Basic Engineering Mathematics, Fifth Edition

Irregular areas and volumes, and mean values 259

Now try the following Practice Exercise

Practice Exercise 110 Areas of irregular

figures (answers on page 352)

1. Plot a graph ofy= 3 x−x^2 by completing
   a table of values ofy fromx=0tox=3.
   Determine the area enclosed by the curve,
   thex-axis and ordinatesx=0andx=3by
   (a) the trapezoidal rule (b) the mid-ordinate
   rule (c) Simpson’s rule.


2. Plot the graph ofy= 2 x^2 +3 betweenx= 0
   andx= 4 .Estimate the area enclosed by the
   curve, the ordinatesx=0andx=4andthe
   x-axis by an approximate method.


3. The velocity of a car at one second intervals is
   given in the following table.

Timet(s) 0 1 2 3 4 5 6

v(m/s) 0 2.0 4.5 8.0 14.0 21.029.0

Velocity

Determine the distance travelled in 6 sec-

onds (i.e. the area under thev/tgraph) using

Simpson’s rule.

4. The shape of a piece of land is shown in
   Figure 28.4. To estimate the area of the land,
   a surveyor takes measurements at intervals
   of 50m, perpendicular to the straight portion
   with the results shown (the dimensions being
   in metres). Estimate the area of the land in
   hectares(1ha= 104 m^2 ).

50 50

140160 200 190 180 130

50 50 50 50

Figure 28.4

5. The deck of a ship is 35m long. At equal inter-
   vals of 5m the width is given by the following
   table.

Width (m) 0 2.85.26.55.8 4.13.02.3

Estimate the area of the deck.

28.2 Volumes of irregular solids

If the cross-sectional areasA 1 ,A 2 ,A 3 ,...of an irreg-

ular solid bounded by two parallel planes are known at

equal intervals of widthd(as shown in Figure 28.5), by

Simpson’s rule

Volume,V=

d

3

[(A 1 +A 7 )+ 4 (A 2 +A 4 +A 6 )

+ 2 (A 3 +A 5 )]

d d d d d d

A 1 A 2 A 3 A 4 A 5 A 6 A 7

Figure 28.5

Problem 3. A tree trunk is 12m in length and has

a varying cross-section. The cross-sectional areas at

intervals of 2m measured from one end are

0.52, 0.55, 0.59, 0.63, 0.72, 0.84 and 0.97m^2.

Estimate the volume of the tree trunk

A sketch of the tree trunk is similar to that shown

in Figure 28.5 above, whered=2m,A 1 = 0 .52m^2 ,

A 2 = 0 .55m^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]=8.13m^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 and 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