Higher Engineering Mathematics

218 GRAPHS

From para. (d) above,

Area=^13 (3)[(0+0)+4(2. 2 + 4. 5 + 2 .4)

+2(3. 3 + 4 .2)]

=(1)[0+ 36. 4 +15]=51.4 m^2

Now try the following exercise.

Exercise 90 Further problems on areas of

irregular figures

1. Plot a graph ofy= 3 x−x^2 by completing
   a table of values ofyfromx=0tox=3.
   Determine the area enclosed by the curve, the
   x-axis and ordinatex=0 andx=3 by (a) the
   trapezoidal rule, (b) the mid-ordinate rule and
   (c) by Simpson’s rule. [4.5 square units]


2. Plot the graph ofy= 2 x^2 +3 betweenx= 0
   andx=4. Estimate the area enclosed by the
   curve, the ordinatesx=0 andx=4, and the
   x-axis by an approximate method.
   [54.7 square units]


3. The velocity of a car at one second intervals
   is given in the following table:
   time
   t(s) 0 1 2 3 4 5 6
   velocity
   v(m/s) 0 2.0 4.5 8.0 14.0 21.0 29.0
   Determine the distance travelled in 6 seconds
   (i.e. the area under thev/t graph) using
   Simpson’s rule. [63.33 m]


4. The shape of a piece of land is shown in
   Fig. 20.4. To estimate the area of the land,
   a surveyor takes measurements at intervals
   of 50 m, perpendicular to the straight portion
   with the results shown (the dimensions being
   in metres). Estimate the area of the land in
   hectares (1 ha= 104 m^2 ). [4.70 ha]

Figure 20.4

5. The deck of a ship is 35 m long. At equal
   intervals of 5 m the width is given by the
   following table:

Width (m) 0 2.8 5.2 6.5 5.8 4.1 3.0 2.3

Estimate the area of the deck. [143 m^2 ]

20.2 Volumes of irregular solids

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

irregular solid bounded by two parallel planes are

known at equal intervals of widthd(as shown in

Fig. 20.5), then by Simpson’s rule:

volume,V=

d

3

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

+A 6 )+2(A 3 +A 5 )]

Figure 20.5

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

has a varying cross-section. The cross-sectional

areas at intervals of 2 m measured from one end

are:

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

Estimate the volume of the tree trunk.

A sketch of the tree trunk is similar to that shown

in Fig. 20.5 above, whered=2m,A 1 = 0 .52 m^2 ,

A 2 = 0 .55 m^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]=8.13 m^3