Higher Engineering Mathematics

IRREGULAR AREAS, VOLUMES AND MEAN VALUES OF WAVEFORMS 219

C

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 × 106 cm^3 and since
1 litre=1000 cm^3 ,

capacity of reservoir=

16400 × 106

1000

litres

=1 6400000

=1.64× 107 litres

Now try the following exercise.

Exercise 91 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 sec-

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

2. To estimate the amount of earth to be removed
   when constructing a cutting the cross-
   sectional area at intervals of 8 m were esti-
   mated 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 Circumference

one end (m) (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 ]

20.3 The mean or average value of a

waveform

The mean or average value,y, of the waveform

shown in Fig. 20.6 is given by:

y=

area under curve

length of base,b

Figure 20.6

If the mid-ordinate rule is used to find the area under

the curve, then:

y=

sum of mid-ordinates

number of mid-ordinates

(

=

y 1 +y 2 +y 3 +y 4 +y 5 +y 6 +y 7

7

for Fig. 20.6

)