Calculus III Chapter 16 - Vector Calculus vector calculus.pdfCalculus III Chapter 16 - Vector...
Transcript of Calculus III Chapter 16 - Vector Calculus vector calculus.pdfCalculus III Chapter 16 - Vector...
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