February 2, 2006
Lecturer: Dr Martin Kurth
Hilary Term 2006
Problem Sheet 3 – Solutions
1. Here you first have to find a function to revolve about the x-axis or the
y-axis to give the described column. For example, you can revolve
1
x + 1,
4
R(x) =
0≤x≤4
about the x-axis.
V
=
Z
4
Z
4
π[R(x)]2 dx
0
=
π
0
4
·
¸2
1
+ 1 dx
4
¸
1 2 1
x + x + 1 dx
=
π
16
2
0
µ
¶¯4
¯
1 3 1 2
= π
x + x + x ¯¯
48
4
0
28
π
=
3
Z
·
2. First we have to find the limits of integration from
√
x + 4 = 18x.
This gives
√ us a = 2 and b = 8 (check if you don’t believe it). Then
R(x) = 18x is the outer radius, while r(x) = x + 4 is the inner radius,
and the volume is
Z b
¡
¢
π [R(x)]2 − [r(x)]2 dx
V =
a
=
Z
8
2
= π
=
π(−x2 + 10x − 16) dx
µ
36π
¶¯8
¯
1
− x3 + 5x2 − 16x ¯¯
3
2
3. Here there are several possibilities to generate the doughnut by revolving
areas about some axis. For example, half the doughnut can be generated
by revolving the area bounded by the x-axis and the graph of
p
f (x) = 1 − (x − 3)2 ,
2≤x≤4
1
about the y-axis. Then the volume of the doughnut is
Z 4
p
V = 2
2πx 1 − (x − 3)2 dx
2
=
=
2
Z
4π
1
−1
Z
p
2π(x + 3) 1 − x2 dx
1
−1
Z
p
2
x 1 − x dx + 12π
1
−1
p
1 − x2 dx.
The first integral is 0, as we are integrating an odd function over a symmetric interval (if you don’t see this, you can calculate it explicitely by
substituting u = x2 ). So we’re left with
V = 12π
Z
1
−1
p
1 − x2 dx.
The integral is half the area of a circle of radius 1, ie the integral is π/2,
and
V
= 6π 2
≈ 59.22.
You have been cheated by your baker, but only slightly.
2