Section 14.8: Stokes’ Theorem
Definition: Let S be an orientable surface. The orientation of S induces the positive orientation of the boundary curve C. This means that if you walk in the positive direction
~ , then the surface will be on your
around C with your head pointing in the direction of N
left.
Theorem: (Stokes’ Theorem)
Let S be an orientable 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 have continuous partial derivatives on an open region in R3 that contains S.
Then
ZZ
I
~
~
~
F · dR =
curlF~ · dS.
C
S
Note: Stokes’ Theorem relates a surface integral over a surface S to a line integral around
the boundary curve of S.
ZZ
~ where F~ (x, y, z) = hxyz, x, exy cos zi
curlF~ · dS,
Example: Use Stokes’ Theorem to evaluate
S
and S is the hemisphere x2 + y 2 + z 2 = 1, z ≥ 0, oriented upward.
The boundary curve C is the circle x2 + y 2 = 1 in the xy-plane, traversed counterclockwise.
This curve can be parameterized by
~
R(t)
= hcos t, sin t, 0i
for 0 ≤ t ≤ 2π. Then
~ 0 (t) = h− sin t, cos t, 0i
R
F~ (t) = h0, cos t, esin t cos t i.
and
By Stokes’ Theorem,
ZZ
~ =
curlF~ · dS
S
I
~
F~ · dR
ZC2π
=
Z0 2π
=
~ 0 (t)dt
F~ (t) · R
cos2 tdt
0
Z
1 2π
(1 + cos 2t)dt
=
2 0
2π
1
1
=
t + sin 2t
2
2
0
= π.
Example: Use Stokes’ Theorem to evaluate
I
~
F~ · dR,
C
where F~ (x, y, z) = h2z, 4x, 5yi and C is the curve of intersection of the plane z = x + 4 and
the cylinder x2 + y 2 = 4 oriented counterclockwise as viewed from above.
The curve C is an ellipse which is not easy to parameterize. Thus, we apply Stokes’ Theorem.
First,
~k
~i
~j
∂
∂
∂
curlF~ =
= h5, 2, 4i.
∂x ∂y ∂z
2z 4x 5y
The surface S is the elliptical region in the plane z = x + 4 bounded by C. If S is oriented
upward, then C has the induced positive orientation. The projection of S onto the xy-plane
is the disk x2 + y 2 ≤ 4. By Stokes’ Theorem,
ZZ
I
~
~
~
curlF~ · dS
F · dR =
C
Z ZS ∂g
∂g
=
−Q
+ R dA
−P
∂x
∂y
D
ZZ
=
(−5 + 4)dA
D
ZZ
dA
= −
D
= −4π.
Example: Use Stokes’ Theorem to evaluate
I
~
F~ · dR,
C
where F~ (x, y, z) = hxz, 2xy, 3xyi and C is the boundary of the part of the plane 3x+y+z = 3
in the first octant, oriented counterclockwise as viewed from above.
First,
curlF~ =
~k
~i
~j
∂
∂
∂
∂x ∂y ∂z
xz 2xy 3xy
= h3x, x − 3y, 2yi.
The surface S is the part of the plane 3x + y + z = 3 in the first octant. If S is oriented
upward, then C has the induced positive orientation. The projection of S onto the xy-plane
is the triangular region given by
D = {(x, y)|0 ≤ x ≤ 1, 0 ≤ y ≤ 3 − 3x}.
By Stokes’ Theorem,
I
~ =
F~ · dR
C
=
=
=
=
=
=
=
=
ZZ
~
curlF~ · dS
Z ZS ∂g
∂g
−P
−Q
+ R dA
∂x
∂y
D
ZZ
(10x − y)dA
D
Z 1 Z 3−3x
(10x − y)dydx
0
0
3−3x
Z 1
1 2
10xy − y
dx
2
0
0
Z 1
1
2
2
30x − 30x −
9 − 18x + 9x dx
2
0
Z 1
9
69
(39x − x2 − )dx
2
2
0
39 23 9
−
−
2
2
2
7
.
2