Irregular areas, volumes and mean values of waveforms 205
Problem 2. A river is 15 m wide. Soundings of
the depth are made at equal intervals of 3 m across
the river and are as shown below.
Depth (m) 0 2.2 3.3 4.5 4.2 2.4 0
Calculate the cross-sectional area of the flow of
water at this point using Simpson’s rule.
From para. (d) above,
Area =
1
3 (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 82 Further problems on areas of
irregular figures
1. Plot a graph of y = 3x − x 2 by completing
a table of values of y from x = 0 to x = 3.
Determine the area enclosed by the curve, the
x-axis and ordinate x = 0 and x = 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 of y = 2x 2 + 3 between x = 0
and x = 4. Estimate the area enclosed by the
curve, the ordinates x = 0 and x = 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 the v/t graph) using
Simpson’s rule.
[63.33 m]
4. The shape of a piece of land is shown in
Fig. 19.4. 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 = 10
4 m
2 ).
[4.70 ha]
50 50 50 50
140 160 200 190 180 130
50 50
Figure 19.4
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 ]
19.2 Volumes of irregular solids
If the cross-sectional areas A 1 , A 2 , A 3 , ... of an irregular
solid bounded by two parallel planes are known at equal
intervals of width d (as shown in Fig. 19.5), 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 )]
A 1
A 2
A 3
A 4
A 5
A 6
A 7
d d
d
d
d
d
d
Figure 19.5
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.
Problem 2. A river is 15 m wide. Soundings of
the depth are made at equal intervals of 3 m across
the river and are as shown below.
Depth (m) 0 2.2 3.3 4.5 4.2 2.4 0
Calculate the cross-sectional area of the flow of
water at this point using Simpson’s rule.
From para. (d) above,
Area =
1
3 (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 82 Further problems on areas of
irregular figures
1. Plot a graph of y = 3x − x 2 by completing
a table of values of y from x = 0 to x = 3.
Determine the area enclosed by the curve, the
x-axis and ordinate x = 0 and x = 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 of y = 2x 2 + 3 between x = 0
and x = 4. Estimate the area enclosed by the
curve, the ordinates x = 0 and x = 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 the v/t graph) using
Simpson’s rule.
[63.33 m]
4. The shape of a piece of land is shown in
Fig. 19.4. 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 = 10
4 m
2 ).
[4.70 ha]
50 50 50 50
140 160 200 190 180 130
50 50
Figure 19.4
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 ]
19.2 Volumes of irregular solids
If the cross-sectional areas A 1 , A 2 , A 3 , ... of an irregular
solid bounded by two parallel planes are known at equal
intervals of width d (as shown in Fig. 19.5), 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 )]
A 1
A 2
A 3
A 4
A 5
A 6
A 7
d d
d
d
d
d
d
Figure 19.5
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.