Understanding Engineering Mathematics

If we rotate a thin strip of area,dxwide, about the axis, then a thin disc is formed,

whose radius isr=y, and thickness ish=dx. The volume of this disc is approximately

‘πr^2 h’=πy^2 dx. So the total volume between the limitsx=aandx=bis obtained by

integration as

∫b

a

πy^2 dx=π

∫b

a

y^2 dx

So, given a functiony=f(x)between valuesx=aandx=b, entirely above the

x-axis, by rotating the area under its curve we generate a volume of revolution deter-

mined by substituting the function and values ofaandbin the above integral. Review

Question 10.1.6 illustrates this.

Solution to review question 10.1.6

The volume of the solid of revolution obtained by rotating the area under

the curvey=f(x)once about thex-axis between the limitsx=a,bis

given by

π

∫b

a

y^2 dx=π

∫b

a

[f(x)]^2 dx

The area bounded by the curvesy= 1 −x^2 ,y=0for− 1 ≤x≤1isthe

positive area enclosed between the curvey= 1 −x^2 and thex-axis. The

curve cuts thex-axis atx=±1 and so the required volume is

π

∫ 1

− 1

( 1 −x^2 )^2 dx= 2 π

∫ 1

0

(x^4 − 2 x^2 + 1 )dx

=

16 π

15

square units

10.3 Reinforcement

10.3.1 The derivative as a gradient and rate of change

➤➤

291 292

➤

A.For the following functions find (a) the rate of change and (b) the slope of the graph
at the points specified.
(i)y=x^3 + 2 x−1, x= 0 , 2

(ii) y=sinxx=0,

π

3

(iii) y=excosxx= 0 , 1

(iv)y=ln(x^2 + 1 )x= 0 , 2

B.Find the point on the curve

y=x^3 + 3 x^2 − 9 x+ 1

where the gradient is−12.