206 Higher Engineering Mathematics
A sketch of the tree trunk is similar to that shown
in Fig. 19.5 above, where d=2m, A 1 = 0 .52m^2 ,
A 2 = 0 .55m^2 , and so on.
Using Simpson’s rule for volumes gives:
Volume=^23 [( 0. 52 + 0. 97 )+ 4 ( 0. 55 + 0. 63
+ 0. 84 )+ 2 ( 0. 59 + 0. 72 )]
=^23 [1. 49 + 8. 08 + 2 .62]=8.13m^3
Problem 4. The areas of seven horizontal
cross-sections of a water reservoir at intervals of
10m are:
210,250,320, 350,290, 230,170m^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]
=16400m^3
16400m^3 = 16400 × 106 cm^3 and since
1litre=1000cm^3 ,
capacity of reservoir=
16400 × 106
1000
litres
= 16400000
=1.64× 107 litres
Now try the following exercise
Exercise 83 Further problemson volumes
of irregular solids
- The areas of equidistantly spaced sections of
the underwater form of a small boat are as
follows:
1.76,2.78,3.10,3.12,2.61,1.24,0.85m^2
Determine the underwater volume if the
sections are 3m apart. [42.59m^3 ]
- To estimate the amount of earth to be removed
when constructing a cutting the cross-
sectional area at intervals of 8m were esti-
mated as follows:
0, 2.8, 3.7, 4.5, 4.1, 2.6, 0m^3
Estimate the volume of earth to be excavated.
[147m^3 ]
- Thecircumferenceofa 12mlonglogoftimber
of varying circular cross-section is measured
at intervals of 2m 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. [20.42m^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