[Chapter16.7] Surface Integrals
2. x2+ y2= 1 , − 1 ≤ z ≤ 1 , f (±1, 0, 0) = 2 , f (0, ±1, 0) = 3 , f (0, 0, ±1) = 4
Each quarter-cylinder has surface areas 14[2π · 1 · 2] = π, and the top and bottom disks have surface area π.
Z Z
S
f (x, y, z)dS = π[f (−1, 0, 0) + f (1, 0, 0) + f (0, −1, 0) + f (0, 1, 0) + f (0, 0, −1) + f (0, 0, 1)] = 18π
10. RR
S
p1 + x2+ y2dS , r(u, v) = u cos v i + u sin v j + v k , 0 ≤ u ≤ 1 , 0 ≤ v ≤ π
ru= cos v i + sin v j, rv= −u sin v i + u cos v j + k
ru× rv= sin v i − cos v j + u k, |ru× rv| =√ 1 + u2
So, Z Z
S
p1 + x2+ y2dS = Z π
0
Z 1 0
p1 + u2p
1 + u2dudv = 4 3π
20. F(x, y, z) = y i + j + z2 k , r(u, v) = u cos v i + u sin v j + v k , 0 ≤ u ≤ 1 , 0 ≤ v ≤ π
F(r(u, v)) = u sin v i + u cos v j + v2 k
Z Z
S
F · dS = Z Z
D
F · (ru× rv)dA = Z π
0
Z 1 0
[u sin2v − u cos22v + uv2]dudv
= Z π
0
Z 1 0
[−u cos(2v) + uv2]dudv = Z π
0
[−1
2cos(2v) +1
2v2]dv = π3 6
1
29. F(x, y, z) = x2 i + y2 j + z2 k , 0 ≤ z ≤p
1 − y2 , 0 ≤ x ≤ 2 On S1: The top surface , z =p
1 − y2 for 0 ≤ x ≤ 2 , − 1 ≤ y ≤ 1 with upward orientation.
Z Z
S1
F · dS = Z 2
0
Z 1
−1
[−x2 (0) − y2 (− y
p1 − y2) + z2]dydx = Z 2
0
Z 1
−1
[ y3
p1 − y2 + 1 − y2]dydx
= Z 2
0
[−p
1 − y2+1
3(1 − y2)3/2+ y −1
3y3]y=1y=−1dx = Z 2
0
4
3 dx = 8 3
On S2: The bottom surface , z = 0 with downward orientation.
Z Z
S2
F · dS = Z 2
0
Z 1
−1
−z2dzdy = Z 2
0
Z 1
−1
0 dzdy = 0
On S3: the front half-disk in the plane x = 2, for −1 ≤ y ≤ 1 , 0 ≤ z ≤p
1 − y2,oriented in the positive x-direction.
Regarding y and z as parameters, we have ry× rz= i, and
Z Z
S3
F · dS = Z 1
−1
Z
√
1−y2 0
x2 dzdy = Z 1
−1
Z
√
1−y2 0
4 dzdy = 2π
On S4: the back half-disk in the plane x = 0 for −1 ≤ y ≤ 1 , 0 ≤ z ≤p
1 − y2,oriented in the negative x-direction.
Regarding y and z as parameters, we use −(ry× rz) = −i, and
Z Z
S4
F · dS = Z 1
−1
Z
√
1−y2
0
x2 dzdy = Z 1
−1
Z
√
1−y2
0
0 dzdy = 0
Thus, Z Z
S
F · dS = 8
3+ 0 + 2π + 0 = 2π + 8 3
38. r(x, y) = x i + y j +p
x2+ y2k , | rx× ry| = s
1 + x2+ y2 x2+ y2 =√
2
m = Z Z
S
[10 −p
x2+ y2]dS = Z Z
1≤x2+y2≤16
[10 −p
x2+ y2]√ 2 dA =
Z 2π 0
Z 4 1
√
2(10 − r)r drdθ = 108√ 2π
47. Let S be a sphere of radius a centered at the origin. Then |r| = a, and F(r) = cr/|r|3= (c/a3)[x i + y j + z k].
A parametric representation for S is r(φ, θ) = a sin φ cos i + a sin φ sin θ j + a cos φ k , 0 ≤ φ ≤ π , 0 ≤ θ ≤ 2π.
rφ= a cos φ cos θ i + a cos φ sin θ j − a sin φ k, rθ= −a sin φ sin θ i + a sin φ cos θ j
and the outward orientation is given by rφ× rθ= a2sin2φ cos θ i + a2sin2φ sin θ j + a2sin φ cos φ k.
Z Z
S
F · dS = Z π
0
Z 2π 0
F(r) · (rφ× rθ)dθdφ = c a3
Z π 0
Z 2π 0
a3(sin3φ + sin φ cos2φ) dθdφ = c Z π
0
Z 2π 0
sin φ dθdφ = 4πc
Thus the flux does not depend on the radius a.
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