MATH 1C HOMEWORK 7.
1. Problem 7.5.4
Evaluate the integral
RR
S
(x + z)dS, where S is the part of the cylinder y 2 + z 2 = 4 with x ∈ [0, 5].
Proof. We parametrize our surface as ϕ(x, θ) = (x, 2 cos θ, 2 sin θ) for x ∈ [0, 5], θ ∈ [0, 2π]. Then we get,
∂ϕ
= (1, 0, 0),
∂x
. Hence dS = ||
∂ϕ
= (0, −2 sin θ, 2 cos θ),
∂θ
∂ϕ ∂ϕ
×
= (0, −2 cos θ, −2 sin θ)
∂x
∂θ
∂ϕ ∂ϕ
×
||dxdθ = 2dxdθ. Using these we compute the surface integral:
∂x
∂θ
RR
R 2π R 5
F dS = 0 0 (x + 2 sin θ) · 2dxdθ
S
R 2π R 5
= 0 0 2x + 4 sin θdxdθ
= 50π
2. Problem 7.5.7
Compute the integral
and x = y.
RR
S
xydS, where S is the surface of the tetrahedron with sides y = 0, z = 0, x+z = 1,
Proof. Let C1 be the face in y = 0, C2 in z = 0, C3 in x + z = 1 and C4 in x = y.
ZZ
xydS = 0
C1
since y = 0. Next C2 inside z = 0 is a triangular region in xy plane enclosed by x = 1, y = 0, x = y. Hence
ZZ
Z
1
Z
xydS =
C2
1
xydxdy =
0
y
1
8
.
√ We parametrize C3 by ϕ(t, y) = (t, y, 1 − t). Therefore the area form dS = ||(1, 0, −1) × (0, 1, 0)||dtdy =
2dtdy. Moreover if we consider x + z = 1 plane as ty plane, C3 is a triangular region enclosed by
t = 1, t = y, y = 0.
√
ZZ
Z 1Z 1 √
2
ty 2dtdy =
xydS =
8
C3
0
y
√ Then C4 can be parameterized by ϕ(t, z) = (t, t, z). dS = ||(1, 1, 0) × (0, 0, 1)||dtdz = ||(1, −1, 0)||dtdz =
2dtdz. C4 is a triangular region enclosed by t = 0, z = 0, t + z = 1 as a tz plane.
√
ZZ
Z 1 Z 1−z √
2
2
xydS =
t 2dtdz =
12
C4
0
0
Therefore the answer is
√
5 2+3
24 .
1
2
MATH 1C HOMEWORK 7.
3. Problem 7.5.14
Evaluate the surface integral
RR
S
z 2 dS, where S is the boundary of the cube C = [−1, 1] × [−1, 1] × [−1, 1].
Proof. The boundary of the cube C has six faces,
C1 = [−1, 1] × [−1, 1] × {−1}
C2 = [−1, 1] × [−1, 1] × {1}
C3 = {−1} × [−1, 1] × [−1, 1]
C4 = {1} × [−1, 1] × [−1, 1]
C5 = [−1, 1] × {−1} × [−1, 1]
C6 = [−1, 1] × {1} × [−1, 1]
Then,
RR
S
z 2 dS =
ZZ
P6
z 2 dS. Since z 2 = 1 on C1 and C2 ,
ZZ
ZZ
ZZ
2
2
z dS +
z dS =
1dS +
1dS = Area(C1 ) + Area(C2 ) = 8.
i=1
RR
Ci
C1
C2
C1
C2
On C3 , C4 , we parametrize the surface φ(y, z) = (±1, y, z). Therefore dS = dydz,
R1 R1
2 −1 −1 z 2 dydz = 38 .
RR
RR
Likewise, C5 z 2 dS + C6 z 2 dS = 83
RR 2
Therefore, S z dS = 40
3 .
RR
C3
z 2 dS +
RR
C4
z 2 dS =
4. Problem 7.5.15
Find the mass of a spherical surface S of radius R such that at each point (x, y, z) ∈ S the mass density
is equal to the distance of (x, y, z) to some fixed point (x0 , y0 , z0 ) ∈ S.
Proof. Since the sphere is symmetric, we can put (x0 , y0 , z0 ) = (0, 0, R). Moreover the sphere is parametrized
by Φ(φ, θ) = (R sin φ cos θ, R sin φ sin θ, R cos φ) for φ ∈ [0, π], θ ∈ [0, 2π]. Then the area form dS =
R2 sin φdφdθ.
p
2 + y 2 + (z − R)2 . i.e.
Now the mass density δ(x, y, z) = (xq
p
δ(R sin φ cos θ, R sin φ sin θ, R cos φ) = R2 sin2 φ + R2 (cos φ − 1)2 = 2R2 − 2R2 cos φ
Then the total mass =
ZZ
Z 2π Z π p
Z πp
√ 3
2
2
2
δdS =
2R − 2R cos φ · R sin φdφdθ = 2R · 2π ·
1 − cos φ sin φdφ
S
0
0
0
√
3
2
16 3
= 2 2πR3 · (1 − cos φ) 2 |π0 =
πR
3
3
5. Problem 7.6.2
Proof. We have,
∂Φ
∂Φ
∂Φ ∂Φ
= (2 cos u, −3 sin u, 0),
= (0, 0, v),
×
= (−3 sin u, −2 cos u, 0)
∂u
∂v
∂u
∂v
Using these we compute the surface integral:
R 1 R 2π
RR
F · dS = 0 0 (2 sin u, 3 cos u, v 2 ) · (−3 sin u, −2 cos u, 0)dudv
S
R 1 R 2π
= 0 0 −6dudv
= −12π
RR
Hence, S F · dS = −12π.
MATH 1C HOMEWORK 7.
3
6. Problem 7.6.8
Proof. We parametrize the surface S = (x, y, z)|x2 + z 2 = 1, 0 ≤ y ≤ 1, 0 ≤ x ≤ 1 by Φ(u, v) = (cos θ, y, sin θ)
for θ ∈ [− π2 , π2 ], y ∈ [0, 1] since 0 ≤ cos θ ≤ 1. Then we have,
∂Φ
∂Φ
∂Φ ∂Φ
= (− sin θ, 0, cos θ),
= (0, 1, 0),
×
= (cos θ, 0, sin θ)
∂θ
∂y
∂y
∂θ
. Moreover note tha
∂Φ ∂Φ
×
||dydθ = ndydθ
dS = ndS = n||
∂y
∂θ
Using these we compute the surface integral:
RR
R π R1 √
F · dS = −2 π 0 ( y, 0, 0) · (cos θ, 0, sin θ)dydθ
S
R π2
R1√
= −2 π cos θdθ · 0 ydy
2
= 2 · 23 = 43
RR
Hence, S F · dS = 43 .
7. Problem 7.6.10
Proof. First of all,
∇ × F = −2zj + (3y − 1)k
and dS = ndS. Then the upward pointing unit normal n = 41 (x, y, z) for (x, y, z) ∈ S.
ZZ
ZZ
ZZ
1
1
(∇ × F) · dS =
(0, −2z, 3y − 1) · (x, y, z)dS =
(yz − z)dS
4
4
S
S
S
We parametrize our surface as Φ(φ, θ) = (4 sin φ cos θ, 4 sin φ sin θ, 4 cos φ) for φ ∈ [0, π2 ], θ ∈ [0, 2π]. Then
the area form dS = 16 sin φdφdθ.
ZZ
π
2
Z
2π
Z
[256 sin2 φ sin θ cos φ − 64 cos φ sin φ]dθdφ
(yz − z)dS =
S
0
0
The first part
Z
0
π
2
Z
2π
256 sin2 φ sin θ cos φdθdφ = 0
0
R 2π
since 0 sin θdθ = 0.
Therefore
ZZ
π
2
Z
Z
0
S
2π
−64 cos φ sin φdθdφ
(yz − z)dS =
Z
0
π
2
sin 2φdφ = −64π
= −32 · 2π
0
Therefore the answer is −64π.