[16 .8] Stokes’ Theorem
2. The boundary curve C is given by x2+y2= 9, z = 0 ,and oriented in the counterclockwise direction when viewed from above.
A vector equation of C isγ(t) = 3 cos t i + 3 sin t j, 0 ≤ t ≤ 2π
∴ γ′(t)= −3 sin t i + 3 cos t j
Also, we have F(γ(t)) = 6 sin t i + (3 cos t)e3 sin t k Thus, by Stokes’ Theorem,
"
S
curl F· dS =
∫
C
F· dr =
∫ 2π
0
F(γ(t)) · γ′(t)dt
=
∫ 2π
0
−18 sin2tdt
= −18 [1
2t− 1 4sin 2t
]2π
0
= −18π
3. To find the boundary curve C we solve the equations z = x2+ y2 and x2+ y2 = 4. Thus, C is the circle given by the equations x2+ y2 = 4, z = 4 and oriented in the counterclockwise direction when viewed from above.
A vector equation of C isγ(t) = 2 cos t i + 2 sin t j + 4 k, 0 ≤ t ≤ 2π
∴ γ′(t)= −2 sin t i + 2 cos t j
Also, we have F(γ(t)) = 64 cos2t i+ 64 sin2t j+ 16 sin t cos t k Therefore, by Stokes’ Theorem,
"
S
curl F· dS =
∫
C
F· dr =
∫ 2π
0
F(γ(t)) · γ′(t)dt
=
∫ 2π
0
(−128 cos2t sin t+ 128 sin2t cos t)dt
= 128 [1
3cos3t+ 1 3sin3t
]2π
0
= 0
6. The boundary curve C is the circle x2+z2 = 1, y = 0 ,and oriented in the counterclockwise direction when viewed from the right.
So a vector equation of C is
γ(t) = cos(−t) i + sin(−t) k = cos t i − sin t k, 0 ≤ t ≤ 2π
∴ γ′(t)= − sin t i − cos t k
Then F(γ(t)) = e0i+ e− cos t sin t j− cos2t sin t k Thus, by Stokes’ Theorem,
"
S
curl F· dS =
∫
C
F· dr =
∫ 2π
0
F(γ(t)) · γ′(t)dt
=
∫ 2π
0
(− sin t + cos3t sin t)dt
= [
cos t− 1 4cos4t
]2π
0
= 0
1
7. Choose the surface S to be the planar region enclosed by C, so S is the portion of the plane x+ y + z = 1 over D = {(x, y)|0 ≤ x ≤ 1, 0 ≤ y ≤ 1 − x}. Since C is oriented counterclockwise, we orient S upward.
∴ curl F = −2z i−2x j−2y k, γx×γy = (−∂z∂x, −∂y∂z, 1) = (1, 1, 1) ⇒ curl F·(γx×γy)= −2x−2y−2z Thus, by Stokes’ Theorem,
∫
C
F· dr =
"
S
curl F· dS
=
"
D
(−2x − 2y − 2z)dA =
∫ 1
0
∫ 1−x
0
(−2)dydx = 2
∫ 1
0
(1− x)dx = −1
9. Choose the surface S to be the disk x2+y2 ≤ 16, z = 5. Since C is oriented counterclockwise when viewed from above, we orient S upward. Thus,we take n= k.
curl F = (xexy− 2x) i − (yexy− y) j + (2z − z) k (curl F)· n = 2z − z = z = 5 on S
Thus, by Stokes’ Theorem,
∫
C
F· dr =
"
S
curl F· dS
=
"
S
5dS = 5(Area of S ) = 5 · π · 42= 80π
13. S is oriented downward, so the boundary curve C is the circle x2+ y2 = 16, z = 4, oriented in the clockwise direction as viewed from above.
A vector equation of C isγ(t) = 4 cos t i − 4 sin t j + 4 k, 0 ≤ t ≤ 2π
∴ γ′(t)= −4 sin t i − 4 cos t j
Also, we have F(γ(t)) = 4 sin t i + 4 cos t j − 2 k Thus,
∫
C
F· dr =
∫ 2π
0
F(γ(t)) · γ′(t)dt
=
∫ 2π
0
(−16 sin2t− 16 cos2t)dt=
∫ 2π
0
(−16)dt = −32π curl F = 2 k
γx× γy = (√−x
x2+y2, √−y
x2+y2, 1) which is directed upward.
But S is oriented downward,
∴
"
S
curl F· dS = −
"
D
curl F· (γx× γy)dA
= −
"
D
2dA= −2π · 42 = −32π
2
14. The boundary curve C is x2 + y2 = 4, z = 1, oriented in the counterclockwise direction when viewed from above.
A vector equation of C isγ(t) = 2 cos t i + 2 sin t j + 1 k, 0 ≤ t ≤ 2π
∴ γ′(t)= −2 sin t i + 2 cos t j
Also, we have F(γ(t)) = −4 sin t i + 2 sin t j + 6 cos t k Thus,
∫
C
F· dr =
∫ 2π
0
F(γ(t)) · γ′(t)dt
=
∫ 2π
0
(8 sin2t+ 4 cos t sin t)dt = [
8 (1
2t− 1 4sin 2t
)
+ 2 sin2t ]2π
0
= 8π
curl F = (−3 − 2y) j + 2z k and D ={
(x, y)|x2+ y2≤ 4} z= g(x, y) = 5 − x2− y2 ∴ γx× γy = (2x, 2y, 1)
∴
"
S
curl F· dS =
"
D
curl F· (γx× γy)dA
=
"
D
[2y(−3 − 2y) + 2z]dA
=
"
D
[−6y − 4y2+ 2(5 − x2− y2)]dA= · · · = 8π
15. The boundary curve C is x2+ z2 = 1, y = 0 ,and oriented in the counterclockwise direction when viewed from the right.
A vector equation of C is
γ(t) = cos(−t) i + sin(−t) k = cos t i − sin t k, 0 ≤ t ≤ 2π
∴ γ′(t)= − sin t i − cos t k Then F(γ(t)) = − sin t i + cos t k Thus,
∫
C
F· dr =
∫ 2π
0
F(γ(t)) · γ′(t)dt
=
∫ 2π
0
(− cos2t)dt= − [1
2t+ 1 4sin 2t
]2π
0
= −π curl F = − i − j − k and S is parametrized by
γ(ϕ, θ) = sin ϕ cos θ i + sin ϕ sin θ j + cos ϕ k, 0 ≤ ϕ ≤ π, 0 ≤ θ ≤ π γϕ× γθ = sin2ϕ cos θ i + sin2ϕ sin θ j + sin ϕ cos ϕ k
∴
"
S
curl F· dS =
"
x2+z2≤1
curl F· (γϕ× γθ)dA
=
∫ π
0
∫ π
0
(− sin2ϕ cos θ − sin2ϕ sin θ − sin ϕ cos ϕ)dϕdθ
=
∫ π
0
(−2 sin2ϕ − π sin ϕ cos ϕ)dϕ
= [1
2sin 2ϕ − ϕ − π 2sin2ϕ
]π
0
= −π
3
17. All of the points (0, 0, 0), (1, 0, 0), (1, 2, 1), (0, 2, 1) lie on the plane z = 12y
∴Let S be the planar region enclosed by the path of the particle, so S is the portion of the plane z= 12y for 0≤ x ≤ 1, 0 ≤ y ≤ 2 with upward orientation.
curl F = 8y i + 2z j + 2y k
Use Stokes’ Theorem and the work is
∫
C
F· dr =
"
S
curl F· dS
=
"
D
(2y− 1
2y)dA=
∫ 1
0
∫ 2
0
(3
2y)dydx=
∫ 1
0
3dx = 3
18. C lies on the surface z= 2xy, so let S be the part of this surface that bounded by C.
C is traversed clockwise when viewed from above, so S is oriented downward.
Let F(x, y, z) = (y+sin x) i+(z2+cos y) j+ x3k∴ ∫
C
(y+sin x)dx+(z2+cos y)dy+ x3dz= ∫
C
F·dr Thus, curl F = −2z i − 3x2j− k, γx× γy = (−2y, −2x, 1)
Let D= {
(x, y)|x2+ y2≤ 1}
∴
∫
C
F· dr = −
"
S
curl F· dS = −
"
D
curl F· (γx× γy)dA
= −
"
D
(8xy2+ 6x3− 1)dA
= −
∫ 2π
0
∫ 1
0
(8r3cosθ sin2θ + 6r3cos3θ − 1)rdrdθ = π
4
