Higher Engineering Mathematics, Sixth Edition

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

1. The areas of equidistantly spaced sections of
   the underwater form of a small boat are as
   follows:

1.76,2.78,3.10,3.12,2.61,1.24,0.85m^2

Determine the underwater volume if the

sections are 3m apart. [42.59m^3 ]

2. 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 ]

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