MATH 101 HOMEWORK 9 - SOLUTIONS
1. Find the mass of the solid ball of radius 4 cm centered at the origin, if the density at a
point P is r 2 g/cm3 , where r is the distance from P to the x axis.
√
We use cylindrical shells. A shell of radius r has height h = 2 16 − r 2 , and the density
on it is r 2 . So the mass is
Z 4
Z 4
p
p
2
2
m=
r · 2πr · 2 16 − r dr =
4πr 3 16 − r 2 dr.
0
0
Let r = 4 sin θ, dr = 4 cos θ dθ, then
Z
Z
π/2
π/2
4π(4 sin θ) · 16 cos θ dθ = 4096π
3
m=
(1 − cos2 θ) cos2 θ sin θ dθ.
2
0
0
Let now u = cos θ, du = − sin θ dθ:
Z
1
u3 u5 1
2
≈ 1715.728
(u − u )du = 4096π( − ) = 4096π ·
3
5 0
15
2
m = 4096π
0
4
2. Find the mass and the centre of mass of the solid lying above the plane z = 0, inside
the sphere x2 + y 2 + z 2 = 100 cm2 , and outside the sphere x2 + y 2 + z 2 = 36 cm2 , if the
density at (x, y, z) is z 2 + 1 g/cm3 .
We first compute the area A(z) of a slice at height z. If 0 ≤ z ≤ 6, the area is
A(z) = πR2 − πr 2 = π(100 − z 2 ) − π(36 − z 2 ) = 64π.
If 6 ≤ z ≤ 10, the area is A(z) = πR2 = π(100 − z 2 ). The mass is
Z
Z
1
m=
(z + 1) · 64π dz +
0
6
1
10
(z 2 + 1) · π(100 − z 2 ) dz
2
0(z + 1)A(z) dz =
0
Z
6
2
6 Z 10
z3
= 64π( + z) +
π(100 + 99z 2 − z 4 ) dz
3
0
6
5 10
z = 4992π + π(100z + 33z 3 − ) = 4992π + 7827.2π = 12819.2π.
5 6
The z-moment is
Z 1
Z 6
Z 10
2
2
0(z + 1)A(z) dz =
z(z + 1) · 64π dz +
z(z 2 + 1) · π(100 − z 2 ) dz
0
0
= 64π(
z4
z 2 6
+ ) +
4
2 0
6
Z
10
π(100z + 99z 3 − z 5 ) dz
6
z 10
99z
− ) = 21888π + 88533.3π = 110421.4π.
4
6 6
Thus the z-coordinate of the center of mass is 110421.4π/m = 8.6138 cm.
= 21888π + +π(50z 2 +
4
6
3. Find the centroid of the region in Figure 7.41(b) in the textbook.
Let (x̄, ȳ) be the centroid in question. By symmetry, x̄ = ȳ.
We divide the region in two parts: the square with vertices (0, 1), (1, 0), (0, √
−1), (−1, −1),
and the triangle with vertices (0, 1), (1, 1), (1, 0). The area of the square is ( 2)2 = 2, and
its centroid is at 0. The triangle has area 1/2 and centroid at ((0+1+1)/3, (1+1+0)/3) =
(2/3, 2/3) (see the formula on page 439). To get x̄, we average them out:
2 · 0 + 12 ·
x̄ =
2 + 1/2
2
3
=
1/3
2
=
,
5/2
15
and as observed above, ȳ = x̄ = 2/15.
4. Suppose that a point P is chosen at random from the planar region 0 ≤ x ≤ π/3,
0 ≤ y ≤ sin x. We define the random variable X to be the x-coordinate of P .
(a) Find the probability density function f for X.
The probability density will be f (x) = c sin x, where c will be chosen so that
1. We have
Z π/3
1
1
π/3
sin x dx = − cos x|0 = − + 1 = ,
2
2
0
so f (x) = 2 sin x.
(b) Find the expectation of X.
Z
E(X) =
π/3
2x sin x dx =
π/3
−2x cos x|0
0
Z
π/3
+
2 cos x dx
0
√
1
3 √
π/3
= −2 · + 2 sin x|0 = −1 + 2 ·
= 3 − 1 ≈ 0.732.
2
2
2
R π/3
0
f (x)dx =