The left-hand side of Stokes theorem can be expressed
∫∫
S
curlF·dS=
∫∫
S
curlF·ˆendS =
∫∫
S
[
x(1 − 2 z) sin θcos φ+ (z^2 −z) cos θ
]
r^2 sin θ dθ dφ
=
∫ 2 π
0
∫θ 0
0
[
rsin^2 θcos^2 φ(1 − 2 rcos θ) + (r^2 cos^3 θ−rcos^2 θ)
]
r^2 sin θ dθ dφ
=
∫ 2 π
0
r^3
12
[
4(−1 + cos^3 θ 0 )− 3 r(−1 + cos^4 θ 0 )
+ 16(2 + cos θ 0 ) cos^2 φsin^4
(
θ 0
2
)
− 6 rcos^2 φsin^4 θ 0
]
dφ
∫∫
S
curlF·dS=πr^3 cos θ 0 (−1 + rcos θ 0 ) sin^2 θ 0
The right-hand side of Stokes theorem can be expressed
∫
C
©F·dr =
∫
C
©yz dx +xz^2 dy +xy dz
Substitute the values of x, y, z on the curve Cusing
x=rsin θ 0 cos φ,
dx =−rsin θ 0 sin φ dφ,
y=rsin θ 0 sin φ,
dy =rsin θ 0 cos φ,
z=rcos θ 0
dz =0
The right-hand side of Stokes theorem then becomes
∫
C
©F·dr =
∫
C
©
[
yz(−rsin θ 0 sin φ) + xz^2 (rsin θ 0 cos φ) + xy (0)
]
dφ
=
∫ 2 π
0
[
rsin θ 0 sin φ(rcos θ 0 )(−rsin θ 0 sin φ) + (rsin θ 0 cos φ)(r^2 cos^2 θ 0 )(rsin θ 0 cos φ)
]
dφ
=πr^3 cos θ 0 (−1 + rcos θ 0 ) sin^2 θ 0
Proof of Stokes Theorem
To prove Stokes theorem one could verify each of the following integral relations
∫∫
S
(
∂F 1
∂z
ˆe 2 ·ˆen−∂F^1
∂y
ˆe 3 ·ˆen
)
dS =
∫
C
©F 1 dx
∫∫
S
(
∂F 2
∂x
ˆe 3 ·ˆen−∂F^2
∂z
ˆe 1 ·ˆen
)
dS =
∫
C
©F 2 dy
∫∫
S
(
∂F 3
∂y
ˆe 1 ·ˆen−∂F^3
∂x
ˆe 2 ·ˆen
)
dS =
∫
C
©F 3 dz.
(8 .44)
Then an addition of these integrals would produce the Stokes theorem as given by
equation (8.40). However, the arguments used in proving the above integrals are
repetitious, and so only the first integral is verified.
