Introduction
In this section we see the generalization of a familiar theorem, Green’s Theorem. Just as before we are interested in an equality that
allows us to go between the integral on a closed curve to the double integral of a surface. Some important definitions to know before
proceeding are: simple closed curve, divergence, flux, curl, and normal vector. Knowing how to calculate the determinant of 2x2 and 3x3
matrices will also help deepen your understanding of divergence and curl.
Theoretical Discussion
Curl: Let F
@
@
= M (x, y, z) i + N(x, y, z) j + P(x, y, z) k
@
and A
= i
@<
+j
<x
@ <
+ k
<y
@ <
<z then the curl of F is simply the determinant of the 3x3
matrix A × F . There are many ways to take the determinant, but the following is an example of cofactor expansion.
A×F
= i
\<
\
\N
\ i@
\
\ <
\<
\
\M
=
x
\ \<
\ − j @\ <
\ \
\M
P \
<
<z
@\ <y
N
P
P
<P
<y
−
<N
@ <P
) − j(
<z
<x
@
<P
<y
−
<N
@ <M
) + j(
<z
<z
= i(
k
<
<z
<
<z
x
@
= i(
@
<
<y
j
−
−
@
\
\
\
\
\
\
\ <
\
\ + k @\ <
\
\
\
\M
x
<
<y
N
\
\
\
\
<M
@ <N
) + k(
<z
<x
−
<M
)
<y
<P
@ <N
) + k(
<x
<x
−
<M
)
<y
= curl F
Stokes' Theorem
Let n be a normal vector (orthogonal, perpendicular) to the surface S that has the vector field F , then the simple closed curve C is
defined in the counterclockwise direction around n . The circulation on C equals surface integral of the curl of F = A × F dotted with n .
∮
F
˝dr =
C
∬
A × F ˝n dJ
S
This theorem fails when a function, vector field, or derivative is not continuous. Green's Theorem out of Stokes
If the counterclockwise circulation C is only in x­y plane, and it defines a region, call it R, with the vector field F then the z direction is
normal to the plane. Thus
∮
F
˝dr =
C
=
=
∬
∬
∬
A × F ˝n dJ
S
A × F ˝k dx dy
R
R
<N
<x
−
<M
<y
As a note,
<N
−
<M
dx dy
<N
<x
<M
<y
−
is the determinant of the 2x2 matrix
\ <
\<
\
\M
<\
<y \
x
N
\
\
Example 1
Evaluate the equation for Stokes' Theorem for the hemisphere 2
2
C : x + y = 0, z = 0
and the field F = yi − xj .
S : x
2
2
+ y
+ z
2
= 9, z
¦
0
, its bounding circle Hints: Remember that a simple way to parameterize a circle is if x 2 + y2 = r 2 then r(? ) = rcos? + rsin? for ? B [0, 2G] . Also, try
drawing a picture of the hemisphere and its bounding circle to understand the theory behind the problem. Should know how to
normalize a vector and what |A f | means. Find the counterclockwise circulation by using the left­hand side of Stokes' Theorem, then
find the curl integral by using the right­hand side of Stokes' Theorem and compare your results.
Solution
The hemisphere looks much like the image below, with the circumference of the pink bottom being the bounding circle C in the xy ­
plane . We can calculate the counterclockwise circulation around C (viewed from above) using the parametrization r( ? ) = (3cos? )i + (3sin? )j, 0 ¥ ? ¥ 2G
: dr = (−3sin? d? )i + (3cos? d? )j
F = yi − xj = (3sin? )i − (3cos? )j
F
˝dr =
−9sin
F
˝dr =
2
?
2
d? − 9cos
?
d?
= −9d?
2
∮
C
∫
G
− 9d?
= −18?
0
This is the evaluated left­hand side of Stokes' Theorem. Now we want to show that the right­hand side is equal by evaluting the curl
integral. For the curl integral of F , we have
A×F
=
(
<P
<y
−
<N
<M
i +
( <z
<z )
−
<P
<N
j +
( <x
<x )
−
<M
k
<y )
from taking the determinant of the 3X3 matrix of the curl (explained in theoretical discussion). If we look at that 3X3 matrix: \i
\<
\
\<
\
\y
x
\
\
j
k
<
<y
<\
<z \
−x
0
\
\
Evaluating the curl, we see that
A×F
= (0 − 0)i + (0 − 0)j + (−1 − 1)k = −2k
A×F
= (0 − 0)i + (0 − 0)j + (−1 − 1)k = −2k
Our outer unit normal vector will be
xi + yj + zk
n =
xi + yj + zk
=
|xi + yj + zk|
xi + yj + zk
=
k
kkk
k
kk
kk
xk
yk
z
+
+
Sk
2
2
2
xi + yj + zk
kkkk
k
kk9si
kkk
kk
sk
n
S9co
?k
+
?k
2
2
=
3
Then
dJ =
Af |
dA
|A f ˝k|
|
3
=
z
dA
Finally, we can put everything together to find that:
A × F ˝ndJ
2z 3
= −
3
z
dA = −2dA
and ∫
∫
S
A × F ˝ndJ
=
∫
∫
2
x
+y
2
¥9
−2dA = −18G
and we see that the circulation around the circle equals the integral of the curl over the hemisphere, as it should.