Stokes' Theorem Examples
Stokes' Theorem relates surface integrals and line integrals.
STOKES' THEOREM
Let F be a vector field. Let W be an oriented surface, and let G be the boundary curve of W , oriented
using the right-hand rule. Then:
( ( curlaFb † . A œ * F † . s
W
G
EXAMPLE 1 Let G be the curve defined by the parametric equations
Bœ!
C œ #  # cos >
D œ #  # sin >
! Ÿ > Ÿ #1
Use Stokes' Theorem to evaluate ( B# /&D .B  B cos C .C  $C .D .
G
SOLUTION
The parametric equations above describe a circle of radius # on the CD -plane:
z
y
x
Let H be the disc whose boundary is the given circle. By Stokes' Theorem:
# &D
# &D
( ˆB / i  B cos C j  $C k‰ † . s œ ( ( curlˆB / i  B cos C j  $C k‰ † . A
G
H
We compute the curl:
â
â i
â
â
# &D
ˆ
‰
curl B / i  B cos C j  $C k œ â `Î`B
â
â # &D
âB /
j
`Î`C
B cos C
So we have to evaluate ( ( ˆ$ i  &B# /&D j  cos C k‰ † . A.
H
â
k â
â
â
# &D
`Î`D ââ œ ˆ$ i  &B / j  cos C k‰
â
$C â
This integral can be evaluated geometrically. The vector . A for the disc points in the positive B
direction. (Stokes' theorem uses the right-hand rule: if you curl the fingers of your right hand in the
direction of G , then your thumb points in the direction of .A.) So:
ˆ$ i  &B# /&D j  cos C k‰ † . A œ $ .E
Therefore:
# &D
( ( ˆ$ i  &B / j  cos C k‰ † . A œ ( ( $ .E œ $ ‚ athe area of the discb œ "#1
H
è
H
EXAMPLE 2 Let W be the surface D œ Ba"  BbC a"  C b for ! Ÿ B Ÿ " and ! Ÿ C Ÿ ". Evaluate the
integral (( Bk † . A, where . A is the upward-pointing normal vector.
W
If we wish to use Stokes' theorem, we must express Bk as the curl of some vector field F.
The formula for the curl is:
â
â
â i
â
j
k
â
â
â
â
`JC `JB
`JD
`JD `JC
`JB
â
curl F œ â `Î`B `Î`C `Î`D ââ œ Œ

ß

ß


`C
`D
`D
`B `B
`C
â
â
â J
JC
JD ââ
â B
SOLUTION
So what we need is:
`JC
`JD

œ!
`C
`D
`JB
`JD

œ!
`D
`B
`JC
`JB

œB
`B
`C
It is not hard to guess that JC œ "# B# , with JB œ JD œ !. Indeed:
â
â
â i
j
k ââ
â
â
â
"
curlŒ B# j œ ââ `Î`B `Î`C `Î`D ââ œ Bk
#
â
â
â !
â
" #
B
!
â
â
#
By Stokes' Theorem, we conclude that ( ( Bk † . A œ *
W
G
" #
B .C, where G is the boundary curve of the
#
surface W .
So what is the boundary of W ? Well, the equation D œ Ba"  BbC a"  C b specifies a surface whose
D-coordinate varies with horizontal position:
z
y
x
The allowed values of B and C are determined by the inequalities ! Ÿ B Ÿ " and ! Ÿ C Ÿ ", which
describe a square on the BC -plane:
y
1
x
1
The sides of this square are along the lines B œ !, C œ !, B œ ", and C œ ". Looking at the equation, we
see that D œ ! for these values of B and C , so the boundary of the surface is just the boundary of the
square on the BC -plane:
y
1
x
1
We can evaluate *
G
*
G
" #
B .C geometrically:
#
" #
" #
" #
" #
" #
B .C œ (
B .C  (
B .C  (
B .C  (
B .C
#
top #
bottom #
left #
right #
The top and bottom sides are horizontal, so .C œ !. Furthermore, B œ ! along the left edge, and B œ "
along the right edge, so:
*
G
" #
"
"
B .C œ !  !  !  (
.C œ
#
#
right #
è
EXAMPLE 3 Let W be the upper hemisphere of the unit sphere B#  C #  D # œ ". Use Stokes' theorem
to evaluate ( ( ˆB$ /C i  $B# /C j‰ † . A, where . A is the upward-pointing normal vector.
W
If we wish to use Stokes' theorem, we must express B$ /C i  $B# /C j as the curl of some
vector field F. The formula for the curl is:
â
â
â i
j
k ââ
â
â
â
`JC `JB
`JD
`JD `JC
`JB
curl F œ ââ `Î`B `Î`C `Î`D ââ œ Œ

ß

ß


`C
`D
`D
`B `B
`C
â
â
â J
JC
JD ââ
â B
SOLUTION
So what we need is:
`JC
`JD

œ B $ /C
`C
`D
`JB
`JD

œ $B# /C
`D
`B
`JC
`JB

œ!
`B
`C
If we guess that JC and JB are zero, it is not too hard to figure out that JD œ B$ /C . Indeed:
â
â
â i
j
k â
â
â
â
â
$ C ‰
ˆ
curl B / k œ â `Î`B `Î`C `Î`D â œ B$ /C i  $B# /C j
â
â
â
â
â !
!
B $ /C â
By Stokes' Theorem, we conclude that ( ( ˆB$ /C i  $B# /C j‰ † . A œ * B$ /C .D , where G is the
W
G
boundary curve of the surface W .
Since W is the upper hemisphere of the unit sphere, G is just the unit circle on the BC -plane. By the
right-hand rule, G is oriented counterclockwise. (It would be clockwise if we had started with the
downward-pointing . A.) Then * B$ /C .D is zero, since D is not changing over the course of the circle.
G
We conclude that ( ( ˆB / i  $B# /C j‰ † . A œ !.
$ C
W
è
EXAMPLE 4 Let W be the surface defined by D œ B#  C # for D Ÿ %. Use Stokes' Theorem to evaluate
#
$
( ( ˆ$BD i  D k‰ † . A, where . A is the upward-pointing normal vector.
W
If we wish to use Stokes' theorem, we must express $BD # i  D $ j as the curl of some
vector field F. The formula for the curl is:
â
â
â i
j
k ââ
â
â
â
`JC `JB
`JD
`JD `JC
`JB
curl F œ ââ `Î`B `Î`C `Î`D ââ œ Œ

ß

ß


`C
`D
`D
`B
`B
`C
â
â
â J
JC
JD ââ
â B
SOLUTION
So what we need is:
`JC
`JD

œ $BD #
`C
`D
`JC
`JB

œ D$
`B
`C
`JB
`JD

œ!
`D
`B
If we guess that JB and JD are zero, it is not hard to figure out that JC œ BD $ . Indeed:
â
â
â i
j
k â
â
â
â
â
curlˆBD $ j‰ œ â `Î`B `Î`C `Î`D â œ $BD # i  D $ k
â
â
â
â
$
â !
BD
! â
By Stokes' Theorem, we conclude that ( ( ˆ$BD # i  D $ k‰ † . A œ * BD $ .C, where G is the
W
G
boundary curve of the surface W .
So what is the boundary of W ? Well, the equation for the surface W can be expressed as D œ <# ,
where D Ÿ %. This appears as a parabola on the <D -plane:
z
z
(2, 4)
C
y
r
x
Therefore, the surface W is a bowl-shaped paraboloid. Its boundary curve G is a counterclockwise circle
on the plane D œ % with radius #, with parameterization:
B œ # cos >
C œ # sin >
Dœ%
! Ÿ > Ÿ #1
(The circle is counterclockwise by the right-hand rule.) Thus:
$
* BD .C œ (
G
!
#1
a# cos >ba%b$ a# cos > .>b
œ #&'(
#1
œ #&'(
#1
cos# > .>
!
!
"
a"  cos #>b .>
#
(double-angle formula)
#1
"
œ "#)”>  sin #>•
#
!
œ #&'1
è