24 Find the area of the region enclosed by y 5 e x, x 5 0 and the tangent to y 5 e x at
x 5 1.
25 Find the area of the region inside the ‘loop’ in the graph of the curve y2 5 x4(x 1 3).
26 Find the area enclosed by the curve y2 5 2x 2 2 4x 4.
27 Find the area of the region enclosed by x 5 3y2 and x 5 12y 2 y2 2 5.
28 Find the area of the region enclosed by y 5 (x 2 2)2 and y 5 x(x 2 4)2.
29 Find a value for m . 0 such that the area under the graph of y 5 e 2x over the
interval [0, m] is 3 square units.
30 Find the area of the region bounded by y 5 x 3 2 4x 2 1 3x and the x-axis.
16.7
Volumes with integrals
Recall that the underlying principle for finding the area of a plane region is
to divide the region into thin strips, approximate the area of each strip by
the area of a rectangle, and then add the approximations and take the limit
of the sum to produce an integral for the area. The same strategy can be
used to find the volume of a solid.
The idea is to divide the solid into thin slabs, approximate the volume of
each slab, add the approximations and take the limit of the sum to produce
an integral of the volume.
Hint: This is an introductory
section that will not be examined.
It is only used to give you an idea
of why we use integrals to find
volumes.
y
Cross-section
with area A(x)
Given a solid whose volume is to be computed, we start by taking crosssections perpendicular to the x-axis as shown in Figure 16.9. Each slab will
be approximated by a cylindrical solid whose volume will be equal to the
product of its base times its height.
S
0
a
x
A area of base
b x
h height
Figure 16.9
If we call the volume of the slab vi and the area of its base A(x), then
vi 5 A(xi)h 5 A(xi)Dxi .
Using this approximation, the volume
of the whole solid can be found by
y
y
Plane at xi 1
xi xi xi 1
n
​ A(xi) Dxi ​.
V  ​∑​  S
i51
Taking the limit as n increases and the
widths of the sub-intervals approach
zero yields the definite integral:
n
∫
b
​​∑​  ​ A(xi) Dxi​5 ​ ​  ​ ​A(x) dx
V 5 n
​ lim
→
i51
a
0
0
xi 1
Plane at xi
xi
The cylinder’s base
a
xi 1
xi
b x
x
819
16
Integral Calculus
Note: If we place the solid along the y-axis and take the cross-sections
perpendicular to that axis, we will arrive at a similar expression for the
volume of the solid, i.e.
∫
n
b
V 5
​ lim ​​∑​ ​A(yi)yi 5 ​ ​  ​A(y)dy
n→
a
i51
Example 38
Find the
volume of the solid formed when the graph of the parabola
___
y5√
​ 2x ​over [0, 4] is rotated around the x-axis through an angle of 2p
radians, as shown in the diagram.
y
y 2x
y
R(x) 2x
y 2x
0
R(x) 2x
x
0
x
4
4
x
x
disc
Solution
___
The cross-section here is a circular disc whose radius is y 5 √
​  2x ​. Therefore,
___
A(x) 5 pR2 5 p (​ √2x ​)2 5 2px.
The volume is then
∫
∫
4
4
[
]
x 2 ​ 4 5 16p cubic units.
V 5 ​ ​ ​ ​A(x) dx 5 ​ ​ ​ ​2p x dx 5 2p ​ ___
2 0
0
0
Example 38 above is a special case of the general process for finding
volumes of the so-called ‘solids of revolution’.
1
If a region is bounded by a closed interval [a, b] on the x-axis and a function
f (x) is rotated about the x-axis, the volume of the resulting solid of revolution
is given by
∫
b
V 5 ​ ​  ​p ( f (x))2 dx​.
y
a
y
f
f(x)
0
0
820
a
x
b
x
x
2 If the region bounded by a closed interval [c, d ] on the y-axis and a function
g (y) is rotated about the y-axis, the volume of the resulting solid of revolution is
given by
∫
c
V 5 ​ ​  ​p(g (y))2dy.​
d
y
y
d
x g(y)
g(y)
∆y {
C
O
x
O
x
Example 39
(x, y)
y
Find the volume of a sphere with
radius R 5 a.
A(x) π(a2 x2)
a
x
x
a
x
x
Solution
If we place the sphere with its centre at the origin, the equation of the circle
will be
_______
x 2 1 y 2 5 a 2 ⇒ y 5 6​√a 2 2 x 2 ​ .
The cross-section of the sphere, perpendicular to the x-axis, is a circular
disc with radius y , so the area is
_______
A (x) 5 p R2 5 p y2 5 p (​√a 2 2 x 2 ​ ) 5 p (a 2 2 x 2).
So, the volume of the sphere is
∫
a
a
[
]
3
V 5 ​ ​ ​​ p (a 2 2 x 2) dx 5 p ​ a2x 2 __
​ x  ​  ​
3 2a
2a
(
5 p (​ 2a
)
(
)
3
3
5 p ​ a 3 2 __
​ a  ​  ​2p ​ 2 a 3 1 __
​ a  ​  ​
3
3
3
)
3
3
2 2 __
​ a ​  ​5 _____
​ 4pa ​.
3
3
821
16
Integral Calculus
Note: If we want to rotate the right-hand region of the circle around the
y-axis, then the cross-section of the sphere, perpendicular to the y-axis is a
circular disc with radius x. Solving the equation for x instead:
______
x 2 1 y 2 5 a2 ⇒ x 5 6​ √a2 2 y 2 ​, and hence the area is
______
​  a2 2 y 2 ​ )2​ 5 p(a2 2 y 2),
A(y) 5 pR2 5 px 2 5 p ​( √
and the volume of the sphere is
∫
[
a
]
3
3
y3 a
​ ​ p (a2 2 y 2)dy​5 p ​​ a2y 2 __
​   ​  ​​ ​5 p ​ a3 2 __
​ a  ​  ​2 p ​ 2a3 1 __
​ a  ​  ​
3 2a
3
3
2a
3
3
a  ​  ​5 ____
​ 4pa ​
5 p ​ 2a3 2 2 ​ __
3
3
V5​
(
(
)
(
)
)
This is the same result as above.
Example 40
Find the
volume of the solid generated when the region enclosed by
___
y5√
​  3x ​, x 5 3 and y 5 0 is revolved about the x-axis.
Solution
∫
3
V 5 ​ ​ ​p(f (x))2dx
y
0
∫
3
4
___
5 p​ ​  ​(√
​  3x ​)2dx
3
0
2
[ ]
2 3
​ x  ​  ​​​ 5 ____
​ 27p
​
5 3p ​​ __
2 0
2
1
1 O
1
1
2
3
2
3
4
Example 41
Find the
volume of the solid generated when the region enclosed by
___
y5√
​  3x ​, y 5 3 and x 5 0 is revolved about the y-axis.
Solution
Here, we first find x as a function of y.
___
y2
​   ​, the interval on the y-axis is [0, 3]
y5√
​  3x ​⇒ x 5 __
3
So, the volume required is
∫ ( )
∫
[ ]
3
3
y2 2
y 5 3 27p
p ​​ ​ ​y 4dy 5 ​ __
p ​ ​​ __
V 5 ​ ​ ​p ​​ __
​   ​  ​​ ​dy 5 ​ __
​   ​  ​​ ​​ 5 ____
​   ​ .
5
3
9 0
9 5 0
0
822
4
x
Washers
Consider the region R between two curves, y 5 f (x) and y 5 g (x), and
from x 5 a to x 5 b where f (x) . g (x). Rotating R about the x-axis
generates a solid of revolution S. How do we find the volume of S?
y
f(x)
B
a
x
b
y
B
f(x)
g(x)
A
O
y
A
O
x
a
g(x)
x
x
b
O
a
Consider an arbitrary point x in the interval [a, b]. The segment AB
represents the difference f (x) 2 g (x). When we rotate this slice, the crosssection perpendicular to the x-axis is going to look like a ‘washer’ whose
area is
x
b
x
Area = (R2 – r2)
A 5 p(R2 2 r 2) 5 p((f (x))2 2 (g (x))2).
R = f(x)
So, the volume of S is
∫
∫
b
b
r = f(x)
V 5 ​ ​  ​A(x)dx 5 p​ ​  ​(( f (x))2 2 (g (x))2)dx.
a
a
Note: If you are rotating about the y-axis, a similar formula applies.
∫
d
V 5 p​ ​  ​((p(y))2 2 (q(y))2)dy
c
Note: To understand the washer more, you can think of it in the following
manner: Let P be the solid generated by rotating the curve y 5 f (x) and Q
be the solid generated by rotating the curve y 5 g (x). Then S can be found
by removing the solid of revolution generated by y 5 g (x) from the solid
of revolution generated by y 5 f (x), as shown.
y
y
y
f(x)
g(x)
P
x
O
O
Q
x
O
x
Therefore, volume of S 5 volume of P 2 volume of Q. And this justifies
the formula:
∫
b
∫
b
∫
b
V 5 p​ ​  ​( f (x))2dx 2 p​ ​  ​(g (x))2dx 5 p​ ​  ​​​( ( f (x))2 2 (g (x))2 )​dx
a
a
a
823
16
Integral Calculus
Example 42
The region in the first quadrant
between f (x) 5 6 2 x 2 and
8  ​is rotated about the
h(x) 5 ​ __
x2
x-axis. Find the volume of the
generated solid.
y
10
5
0
x
5
10
1
Solution
The rotated region is shown in the diagram. f (x) is larger than h(x)in this
interval. Moreover, the two curves intersect at:
__
__
​  2 ​, x 5 2
​ 8  ​5 6 2 x 2 ⇒ x 5 √
x2
Hence, the volume of the solid of revolution is
∫(
( ))
​ 64 ​ )​dx
5 p​∫ ​ ​​ ​( x 2 12x 1 36 2 ___
x
2
2
​ 82 ​  ​​ ​  ​dx
V 5 p​ __​ ​​ ​ (6 2 x 2)2 2 ​​ __
x
​ √2 ​
2
__
4
2
4
​ √2 ​
[
]
2
5
​ x  ​2 4x 3 1 36x 1 ___
​ 643 ​  ​​ ​_
5 p ​​ __
5
3x √
​  2 ​
__
736 2 512​ √ 2 ​
​ p.
5 ​ ___________
15
An alternative method: Volumes by cylindrical shells
Consider the region R under the curve y 5 f (x). Rotate R about the y-axis. We divide
R into vertical strips of width x each as shown. When we rotate a strip around the
y-axis, we generate a cylindrical shell of x thickness and height f (x). To understand
how we get the volume, we can cut the shell vertically as shown and ‘unfold’ it. The
resulting rectangular parallelpiped has length 2px, height f (x) and thickness x.
x
f(x)
x
{
So, the volume of this shell is
y
x
vi 5 length 3 height 3 thickness
5 (2px) 3 f (x) 3 x.
2πx
The volume of the whole solid is the sum of the volumes of these shells as the
number of shells increases, and consequently
f(x)
v
V 5
​ lim ​​∑​ ​vi 5
​  lim ​ ∑(2px) 3 f (x) 3 x
n→
∫
i51
f(x)
b
5 2p​ ​  ​​xf (x)dx.
a
x → 0
2πx
∆x
In many problems involving rotation about the y-axis, this would be more accessible than the disc/washer method.
824
x
Example 43
3
Find the volume of the solid generated
when we rotate the region under
f (x) 5 ______
​  2 2 ​, x 5 0 and x 5 3 around the y-axis.
11x
y
2
1
0
Solution
x
Using the shell method, we have
1
4 3 2 1 0
∫
5 2p​∫ ​ ​______
​​  2x  ​dx 5 2p​∫​  ​___
​​ du
u​
11x
3
​  2 2 ​dx
V 5 2p​ ​ ​x 3 ______
11x
0
3
0
1
2
3
4
10
2
1
10
5 2p[ln u]​
1 ​ 5 2p ln 10.
Exercise 16.7
In questions 1–19, find the volume of the solid obtained by rotating the region
bounded by the given curves about the x-axis. Sketch the region, the solid and a
typical disc.
x
​   ​, y 5 0, x 5 2, x 5 3  2 y 5 2 2 x 2, y 5 0
1 y 5 3 2 __
3
_______
​  16 2 x 2 ​, y 5 0, x 5 1, x 5 3  4 y 5 __
​ 3 ​, y 5 0, x 5 1, x = 3
3 y5√
x
____
5 y 5 3 2 x, y 5 0, x 5 0  6 y 5 √
​  sin x ​, y 5 0, 0 < x < p
_____
p ​< x < __
7 y5√
​ cos x ​, y 5 0, 2 ​ __
​ p ​  8 y 5 4 2 x 2, y 5 0
2
3
9 y 5 x 3 1 2x 1 1, y 5 0, x 5 1
10 y 5 24 x 2 x 2, y 5 x 2
11 y 5 sec x, x 5 __
​ p ​, x 5 __
​ p ​, y 5 0
4
3
12 y 5 1 2 x 2, y 5 x 3 1 1
​  36 2 x 2 ​, y 5 4
13 y 5 √
14 x 5 √
​  y ​, y 5 2x
15 y 5 sin x, y 5 cos x, x 5 __
​ p ​, x 5 __
​ p ​
4
2
16 y 5 2x 2 1 4, y 5 x, x 5 1, x 5 3
17 y 5 ​  x 1 1 ​, y 5 0, x 5 1, x 5 3
18 y 5 16 2 x, y 5 3x 1 12, x 5 21
_______
______
√ 4
__
1 ,​ y 5 __
​ x
​ 5 ​2 x
19 y 5 __
2
y
4
20 Find the volume resulting from a
rotation of this region about
a) the x-axis
b) the y-axis.
y 2x
3
y 3 x2
2
1
2
1
O
1
2 x
1
825
16
Integral Calculus
In questions 21–31, find the volume of the solid obtained by rotating the region
bounded by the given curves about the y-axis. Sketch the region, the solid and a
typical disc/shell.
21 y 5 x 2, y 5 0, x 5 1, x 5 3
______
​  9 2x 2 ​, x 5 0
22 y 5 x, y 5 √
23 y 5 x 3 2 4x 2 1 4x, y 5 0
___
​  3x ​, x 5 5, x 5 11, y 5 0
24 y 5 √
​  2 2 ​
25 y 5 x 2, y 5 ______
11x
______
​  x 2 1 2 ​, x 5 3, y 5 0, x 5 0
26 y 5 √
7x  ​, x 5 3, y 5 0
​  ______
27 y 5 _______
√
​  x 3 1 7 ​
​ p ​, x 5 __
​ p ​
28 y 5 sin x, y 5 cos x, x 5 __
4
2
29 y 5 2x 2 1 4, y 5 x, x 5 1, x 5 3
__
30 y 5 sin(x 2), y 5 0, x 5 0, x 5 √
​  p ​
31 y 5 5 2 x 3, y 5 5 2 4x
16.8
Modelling linear motion
In previous sections of this text, we have examined problems involving
displacement, velocity and acceleration of a moving object. In different
sections of Chapter 13, we applied the fact that a derivative is a rate of
change to express velocity and acceleration as derivatives. Even though our
earlier work on motion problems involved an object moving in one, two or
even three dimensions, our mathematical models considered the object’s
motion occurring only along a straight line. For example, projectile motion
(e.g. a ball being thrown) is often modelled by a position function that
simply gives the height (displacement) of the object. In that way, we are
modelling the motion as if it were restricted to a vertical line.
In this section, we will again analyze the motion of an object as if its
motion takes place along a straight line in space. This can only make sense
if the mass (and thus, size) of the object is not taken into account. Hence,
the object is modelled by a particle whose mass is considered to be zero.
This study of motion, without reference either to the forces that cause it or
to the mass of the object, is known as kinematics.
Displacement and total distance travelled
Recall from Chapter 13 that given time t, displacement s, velocity v and
acceleration a, we have the following:
( )
2
​ dv ​, and a 5 __
​ d  ​ ​ __
​ ds ​  ​5 ___
​ d 2s ​
v 5 __
​ ds ​, a 5 __
dt
dt
dt dt
dt
826