which, upon comparing to equation (4.49), proves the Theorem.
QED
172
CHAPTER 4. LINE AND SURFACE INTEGRALS
Note: The condition in Stokes’ Theorem that the surface Σ have a (continuously vary-
ing) positive unit normal vector n and a boundary curve C traversed n-positively can be
expressed more precisely as follows: if r( t) is the position vector for C and T( t) = r′( t)/ r′( t) is the unit tangent vector to C, then the vectors T, n, T × n form a right-handed system.
Also, it should be noted that Stokes’ Theorem holds even when the boundary curve C is
piecewise smooth.
Example 4.14. Verify Stokes’ Theorem for f( x, y, z) = z i + x j + yk when Σ is the paraboloid z = x 2 + y 2 such that z ≤ 1 (see Figure 4.5.5).
z
Solution: The positive unit normal vector to the surface
C
z = z( x, y) = x 2 + y 2 is
1
n
− ∂z i
j
∂x − ∂z
∂ y + k
−2 x i − 2 yj + k
n =
=
,
2
2
1 + 4 x 2 + 4 y 2
1 + ∂z
∂x
+ ∂z
∂ y
Σ
and curl f = (1 − 0)i + (1 − 0)j + (1 − 0)k = i + j + k, so
y
0
(curl f ) · n = (−2 x − 2 y + 1)/ 1 + 4 x 2 + 4 y 2 .
x
Figure 4.5.5
z = x 2 + y 2
Since Σ can be parametrized as r( x, y) = x i + yj + ( x 2 + y 2)k for
( x, y) in the region D = {( x, y) : x 2 + y 2 ≤ 1}, then
∂r
∂r
(curl f ) · n dσ =
(curl f ) · n
×
d A
∂x
∂y
Σ
D
−2 x − 2 y + 1
=
1 + 4 x 2 + 4 y 2 d A
1 + 4 x 2 + 4 y 2
D
=
(−2 x − 2 y + 1) d A , so switching to polar coordinates gives
D
2 π
1
=
(−2 r cos θ − 2 r sin θ + 1) r dr dθ
0
0
2 π
1
=
(−2 r 2 cos θ − 2 r 2 sin θ + r) dr dθ
0
0
2 π
r=1
=
− 2 r 3 cos θ
sin θ
dθ
3
− 2 r 3
3
+ r 22
0
r=0
2 π
=
− 2 cos θ
sin θ
dθ
3
− 23
+ 12
0
2 π
= − 2 sin θ
cos θ
θ
3
+ 23
+ 12
= π .
0
4.5 Stokes’ Theorem
173
The boundary curve C is the unit circle x 2 + y 2 = 1 laying in the plane z = 1 (see Figure
4.5.5), which can be parametrized as x = cos t, y = sin t, z = 1 for 0 ≤ t ≤ 2 π. So
2 π
f · dr =
((1)(−sin t) + (cos t)(cos t) + (sin t)(0)) dt
C
0
2 π
1 + cos2 t
1 + cos2 t
=
−sin t +
dt
here we used cos2 t =
0
2
2
t
sin 2 t 2 π
= cos t + +
= π .
2
4
0
So we see that
f
(curl f )
C · dr =
· n dσ, as predicted by Stokes’ Theorem.
Σ
The line integral in the preceding example was far simpler to calculate than the surface
integral, but this will not always be the case.
Example 4.15. Let Σ be the elliptic paraboloid z = x 2
for z
4 + y 2
9
≤ 1, and let C be its boundary
curve. Calculate
f
C
· dr for f( x, y, z) = (9 xz + 2 y)i + (2 x + y 2)j + (−2 y 2 + 2 z)k, where C is traversed counterclockwise.
Solution: The surface is similar to the one in Example 4.14, except now the boundary curve C
is the ellipse x 2
4 + y 2
9 = 1 laying in the plane z = 1. In this case, using Stokes’ Theorem is easier
than computing the line integral directly. As in Example 4.14, at each point ( x, y, z( x, y)) on the surface z = z( x, y) = x 2
the vector
4 + y 2
9
− ∂z i
j
∂x − ∂z
∂ y + k
− x i − 2 y j + k
n =
=
2
9
,
2
2
1 + ∂z
1 + x 2
∂x
+ ∂z
∂ y
4 + 4 y 2
9
is a positive unit normal vector to Σ. And calculating the curl of f gives
curl f = (−4 y − 0)i + (9 x − 0)j + (2 − 2)k = −4 yi + 9 x j + 0k ,
so
(−4 y)(− x ) + (9 x)(− 2 y ) + (0)(1)
2 x y − 2 xy + 0
(curl f ) · n =
2
9
=
= 0 ,
1 + x 2
1
4 + 4 y 2
9
+ x 24 + 4 y 2
9
and so by Stokes’ Theorem
f · dr =
(curl f ) · n dσ =
0 dσ = 0 .
C
Σ
Σ
174
CHAPTER 4. LINE AND SURFACE INTEGRALS
In physical applications, for a simple closed curve C the line integral
f
C · dr is often called
the circulation of f around C. For example, if E represents the electrostatic field due to a
point charge, then it turns out8 that curl E = 0, which means that the circulation
E
C
· dr = 0
by Stokes’ Theorem. Vector fields which have zero curl are often called irrotational fields.
In fact, the term curl was created by the 19th century Scottish physicist James Clerk
Maxwell in his study of electromagnetism, where it is used extensively. In physics, the
curl is interpreted as a measure of circulation density. This is best seen by using another
definition of curl f which is equivalent9 to the definition given by formula (4.46). Namely, for a point ( x, y, z) in R3,
1
n · (curl f)( x, y, z) = lim
f · dr ,
(4.50)
S→0 S C
where S is the surface area of a surface Σ containing the point ( x, y, z) and with a simple
closed boundary curve C and positive unit normal vector n at ( x, y, z). In the limit, think of
the curve C shrinking to the point ( x, y, z), which causes Σ, the surface it bounds, to have
smaller and smaller surface area. That ratio of circulation to surface area in the limit is
what makes the curl a rough measure of circulation density (i.e. circulation per unit area).
An idea of how the curl of a vector field is
y
related to rotation is shown in Figure 4.5.6.
Suppose we have a vector field f( x, y, z) which
is always parallel to the x y-plane at each
f
point ( x, y, z) and that the vectors grow larger
the further the point ( x, y, z) is from the y-
axis. For example, f( x, y, z) = (1+ x 2)j. Think
of the vector field as representing the flow
of water, and imagine dropping two wheels
with paddles into that water flow, as in Fig-
ure 4.5.6. Since the flow is stronger (i.e. the
x
magnitude of f is larger) as you move away
0
from the y-axis, then such a wheel would ro-
tate counterclockwise if it were dropped to
Figure 4.5.6
Curl and rotation
the right of the y-axis, and it would rotate
clockwise if it were dropped to the left of the y-axis. In both cases the curl would be nonzero
(curl f( x, y, z) = 2 x k in our example) and would obey the right-hand rule, that is, curl f( x, y, z) points in the direction of your thumb as you cup your right hand in the direction of the rotation of the wheel. So the curl points outward (in the positive z-direction) if x > 0 and points
inward (in the negative z-direction) if x < 0. Notice that if all the vectors had the same di-
rection and the same magnitude, then the wheels would not rotate and hence there would
be no curl (which is why such fields are called irrotational, meaning no rotation).
8See Ch. 2 in REITZ, MILFORD and CHRISTY.
9See SCHEY, p. 78-81, for the derivation.
4.5 Stokes’ Theorem
175
Finally, by Stokes’ Theorem, we know that if C is a simple closed curve in some solid
region S in R3 and if f( x, y, z) is a smooth vector field such that curl f = 0 in S, then
f · dr =
(curl f ) · n dσ =
0 · n dσ =
0 dσ = 0 ,
C
Σ
Σ
Σ
where Σ is any orientable surface inside S whose boundary is C (such a surface is some-
times called a capping surface for C). So similar to the two-variable case, we have a three-
dimensional version of a result from Section 4.3, for solid regions in R3 which are simply
connected (i.e. regions having no holes):
The following statements are equivalent for a simply connected solid region S in R3:
(a) f( x, y, z) = P( x, y, z)i + Q( x, y, z)j + R( x, y, z)k has a smooth potential F( x, y, z) in S
(b)
f · dr is independent of the path for any curve C in S
C
(c)
f · dr = 0 for every simple closed curve C in S
C
∂R
∂Q ∂P
∂R
∂Q
∂P
(d)
=
,
=
, and
=
in S (i.e. curl f = 0 in S)
∂y
∂z
∂z
∂x
∂x
∂y
Part (d) is also a way of saying that the differential form P dx + Q d y + R dz is exact.
Example 4.16. Determine if the vector field f( x, y, z) = xyz i+ xz j+ xyk has a potential in R3.
Solution: Since R3 is simply connected, we just need to check whether curl f = 0 throughout
R3, that is,
∂R
∂Q
∂P
∂R
∂Q
∂P
=
,
=
,
and
=
∂y
∂z
∂z
∂x
∂x
∂y
throughout R3, where P( x, y, z) = xyz, Q( x, y, z) = xz, and R( x, y, z) = xy. But we see that
∂P
∂R
∂P
∂R
= xy ,
= y
⇒
=
for some ( x, y, z) in R3.
∂z
∂x
∂z
∂x
Thus, f( x, y, z) does not have a potential in R3.
Exercises
A
For Exercises 1-3, calculate
f ( x, y, z) ds for the given function f ( x, y, z) and curve C.
C
1. f ( x, y, z) = z;
C : x = cos t, y = sin t, z = t, 0 ≤ t ≤ 2 π
176
CHAPTER 4. LINE AND SURFACE INTEGRALS
x
2. f ( x, y, z) =
+ y + 2 yz;
C : x = t 2, y = t, z = 1, 1 ≤ t ≤ 2
y
3. f ( x, y, z) = z 2;
C : x = t sin t, y = t cos t, z = 2 2 t 3/2, 0
3
≤ t ≤ 1
For Exercises 4-9, calculate
f
C · dr for the given vector field f( x, y, z) and curve C.
4. f( x, y, z) = i − j + k;
C : x = 3 t, y = 2 t, z = t, 0 ≤ t ≤ 1
5. f( x, y, z) = yi − x j + z k;
C : x = cos t, y = sin t, z = t, 0 ≤ t ≤ 2 π
6. f( x, y, z) =