February 9, 2006
Lecturer: Dr Martin Kurth
Hilary Term 2006
Problem Sheet 4 – Solutions
1. First we have to find the points a and b where the graph of f intersects
the x-axis, ie we have to set
1 − x2 = 0,
which means a = −1 and b = 1. Now we use the “shell formula” for the
volume of solids generated by revolving the area between the x-axis and
the graph of a function f between x = a and x = b about a vertical line
x = x0 :
Z
b
2π(x − x0 )f (x)dx.
V =
a
This gives in our case:
V
=
=
Z
1
2π(x + 2)(1 − x2 )dx
−1
Z 1
2π(−x3 − 2x2 + x + 2)dx
−1
µ
=
2π
=
16
π
3
¶¯1
¯
2 3
− x + 2x ¯¯
3
−1
2. The easiest way of doing this, is to use
f −1 (y) = y 3/2
and calculate an integral over y. The integral starts at
f (1) = 1
and ends at
f (8) = 4.
We get the curve length
L
4
=
Z
4
=
Z
1
1
=
Z
1
4
q
2
1 + ((f −1 )0 (y)) dy
s
1+
r
9
1 + ydy
4
1
µ
3√
y
2
¶2
dy
=
=
=
=
3
2
µ
4
r
4
+ ydy
9
1
¶3/2 ¯¯4
4
¯
+y
¯
¯
9
1
´
³
1
3/2
40 − 133/2
27
7.634
Z
3. The simplest way of doing this is to choose a different coordinate system,
such that one of the points is at (0, 0) and the other point is at (c, 0), with
p
c = (x2 − x1 )2 + (y2 − y1 )2 .
The equation for the line connecting the points is
g(x) = 0,
and we have
g 0 (x) = 0,
and thus we get the line length
Z cp
1 + (g 0 (x))2 dx
Lg =
0
Z c
=
1dx
0
= c,
as required. Now take any other smooth function f . If its graph is not
the straight line connecting the two points, there must be a point x0 in
(0, c) with
f 0 (x0 ) 6= 0.
As f is smooth f 0 is continuous, and there exists a δ > 0 such that
f 0 (x) 6= 0 for all x with |x − x0 | < δ. Choosing for example a = x0 − δ/2
and b = x0 + δ/2 we get that f 0 (x) 6= 0, ie
(f 0 (x))2 > 0
for all x in [a, b]. Now we have for the length of the graph of f
Z cp
Lf =
1 + (f 0 (x))2 dx
0
=
Z
0
a
p
1 + (f 0 (x))2 dx +
Z
b
a
p
1 + (f 0 (x))2 dx +
As (f 0 (x))2 ≥ 0 for all x, we have
p
1 + (f 0 (x))2 ≥ 1,
2
Z
b
c
p
1 + (f 0 (x))2 dx
and by the domination rule
Z ap
Z
1 + (f 0 (x))2 dx ≥
0
and
Z
b
This gives us
c
a
1dx = a
0
p
1 + (f 0 (x))2 dx ≥
Lf ≥ a +
b
Z
a
p
Z
b
c
1dx = c − b.
1 + (f 0 (x))2 dx + (c − b).
Furthermore, the Mean Value Theorem tells us that there is a point ξ in
[a, b] with
Z bp
p
1
1 + (f 0 (x))2 dx = 1 + (f 0 (ξ))2 ,
b−a a
and as (f 0 (x))2 > 0 for all x in [a, b], we have
1
b−a
ie
Z
a
and we get
b
Z
p
a
b
p
1 + (f 0 (x))2 dx > 1,
1 + (f 0 (x))2 dx > b − a,
Lf > a + (b − a) + (c − b) = c.
3