Higher Engineering Mathematics, Sixth Edition

378 Higher Engineering Mathematics

In this case,

r.m.s. value=

1

√

2

× 100 = 70 .71V]

Now try the following exercise

Exercise 148 Further problems on mean

and r.m.s. values

1. The vertical heighthkm of a missile varies
   with the horizontal distancedkm, and is given
   byh= 4 d−d^2. Determine the mean height of
   the missile fromd=0tod=4km.
   [2^23 km].


2. The distances of pointsyfrom the mean value
   of a frequency distribution are related to the
   variatexby the equationy=x+

1

x

. Deter-
mine the standard deviation (i.e. the r.m.s.
value), correct to 4 significant figures for
values ofxfrom 1 to 2. [2.198]
3. A currenti=25sin100πtmA flows in an
electrical circuit. Determine, using integral
calculus, its mean and r.m.s. values each cor-
rect to 2 decimal places over the ranget= 0
tot=10ms. [15.92mA, 17.68mA]
4. A wave is defined by the equation:

v=E 1 sinωt+E 3 sin3ωt

whereE 1 ,E 3 andωare constants.

Determine the r.m.s. value ofvover the

interval 0≤t≤

π

ω

.

⎡

⎣

√

E 12 +E 32

2

⎤

⎦

38.4 Volumes of solidsof revolution

With reference to Fig. 38.6, the volume of revolution,

V, obtained by rotating areaAthrough one revolution

about thex-axis is given by:

V=

∫b

a

πy^2 dx

If a curvex=f(y)is rotated 360◦about they-axis

between the limitsy=candy=dthen the volume

0 x 5 ax 5 b

y 5 f(x)

y

x

A

Figure 38.6

generated,V, is given by:

V=

∫d

c

πx^2 dy

Problem 5. The curvey=x^2 +4 is rotated one

revolution about thex-axis between the limitsx= 1

andx=4. Determine the volume of solid of

revolution produced.

Revolving the shaded area shown in Fig. 38.7, 360◦

about thex-axis produces a solidof revolutiongiven by:

Volume=

∫ 4

1

πy^2 dx=

∫ 4

1

π(x^2 + 4 )^2 dx

=

∫ 4

1

π(x^4 + 8 x^2 + 16 )dx

=π

[

x^5

5

+

8 x^3

3

+ 16 x

] 4

1

=π[( 204. 8 + 170. 67 + 64 )

−( 0. 2 + 2. 67 + 16 )]

=420.6πcubic units

012 x

4

5

30

(^20) A B
D C
10
y
3 4 5
y 5 x^214
Figure 38.7