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
- 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]. - 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