Math 212 Spring 2008: Solutions: HW #12
Instructor: S. Cautis
1. section 7.5 #2
The hardest part of this question is to figure out how to parametrize the
triangle S. The answer is that S is made up of all points of the form
a(1, 0, 0) + b(0, 2, 0) + c(0, 1, 1) = (a, 2b + c, c)
where a + b + c = 1 and 0 ≤ a, b, c ≤ 1. Thus replacing a by 1 − b − c we
see that the triangle is parametrized by
(1 − b − c, 2b + c, c)
where 0 ≤ b, c ≤ 1 and 0 ≤ a ≤ 1 which is equivalent to 0 ≤ 1 − b − c ≤ 1
or equivalently 0 ≤ b + c ≤ 1. Thus the region over which b, c vary is the
triangle in the b, c plane with vertices (0, 0), (0, 1), (1, 0).
Now Tb = (−1, 2, 0)√ and Tc = (−1, 1, 1) so that Tb × Tc = (2, 1, 1) and
hence ||Tb × Tc || = 6. Thus the integral we want is
Z Z
S
xyzdS
=
Z
0
1
Z
√ Z
=
6
1−c
0
1
0
√ Z
6
=
0
√
(1 − b − c)(2b + c)c 14dbdc
Z
1−c
Z
1−c
(2bc + c2 − 2b2 c − bc2 − 2bc2 − c3 )dbdc
0
1
0
(2bc + c2 − 2b2 c − 3bc2 − c3 )dbdc
√ Z 1 2
2
3
6
[b c + bc2 − b3 c − b2 c2 − bc3 ]1−c
=
0 dc
3
2
0
√ Z 1
2
3
6
[(1 − c)2 c + (1 − c)c2 − (1 − c)3 c − (1 − c)2 c2 − (1 − c)c3 ]dc
=
3
2
0
√ Z 1
3
2
6
=
c(1 − c)(1 − c + c − (1 − 2c + c2 ) − (c − c2 ) − c2 )dc
3
2
0
√ Z 1
1
1 1
=
6
c(1 − c)( − c − c2 )dc
3 6
6
0
√ Z 1
6
=
(2c − 3c3 + c4 )dc
6 0
√
6 2 3 4 1 51
[c − c + c ]0
=
6
4
5
√
6
3 1
=
[1 − + ]
6√
4 5
3 6
=
40
2. section 7.5, #8
There are six faces. The front face parametrized by (1, u, v) where −1 ≤
u, v ≤ 1 so that Tu = (0, 1, 0) and Tv = (0, 0, 1) and so the normal vector
is (1, 0, 0) which we could have guessed anyway. So the integral over this
face is
Z Z
1
1
−1
−1
v 2 dudv = 4/3
Similarly the integral over the back face is
Z 1Z 1
v 2 dudv = 4/3
−1
−1
The right face is parametrized by (u, 1, v) so the integral is
Z
1
Z
−1
1
v 2 dudv = 4/3
−1
and similarly for the left face.
The top face is parametrized by (u, v, 1) so the integral is
Z
1
−1
Z
1
1dudv = 4
−1
and similarly for the bottom face.
Adding these 6 we get
4
3
·4+4·2 =
40
3 .
3. section 7.5, #10
√
We can parametrize the hemisphere by (u, v, R2 − u2 − v 2 ) where 0 ≤
u2 + v 2 ≤ R2 . Even
√ better we can parametrize it in polar coordinates
as (r cos θ, r sin θ, R2 − r2 ) where 0 ≤ r ≤ R and 0 ≤ θ ≤ 2π. Then
Tr = (cos θ, sin θ, √R−r
) and Tθ = (−r sin θ, r cos θ, 0). Thus
2 −r 2
−r
−r
· r cos θ, √
Tr × Tθ = ( √
· r sin θ, r2 )
2
2
R2 − r 2
R −r
Hence
||Tr × Tθ || =
r
r2
r2
· r2 cos2 θ + 2
· r2 sin2 θ + r2 = r
2
2
R −r
R − r2
r
Thus the integral we want to calculate is
Z
0
R
Z
0
2π
R
r ·r· √
dθdr = 2Rπ
2
R − r2
2
Z
0
R
r3
√
dr
R2 − r 2
R2
− r2
R2
Now substituting r = R sin α so that dr = R cos αdα we get
Z π/2 3 3
Z R
Z π/2
R sin α
r3
√
dr =
(R cos αdα) = R3
sin3 αdα
R cos α
R2 − r 2
0
0
0
R π/2
Now 0 sin3 αdα = 2/3 where we used integration by parts (or looked
it up). Thus the final answer is 2Rπ · R3 · 2/3 = 4/3πR4 .
4. section 7.6, #4
The surface is the side of a cylinder which we can parametrize as (cos θ, t, sin θ)
where 0 ≤ t ≤ 1 and −π/2 ≤ θ ≤ π/2 (since x ≥ 0). Now Tθ =
(− sin θ, 0, cos θ) and Tt = (0, 1, 0) so that Tr × Tt heta = (cos θ, 0, sin θ).
Then the integral we want to calculate is
Z 1 Z π/2
Z 1 Z π/2
√
√
t cos θdθdt
( t, 0, 0) · (cos θ, 0, sin θ)dθdt =
0
0
−π/2
= 2
Z
1
−π/2
√
tdt
0
= 4/3
5. section 7.6, #6
√
The surface is parametrized by (u, v, 16 − u2 − v 2 ) where u2 + v 2 ≤ 16
(i.e. the disk D of radius 4). Meanwhile, a quick calculation gives that
∇ × F = (0, −2z, 3y − 1)
Now using the formula for the integral of a vector field over the graph of
a function (p.495) we get the integral
Z Z
Z Z
p
v
(−2 16 − u2 − v 2 ) √
v − 1dudv
+ 3v − 1dudv =
16 − u2 − v 2
D
D
Z 2π Z 4
=
(r sin θ − 1)rdrdθ
0
=
Z
0
2π
0
=
43 /3 sin θ − 8dθ
−16π
6. section 7.6, #10
The side of the cylinder is parametrized by (cos θ, sin θ, z) where 0 ≤ z ≤ 1
and 0 ≤ θ ≤ 2π. Then Tθ = (− sin θ, cos θ, 0) while Tz = (0, 0, 1) so that
Tθ × Tz = (cos θ, sin θ, 0) which is also the normal vector n.
Hence
Z Z
S
F · ndA =
Z
=
Z
2π
0
2π
0
Z
Z
1
(1, 1, z) · (cos θ, sin θ, 0)dzdθ
0
1
cos θ + sin θdzdθ = 0
0
Now the top of the cylinder is parametrized by (u, v, 1) and the integral
is over the unit disk D:
Z 2π Z 1
Z Z
r4 rdrdθ = 2π/6
(1, 1, (u2 + v 2 )2 ) · (0, 0, 1)dudv =
D
0
0
The bottom of the cylinder is parametrized by (u, v, 0) and the integral is
Z Z
(1, 1, 0) · (0, 0, −1)dudv = 0
D
So the total integral is equal to 0 + 2π/6 + 0 = π/3.
7. section 7.6, #16a
√
The cone is parametrized by (u, v, u2 + v 2 ). Using again the expression
for the integral of a vector field over the graph of a function we get
Z Z
−1dudv = −π
D
since F1 = F2 = 0.
8. section 7.6 #18
Here D denote
the unit disk. The hemisphere is the graph of the function
√
f (u, v) = 1 − u2 − v 2 over D.
a) Using again the expression for the integral of a vector field over a graph
we get
Z 1 Z 2π
Z Z
r2
u2 + v 2
√
√
rdθdr
dudv =
2
2
1 − r2
1−u −v
0
0
D
Z 1
r3
√
= 2π
dr
1 − r2
0
Z 1 p
r
= 2π
−r 1 − r2 + √
dr
1 − r2
0
= 2π[1/3(1 − r2 )3/2 − (1 − r2 )1/2 ]10
= 4π/3
b) Similarly we get
Z Z
D
2uv
√
dudv
1 − u2 − v 2
1
Z
2π
=
Z
1
Z
2π
=
Z
=
0
0
0
0
0
2r2 cos θ sin θ
√
rdθdr
1 − r2
r3 sin(2θ)
√
dθdr
1 − r2
c) In part (a) ∇ × F = (0, 0, 0) so the integral is zero while
Z
C
F · ds =
Z
0
2π
(cos θ, sin θ, 0) · (− sin θ, cos θ, 0)dθ = 0
In part (b) ∇ × F = (0, 0, 0) so the first integral is zero. The second
integral is
Z
0
2π
(sin θ, cos θ, 0) · (− sin θ, cos θ, 0)dθ =
Z
0
2π
− sin2 θ + cos2 θdθ = 0