MATH 101 HOMEWORK 9 – SOLUTIONS
1. Find the mass of a circular cone of base radius 10 cm and height 20 cm, if its density
at a point P is 1 + r/100, where r is the distance from P to the axis of the cone.
The density is constant on cylindrical shells. The area of a shell at distance r from the
axis is 2πrh, where h/(10 − r) = 20/10 = 2, i.e. h = 2(10 − r). Thus
Z
10
m=
0
r
)dr = 4π
2πr · 2(10 − r) · (1 +
100
= 4π(5r 2 −
Z
10
(10r −
0
r3
9r 2
−
)dr
10
100
r 4 10
3r 3
−
) = 700π ≈ 2199.11
10
400 0
2. Find the centroids of the following planar regions.
(a) The region enclosed between the parabola y = 3x − x2 and the line y = 2x − 6.
We first find the intersection points: 3x − x2 = 2x − 6, x2 − x − 6 = 0, x = −2, 3. Hence
the centroid has coordinates x̄ = Mx /m, ȳ = My /m, where
Z
Z
x2
125
x3 3
m=
(3x − x − (2x − 6))dx =
(6 + x − x )dx = 6x +
− =
,
2
3 −2
6
−2
−2
3
3
2
2
1
Z
Mx =
Z
3
(3x − x − (2x − 6))xdx =
3
2
−2
1
My =
2
Z
−2
(6x + x2 − x3 )dx = 3x +
x3
125
x4 3
− =
,
3
4 −2
12
Z
1 3
((3x − x ) − (2x − 6) )xdx =
(5x2 − 6x3 + x4 + 24x − 36)dx
2
−2
−2
3
1 5x3
625
3x4
x5
=
−
+
+ 12x2 − 36x = −
,
2 3
2
5
12
−2
3
2 2
2
so that
x̄ =
5
1
, ȳ = − .
2
2
(b) The region in Figure 7.41(d) in the textbook (page 442).
By symmetry, x̄ = 0. To find ȳ, we write
ȳ =
m1 ȳ1 + m2 ȳ2 + m3 ȳ3
,
m1 + m2 + m3
where m1 , ȳ1 denote the area of the large half-disc and the y-coordinate of its centroid, and
m2 , m3 , ȳ2 , ȳ3 are the areas and y-coordinates of the centroids of the two small half-discs.
We have m1 = π/2, m2 = m + 3 = π/8, and from the example in class, y1 = 4/3π, y2 =
y3 = −2/3π. Hence
π
2
· 4 + 2 · π8 · 3π
2
ȳ = 2 3ππ
=
.
π
3π
2 +2· 8
3. A random variable X has probability density f (x) = Cx2 on [0, 1], for some value of C.
Find this C. For this C, find the mean value of X.
Z
1
1
x3
C
= 1, so that C = 3, f (x) = 3x2 . Then
Cx dx = C
=
30
3
2
We want
0
Z
1
x · 3x2 dx =
E(X) =
0
2
3x4 1
3
= .
4 0
4