0) = (4 + 1)− 0 = 5.

27

Another way to evaluate line integrals of vector fields in R2 is using Green’s Theorem:∫C

P dx+Q dy =

∫∫D

(∂Q

∂x− ∂P

∂y

)dA,

where C = ∂D, positively oriented (counterclockwise). This is especially useful when C isn’t

a smooth curve but D is a “nice” region.

Example 34. Let C be the triangle with vertices (0, 0), (1, 0), and (0, 1), oriented counter-

clockwise. We’ll evaluate∫Ce2x+y dx+ e−y dy.

This notation is just another way to write∫CF · dr where F = 〈e2x+y dx, e−y〉. The region

D enclosed by C is D = {(x, y) : 0 ≤ x ≤ 1, 0 ≤ y ≤ 1− x}. Hence, by Green’s Theorem,∫C

e2x+y dx+ e−y dy =

∫∫D

(0− e2x+y

)dA = −

∫ 1

0

∫ 1−x

0

e2x+y dy dx

= −∫ 1

0

[e2x+y

]1−x0

dx = −∫ 1

0

ex+1 − e2x dx

= −[ex+1 − 1

2e2x]10

= e− 1

2− 1

2e2.

Surface integrals Given a (scalar) function f(x, y, z) defined on a surface s with parame-

terization r(u, v), (u, v) ∈ D, the surface integral of f over S is∫∫S

f(x, y, z) dS =

∫∫D

f(r(u, v))|ru × rv| dA.

If F is a vector field defined on S, then the surface integral of F over C (flux of F across S) is∫∫S

F · dS =

∫∫D

F · n dS =

∫∫D

F(r(u, v)) · (ru × rv) dA.

Surface integrals also give another way to evaluate line integrals.

Example 35. Let S be the part of the paraboloid z = 5− x2− y2 that lies above the plane

z = 1, oriented upward, and let F = −2yzi + yj + 3xk. We will evaluate∫∫

ScurlF · dS.

An easy computation shows that curlF = −(3 + 2y)j + 2zk. One could use the special

case of surface integrals when S is the graph of the function z = g(x, y), or we could just

parameterize with

r(u, v) = 〈u, v, 5− u2 − v2〉 (u, v) ∈ D.

Hence, ru × rv = 〈2u, 2v, 1〉 and so

curlF(r(u, v)) · (ru × rv) = 〈0,−(3 + 2v), 2(5− u2 − v2)〉 · 〈2u, 2v, 1〉 = 10− 2u2 − 6v2 − 6v.28

We evaluate the integral by switching to cylindrical coordinates,∫∫S

curlF · dS =

∫∫D

10− 2u2 − 6v2 − 6v dA

=

∫ 2

0

∫ 2π

0

(10− 2r cos2(x)− 6r sin2(x)− 6r sin(x)

)r dθ dr = 8π

On the other hand, if F is a vector field defined on a “nice” oriented surface S and C its

boundary curve, then by Stokes’ Theorem:∫C

F · dr =

∫∫S

curlF · dS.

Example 36. Keeping the setup of the previous example. The boundary curve C is a

circle of radius 2, centered at the origin, in the plane z = 1. We parameterize as r(t) =

{2 cos t, 2 sin t, 1}, 0 ≤ t ≤ 2π. Then,∫C

F · dr =

∫ 2π

0

F(r(t)) · r′(t) dt =

∫ 2π

0

8 sin2 t+ 4 sin t cos t dt = 8π.

Suppose in the previous example we wanted to evaluate∫∫

SF · dS. We could parameterize

the surface using cylindrical coordinates:

r(r, θ) = {r cos θ, r sin θ, 5− r2}, 0 ≤ r ≤ 2, 0 ≤ θ ≤ 2π.

Now, rr × rθ = 2r2 cos θi + 2r2 sin θj + rk and so

F(r(r, θ)) · (rr × rθ) = (it’s not pretty).

As an alternative, we can evaluate using the Divergence Theorem. Let E be a simple solid

region and S the boundary region of E, then∫∫S

F · dS =

∫∫∫E

divF dV.

Example 37. With the setup of the previous example, we have divF = 1. Then,∫∫S

F · dS =

∫∫∫E

divF dV =

∫∫D

∫ 5−x2−y2

1

1 dz dA =

∫∫D

4− x2 − y2 dA.

The region D can be expressed in polar coordinates D = {(r, θ) : 0 ≤ r ≤ 2, 0 ≤ θ ≤ 2π}.Hence, ∫∫

S

F · dS =

∫∫D

4− x2 − y2 dA =

∫ 2

0

∫ 2π

0

(4− r2)r dθ dr

= 2π

∫ 2

0

4r − r3 dr = 2π

[2r2 − 1

4r4]20

= 8π.

29