Answer Key for Quiz 7 (section A)
1. Since 0 = x(1 − x) implies that either x = 0 or x = 1, the curve y = x(1 − x) intersects the x-axis at (0, 0)
and (1, 0), and is above the x-axis between these points:
A vertical strip of this region at a point x generates a circular disk of radius x(1 − x) when revolved around
the x-axis. Since the area of such a disk is πx2 (1 − x)2 , the volume we get is obtained by multiplying by dx
and integrating over the possible values of x, namely 0 ≤ x ≤ 1. This gives
Z 1
V =
πx2 (1 − x)2 dx.
0
No substitution helps this much, so we just multiply it out:
Z 1
¡
¢
V =π
x2 1 − 2x + x2 dx
Z
0
1
=π
¡
¢
x2 − 2x3 + x4 dx
0
¶¯1
x4
x5 ¯¯
x3
−2
+
=π
3
4
5 ¯0
¶
µ
π
π
1 1 1
− +
=
(10 − 15 + 6) =
.
=π
3 2 5
30
30
µ
To do this with cylindrical shells is possible, but more difficult. A horizontal strip of the region generates a
hollow cylinder of radius y and some height when revolved around the x-axis, and to get the height takes
some work. We have
r
µ
¶2
µ
¶2
1
1
1
1
1
1
2
y =x−x = − x−
so
x−
= − y and therefore x = ±
− y.
4
2
2
4
2
4
The horizontal strips go from the left half of the parabola to the right half, so the one at a particular y has
height
Ã
!
r
r
r
1
1
1
1
1
h= +
−y−
−
−y =2
− y.
2
4
2
4
4
Then the cylindrical shells have area
à r
2πrh = 2πy 2
!
1
−y .
4
Multiplying by dy and integrating over the possible y’s we get
Z 14 r
1
y
− y dy.
V = 4π
4
0
To do this it helps to substitute u = 14 − y, so that du = −dy and therefore dy = −du. If y = 0 then
u = 41 = 0 = 14 , and if y = 14 then u = 41 − 14 = 0. Also, if u = 14 − y then y = 14 − u, so the integral becomes
¶
¶
Z 0µ
Z 14 µ
√
1
1
1
V = 4π
−u
u (−du) = 4π
− u u 2 du
1
4
4
0
4
µ
¶¯ 1
Z 41 ³
´
1
3
2 3
8 5 ¯¯ 4
2
2
2
2
u − 4u du = π
=π
u − u ¯
3
5
0
0
µ
¶
µ
¶
2 1 8 1
1
1
π
π
=π
−
=π
−
=
(5 − 3) =
.
3 8 5 32
12 20
60
30
2. The curves y = x1/4 and y = x4 intersect at (0, 0) and (1, 1), so the region of interest looks like
A vertical strip at a given x generates a washer with outer radius x1/4 and inner radius x4 when revolved
around the x-axis. The area of such a washer is
³ 1 ´2
´
³ 1
¡ ¢2
π x 4 − π x4 = π x 2 − x8 .
Multiplying by dx and integrating over all the possible values of x we get
Z 1³
´
1
V =π
x 2 − x8 dx
0
µ
¶¯1
2 3
1 9 ¯¯
=π
x2 − x ¯
3
9
0
¶
µ
5π
2 1
=π
−
=
.
3 9
9
This also works about as well with cylindrical shells. For a given y, a horizontal strip generates a hollow
1/4
4
1/4
cylinder with radius y and height
then x = y 4 and vice versa. The surface area
³ 1 y ´− y , since if y = x
of such a shell is 2πrh = 2πy y 4 − y 4 . Multiplying by dy and integrating over all the possible values of y
we get
Z 1 ³
´
1
V = 2π
y y 4 − y 4 dy
Z
0
1
= 2π
0
µ
³
´
5
y 4 − y 5 dy
¶¯1
¯
4 9
1
y 4 − y 6 ¯¯
9
6
¶
µ0
¶
µ
8
3
5π
4 1
−
= 2π
−
=
.
= 2π
9 6
18 18
9
= 2π