206 Higher Engineering Mathematics
A sketch of the tree trunk is similar to that shown
in Fig. 19.5 above, where d = 2 m, A 1 = 0.52 m 2 ,
A 2 = 0.55 m 2 , and so on.
Using Simpson’s rule for volumes gives:
Volume =
2
3 [(0.52 + 0.97) + 4(0.55 + 0.63
+ 0.84) + 2(0.59 + 0.72)]
=
2
3 [1.49 + 8.08 + 2.62] = 8.13 m
3
Problem 4. The areas of seven horizontal
cross-sections of a water reservoir at intervals of
10 m are:
210, 250, 320, 350, 290, 230, 170 m 2
Calculate the capacity of the reservoir in litres.
Using Simpson’s rule for volumes gives:
Volume =
10
3
[(210 + 170) + 4(250 + 350
+ 230) + 2(320 + 290)]
=
10
3
[380 + 3320 + 1220]
= 16400 m
3
16400 m 3 = 16400 × 10 6 cm 3 and since
1 litre = 1000 cm
3 ,
capacity of reservoir =
16400 × 10 6
1000
litres
= 1 6400000
= 1.64 × 10
7 litres
Now try the following exercise
Exercise 83
Further problems on volumes
of irregular solids
1. The areas of equidistantly spaced sections of
the underwater form of a small boat are as
follows:
1.76, 2.78, 3.10, 3.12, 2.61, 1.24, 0.85 m 2
Determine the underwater volume if the
sections are 3 m apart.
[42.59 m 3 ]
2. To estimate the amount of earth to be removed
when constructing a cutting the crosssectional area at intervals of 8 m were estimated as follows:
0, 2.8, 3.7, 4.5, 4.1, 2.6, 0 m
3
Estimate the volume of earth to be excavated.
[147 m 3 ]
3. The circumference of a 12 m long log of timber
of varying circular cross-section is measured
at intervals of 2 m along its length and the
results are:
Distance from
one end (m)
Circumference
(m)
0
2.80
2
3.25
4
3.94
6
4.32
8
5.16
10
5.82
12
6.36
Estimate the volume of the timber in cubic
metres.
[20.42 m 3 ]
19.3 The mean or average value of a
waveform
The mean or average value, y, of the waveform shown
in Fig. 19.6 is given by:
y =
area under curve
length of base, b