The surface integral portion of Stokes theorem can therefore by expressed as
∫∫
S
curl F d S=
∫∫
S
(2 xy − 3 x^2 )ˆe 3 ·ˆendS =
∫∫
S
(2 xy − 3 x^2 )dx dy
=
∫ 1
− 1
∫y=+ √ 1 −x 2
y=−√ 1 −x^2
x(2 y− 3 x)dy dx =
∫ 1
− 1
x(y^2 − 3 xy )
+√ 1 −x^2
−√ 1 −x^2
dx
=
∫ 1
− 1
{
x
[
(1 −x^2 )− 3 x
√
1 −x^2
]
−x
[
(1 −x^2 ) + 3x
√
1 −x^2
]}
dx
=
∫ 1
− 1
− 6 x^2
√
1 −x^2 dx =−
3 π
4.
For the line integral portion of Stokes theorem one should observe the boundary
of the surface Sis the circle
x= cos θ y = sin θ z = 0 0 ≤θ≤ 2 π.
Consequently, there results
∫
C
©F·dr =
∫
C
© 3 x^2 y dx +x^2 y dy +z dz
=
∫ 2 π
0
−3 cos^2 θsin^2 θ dθ + cos^3 θsin θ dθ =−^34 π.
Think of the unit circle x^2 +y^2 = 1 with a rubber sheet over it. The hemisphere
in this example is assumed to be formed by stretching this rubber sheet. All two-
sided surfaces that result by deforming the rubber sheet in a continuous manner are
surfaces for which Stokes theorem is applicable.
Related Integral Theorems
Let φdenote a scalar field and F a vector field. These fields are assumed to be
continuous with continuous derivatives. For the volumes, surfaces, and simple closed
curves of Stokes theorem and the divergence theorem, there exist the additional
integral relationships
∫∫∫
V
curl F dV =
∫∫∫
V
∇× F dV =
∫∫
S
eˆn×F dS (8 .51)
∫∫∫
V
grad φ dV =
∫∫∫
V
∇φ dV =
∫∫
S
φˆendS (8 .52)
∫∫
S
ˆen×grad φ dS =
∫∫
S
ˆen×grad φ dS =−
∫∫
S
grad φ×dS=
∫
C
©φ dr. (8 .53)
