Math 217: Some Assignment 7 Solutions
15.3 # 52: The region of integration in the xy-plane is (sketch it)
{(x, y) | 0 ≤ x ≤ 1, x ≤ y ≤ 1} = {(x, y) | 0 ≤ y ≤ 1, 0 ≤ x ≤ y}
so
1
Z
Z
1
e
0
x/y
1
Z
y
Z
dydx =
e
x
0
x/y
1
Z
dxdy =
0
Z
x=y
yex/y |x=0
dy
1
y(e − 1)dy =
=
0
0
1
(e − 1)
2
# 56:
Z Z
Z
0
Z
y−y 3
ydA =
y dx dy +
−1
D
Z
−1
0
0
Z
−1
Z
√
y dx dy
y−1
√
y + 1)dy
0
0
2
1
Z
4
(y 2 − y 4 − y 3/2 + y)dy
(y − y + y)dy +
=
y−y 3
y(y − y 3 −
y(y − y + 1)dy +
=
Z
1
Z
3
1
−1
0
= (y 3 /3 − y 5 /5 + y 2 /2)|0−1 + (y 3 /3 − y 5 /5 − 2y 5/2 /5 + y 2 /2)10
= 1/3 − 1/5 − 1/2 + 1/3 − 1/5 − 2/5 + 1/2 = −2/15
15.4 # 14,
# 28: (a) Draw a picture! We can describe the upper-half of the solid in question in
cylindrical (polar plus z) coordinates as
q
{(r, θ, z) | 0 ≤ θ ≤ 2π, r1 ≤ r ≤ r2 , 0 ≤ z ≤ r22 − r2 }
so the volume is
Z
2π
V =2
Z
r2
dθ
0
r1
q
r22 − r2 rdr = 4π(−1/3)[r22 − r2 ]3/2 |rr21 = (4/3)π[r22 − r12 ]3/2 .
p
p
(b) The cylindrical hole intersects thepsphere when r1 = r = r22 − z 2 so z = r22 − r12 .
Thus the height of the ring is H := 2 r22 − r12 , and V = πH 3 /6.
15.5 # 10: the mass is
√
Z Z
m=
D
Z
1
√
Z
xdA =
0
x2
x
√
Z
xdydx =
0
1
1
√ √
x2
2x7/2 1
3
x( x − x2 )dx =
−
|0 =
,
2
7
14
and similarly,
Z Z
My =
x
3/2
1
Z
dA =
D
0
√
2x9/2 1
x3
1
−
|0 = ,
x3/2 ( x − x2 )dx =
3
9
9
while
√
Z Z
Z
1
√
Z
x
y xdA =
Mx =
√
Z
y xdydx =
D
0
x2
0
1
√ y 2 y=√x
x |y=x2 dx =
2
Z
0
1
√
x
(x − x4 )dx
2
x5/2
x11/2 1
6
−
|0 =
.
5
11
55
=
So the centre of mass is
(x̄, ȳ) = (My /m, Mx /m) = (
14 28
, ).
27 55
15.7 # 42: First, draw a picture! By symmetry, we must have x̄ = z̄. First, the mass
Z Z Z
Z 1
Z 1−y
Z 1−y−x
m=
ydV =
ydy
dx
dz
E
Z
1
=
=
0
0
0
1
Z
x=1−y
y[(1 − y)x − x2 /2]|x=0
dy
(1 − y − x)dx =
ydy
0
Z
0
1−y
Z
0
1
y[(1 − y)2 /2]dy = (1/2)
Z
0
1
[y 3 − 2y 2 + y]dy = (1/2)[1/4 − 2/3 + 1/2] = 1/24
0
(this is just one possible way to set up the integral). The integral for Mxz is the same, but
with y replaced by y 2 , so
Z 1
Mxz = (1/2)
[y 4 − 2y 3 + y 2 ]dy = (1/2)[1/5 − 1/2 + 1/3] = 1/60.
0
Thus ȳ = (1/60)/(1/24) = 2/5. Now
Z 1
Z 1−y
Z
Mxy = Myz =
ydy
xdx
0
Z
0
1−y−x
Z
1
dz =
0
ydy
0
1
2
ydy[(1 − y)x /2 − x
=
Z
3
dy
/3]|x=1−y
x=0
1−y
[x(1 − y) − x2 ]dx
0
Z
=
0
1
y[(1 − y)3 /6]dy
0
Z
= (1/6)
1
[−y 4 + 3y 3 − 3y 2 + y]dy = (1/6)[−1/5 + 3/4 − 1 + 1/2]
0
= (1/6)(1/20) = 1/120,
so x̄ = z̄ = (1/120)/(1/24) = 1/5. Thus the centre of mass is (1/5, 2/5, 1/5).
Nov. 5
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