[16 .7] Surface Integrals
3. Use two planes x= 0 and z = 0 to divide H into four patches of equal size, each with surface area equal to 18 the surface area of a sphere with radius √
50. So∆S = 18 · 4π(√
50)2 = 25π. Choose P∗i j = (±3, ±4, 5) for the four patches. Then
"
H
f (x, y, z)dS ≈ f (3, 4, 5)∆S + f (3, −4, 5)∆S + f (−3, 4, 5)∆S + f (−3, −4, 5)∆S
= (7 + 8 + 9 + 12) × 25π = 900π
8. γ(u, v) = ⟨2uv, u2− v2, u2+ v2⟩, D = {(u, v)|u2+ v2 ≤ 1}
γu= ⟨2v, 2u, 2u⟩, γv = ⟨2u, −2v, 2v⟩
⇒ γu× γv = ⟨8uv, 4u2− 4v2, −4v2− 4u2⟩ ⇒ |γu× γv| = 4√
2(u2+ v2)
∴
"
S
(x2+ y2)dS =
"
D
[(2uv)2+ (u2− v2)2]|γu× γv|dA
=
"
D
(u2+ v2)2· 4√
2(u2+ v2)dA= 4√ 2
"
D
(u2+ v2)3dA
= 4√ 2
∫ 2π 0
∫ 1 0
r3· rdrdθ = √ 2π
15. Parametrize S byγ(x, z) = ⟨x, x2+ z2, z⟩, D = {(x, z)|x2+ z2≤ 4}
γx = ⟨1, 2x, 0⟩, γz= ⟨0, 2z, 1⟩
⇒ γx× γz = ⟨2x, −1, 2z⟩ ⇒ |γx× γz| = √
1+ 4(x2+ z2)
∴
"
S
ydS =
"
D
(x2+ z2)√
1+ 4(x2+ z2)dA
=
∫ 2π
0
∫ 2
0
r2√
1+ 4r2rdrdθ =
∫ 2π
0
dθ
∫ 2
0
r2√
1+ 4r2rdr
= 2π
∫ 17 1
1
4(u− 1)√ u· 1
8du [u= 1 + 4r2 ⇒ rdr = 1 8du]
= 1
16π
∫ 17
1
(u32 − u12)= π 16
[2 5u52 − 2
3u32 ]17
1
= π
60(391√
17+ 1)
20. Let S1be the lateral surface, S2the top disk, and S3 the bottom disk.
On S1 :
γ(θ, z) = ⟨3 cos θ, 3 sin θ, z⟩, 0 ≤ θ ≤ 2π, 0 ≤ z ≤ 2 γθ = ⟨−3 sin θ, 3 cos θ, 0⟩, γz= ⟨0, 0, 1⟩
⇒ γθ× γz= ⟨3 cos θ, 3 sin θ, 0⟩ ⇒ |γθ× γz| = 3
∴ !
S1
(x2+ y2+ z2)dS =∫2π
0
∫2
0 3(9+ z2)dzdθ = 124π On S2 :
γ(r, θ) = ⟨r cos θ, r sin θ, 2⟩, 0 ≤ r ≤ 3, 0 ≤ θ ≤ 2π
⇒ |γr× γθ| = r
∴ !
S2
(x2+ y2+ z2)dS =∫2π
0
∫3
0(r2+ 4)rdrdθ = 1532 π
1
On S3 :
γ(r, θ) = ⟨r cos θ, r sin θ, 0⟩, 0 ≤ r ≤ 3, 0 ≤ θ ≤ 2π
⇒ |γr× γθ| = r
∴ !
S3
(x2+ y2+ z2)dS =∫2π
0
∫3
0(r2)rdrdθ = 812π Thus,!
S
(x2+ y2+ z2)dS = 124π + 1532 π +812π = 241π 23. F(x, y, z) = xy i + yz j + zx k
z= 4 − x2− y2,D= {(x, y)|0 ≤ x ≤ 1, 0 ≤ y ≤ 1} ∴ γx× γy = −∂x∂z i− ∂y∂z j+ k = 2x i + 2y j + k Thus,
"
S
F· dS =
"
D
F· (γx× γy)dA
=
∫ 1
0
∫ 1
0
[2x2y+ 2y2(4− x2− y2)+ x(4 − x2− y2)]dydx= · · · = 713 180 25. F(x, y, z) = x i − z j + y k
z= √
4− x2− y2,D= {(x, y)|0 ≤ x ≤ 2, 0 ≤ y ≤ √ 4− x2}
∴ γx× γy = −∂x∂z i− ∂y∂z j+ k = √ x
4−x2−y2 i+ √ y
4−x2−y2 j+ k Since S is oriented downward,
∴
"
S
F· dS = −
"
D
F· (γx× γy)dA
= −
"
D
( x2
√4− x2− y2 − zy
√4− x2− y2 + y)dA = −
"
D
x2
√4− x2− y2dA
= −
∫ π2
0
∫ 2 0
r2cos2θ
√4− r2rdrdθ = −
∫ π2
0
cos2θdθ
∫ 2 0
r3
√4− r2dr
= −
∫ π2
0
(1 2 + 1
2cos 2θ)dθ
∫ 0 4
−1
2(4− u)u−12du [u= 4 − r2 ⇒ −1
2du= rdr]
= −1 2
[1 2θ + 1
4sin 2θ ]π2
0
· (−1 2)
[ 8√
u− 2 3u32
]0 4
= −4 3π
29. Let S1, S2, S3, S4, S5,and S6be the faces of the cube in the plane x = 1, y = 1, z = 1, x = −1, y =
−1, and z = −1 respectively.
On S1 : F= i + 2y j + 3z k, γy× γz= i ∴!
S1
F· dS =∫1
−1
∫1
−1dydz= 4 On S2 : F= x i + 2 j + 3z k, γz× γx = j ∴!
S2
F· dS = ∫1
−1
∫1
−1dxdz= 8 On S3 : F= x i + 2y j + 3 k, γx× γt = k ∴!
S3
F· dS = ∫1
−1
∫1
−1dxdy= 12 On S4 : F= −1 i + 2y j + 3z k, γz× γy = −i ∴!
S4
F· dS = 4 On S5 : F= x i − 2 j + 3z k, γx× γz = −j ∴!
S5
F· dS = 8 On S6 : F= x i + 2y j − 3 k, γy× γx = −k ∴!
S6
F· dS = 12 Thus,!
S
F· dS = ∑6
i=1
!
Si
F· dS = 48
2
31. Let S1be the top surface, S2the bottom surface, S3 the front half-disk in the plane x= 2,and S4
the back half-disk in the plane x= 0.
On S1 :
The surface is z= √
1− y2for 0 ≤ x ≤ 2, −1 ≤ y ≤ 1 with upward orientation.
∴ !
S1
F· dS =∫2
0
∫1
−1[−x2· (0) − y2(−√y
1−y2)+ z2]dydx= · · · = 83 On S2 :
The surface is z= 0 for 0 ≤ x ≤ 2, −1 ≤ y ≤ 1 with downward orientation.
∴ !
S2
F· dS =∫2
0
∫1
−1(−z2)dydx=∫2
0
∫1
−1(0)dydx= 0 On S3 :
The surface is x= 2 for −1 ≤ y ≤ 1 , 0 ≤ z ≤ √
1− y2oriented in the positive x-direction.
Use y and z as parameters, soγy× γz= i
∴ !
S3
F· dS =∫1
−1
∫ √1−y2
0 x2dzdy=∫1
−1
∫ √1−y2
0 4dzdy= 4A(S3)= 2π On S4 :
The surface is x= 0 for −1 ≤ y ≤ 1 , 0 ≤ z ≤ √
1− y2oriented in the negative x-direction.
Regarding y and z as parameters, and use−(γy× γz)= −i
∴ !
S4
F· dS =∫1
−1
∫ √1−y2
0 x2dzdy=∫1
−1
∫ √1−y2
0 (0)dzdy= 0 Thus,!
S
F· dS = 2π + 83
32. Let S1be the triangular face with vertices (1,0,0),(0,1,0),(0,0,1), S2the face of the tetrahedron in the xy-plane, S3the face in the xz-plane , and S4the face o in the yz-plane.
On S1 :
The surface is z= 1 − x − y for 0 ≤ x ≤ 1, 0 ≤ y ≤ 1 − x with upward orientation.
∴ !
S1
F· dS =∫1
0
∫1−x
0 [−y(−1) − (z − y)(−1) + x]dydx = · · · = 13 On S2 :
The surface is z= 0 for 0 ≤ x ≤ 1, 0 ≤ y ≤ 1 − x with downward orientation.
∴ !
S2
F· dS =∫1
0
∫1−x
0 (−x)dydx = −16 On S3 :
The surface is y= 0 for 0 ≤ x ≤ 1, 0 ≤ z ≤ 1 − x oriented in the negative y-direction.
Regarding x and z as parameters, and useγx× γz= −j
∴ !
S3
F· dS =∫1
0
∫1−x
0 −(z − y)dzdx = · · · = −16 On S4 :
The surface is x= 0 for 0 ≤ y ≤ 1, 0 ≤ z ≤ 1 − y oriented in the negative x-direction.
Regarding y and z as parameters, and use−(γy× γz)= −i
∴ !
S4
F· dS =∫1 0
∫1−y
0 (−y)dzdy = · · · = −16 Thus,!
S
F· dS = 13 − 16 − 16 − 16 = −16
3
40. S is given byγ(x, y) = x i + y j + √
x2+ y2k, D= {(x, y)|1 ≤ x2+ y2 ≤ 16}
|γx× γy| = √
1+ (√x
x2+y2)2+ (√y
x2+y2)2= √ 2
∴ m =
"
S
(10− z)dS
=
"
S
(10− √
x2+ y2)dS
=
"
D
(10− √
x2+ y2)√ 2dA
= √
2
∫ 2π 0
∫ 4 1
(10− r)rdrdθ = 108√ 2π
43. The rate of the flow through the cylinder is the flux!
S
ρv · ndS =!
S
ρv · dS
S is given by γ(u, v) = 2 cos u i + 2 sin u j + v k , D = {(u, v)|0 ≤ u ≤ 2π, 0 ≤ v ≤ 1}
⇒ γu× γv = 2 cos u i + 2 sin u j
v· (γu× γv)= v(2 cos u) + 4 sin2u(2 sin u) Thus, the rate of the flow is
"
S
ρv · ndS = ρ
"
D
v· γu× γvdvdu
= 870
∫ 2π
0
∫ 1
0
(2v cos u+ 8 sin3u)dvdu= · · · = 0
4
