9.14 Stokes’ Theorem
i S : a piecewise smooth orientable surface
i C : a piecewise smooth simple closed curve that
bounds S , oriented counter-clockwise as veiwed from above
i n : the unit normal to S , defines orientation of S
i F : F = P ( x, y, z ) , Q ( x, y, z ) , R ( x, y, z ) is a vector field
with P, Q, R, and all first partial derivatives continuous in
a region of 3-space containing S
F
⋅
d
r
=
curl
F
⋅
n
dS
(
)
∫∫
∫
C
S
3 − dim line ⇔ surface
integral
integral
Use Stokes' Theorem to evaluate
∫∫ ( curl F ) ⋅ n dS
S
By finding
Evaluate
∫ F ⋅ dr = ∫∫ ( curl F ) ⋅ n dS
C
S
Use Stokes' Theorem to evaluate ∫ F ⋅ dr
C
By finding
Evaluate
∫ F ⋅ dr = ∫∫ ( curl F ) ⋅ n dS
C
S
Example 1 Use Stokes' Theorem to evaluate
∫∫ ( curl F ) ⋅ n dS
when F = y 2 z , xz , x 2 y 2 and S is
C
S
that part of the paraboloid z = x 2 + y 2 that lies in the
cylider x 2 + y 2 = 1, oriented upward.
⇒ Find
x = cos t 

y = sin t  0 ≤ t ≤ 2π
z = 1 
∫ F ⋅ dr
C
Parametrize C :
r = cos t ,sin t ,1 ⇒ dr = − sin t , cos t , 0 dt
F on C : sin 2 t , cos t , cos 2 t sin 2 t
F ⋅ dr = ( − sin t + cos t ) dt
3
⇒ ∫ F ⋅ dr =
C
2π
=
∫
0
f ( x ) = − sin 3 x
2
2π
∫ ( − sin
3
t + cos t ) dt
2
0
1
2
cos tdt =
2
2π
∫ (1 + cos ( 2t ) ) dt = π
0
f ( x ) = cos ( 2 x )
Example 2 Use Stokes' Theorem to evaluate
∫ F ⋅ dr
⇒ Find
when F = z 2 , y 2 , xy and C is the
∫∫ ( curl F ) ⋅ n dS
S
C
C
triangle defined by (1,0,0 ) , ( 0,1, 0 ) , and ( 0,0,2 ) .
S
S : plane, we need to find the equation
using a point and the normal vector to the plane
We can get the normal vector by
taking the cross product of two vectors
in the plane.
S : 2 x + 2 y + z = d plug in any point to find d
Vector from (1, 0, 0 ) to ( 0,1, 0 )
(1, 0, 0 ) → 2 (1) + 2 ( 0 ) + ( 0 ) = d = 2
v1 = 0 − 1,1 − 0, 0 − 0 = −1,1, 0
⇒ 2 x + 2 y + z = 2 so S : z = 2 − 2 x − 2 y
Vector from (1, 0, 0 ) to ( 0, 0, 2 )
g = 2x + 2 y + z − 2 = 0
v2 = 0 − 1, 0 − 0, 2 − 0 = −1, 0, 2
i j k
v = v1 × v2 = −1 1 0 = 2, 2,1
−1 0 2
For a plane we could have used v
and made it a unit vector to get n
∇g = 2, 2,1
n=
∇g = 4 + 4 + 1 = 3
∇g 1
= 2, 2,1
∇g
3
2
dS = 1 + ( z x ) + ( z y ) dA
2
2
2
dS = 1 + ( −2 ) + ( −2 ) dA
dS = 3dA
Example 2
( cont.)
F = z 2 , y 2 , xy
( curl F ) =
curl F =
x, 2 z − y , 0
i
j
k
∂
∂x
∂
∂y
∂
∂z
z2
y2
xy
n=
1
2, 2,1
3
= x, 2 z − y , 0
y = 1− x
dS = 3dA
∫∫ ( curl F ) ⋅ n dS = ∫∫ ( 2 x + 4 z − 2 y ) dA
S
R
1 1− x
=∫
∫ 2 x + 4 ( 2 − 2 x − 2 y ) − 2 y  dydx
0 0
1 1− x
=∫
1
∫ [8 − 6 x − 10 y ] dydx = ∫ (8 − 6 x ) y − 5 y
0 0
1
2 1− x
 dx
0
0
1
= ∫ ( 8 − 6 x )(1 − x ) − 5 (1 − x )  dx = ∫ ( x 2 − 4 x + 3) dx


2
0
=
(
3
x
3
0
2
− 2 x + 3x
1
)
0
= −2+3
1
3
=
4
3