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Chapter 28

Irregular areas and volumes,
and mean values
28
...
1

(c) the mid-ordinate rule, or
(d) Simpson’s rule
...

(a) A planimeter is an instrument for directly measuring small areas bounded by an irregular curve
...
A pointer on the
planimeter is used to trace around the boundary
of the shape
...


(iii) Area PQRS


y1 + y7
=d
+ y2 + y3 + y4 + y5 + y6
...
1,

D

A

(i) Divide base PS into any number of equal
intervals, each of width d (the greater the
number of intervals, the greater the accuracy)
...

DOI: 10
...
00028-4

C

d

d

d

d

d

d

Figure 28
...
2,

258 Basic Engineering Mathematics
30

(i) Divide base PS into an even number of intervals, each of width d (the greater the number
of intervals, the greater the accuracy)
...

d
(iii) Area PQRS = [(y1 + y7 ) + 4(y2 + y4 + y6 )
3
+ 2(y3 + y5 )]
In general, Simpson’s rule states

 

1 width of
first + last
Area =
ordinate
3 interval




sum of even
sum of remaining
+4
+2
ordinates
odd ordinates

0

Problem 1
...

3

4

5

6

1

2

24
...
25

17
...
0

12
...
75

5
...
0

3
4
5
Time (seconds)

6

Figure 28
...
Mid-ordinates are erected as shown in
Figure 28
...
The length of each
mid-ordinate is measured
...
25 + 4
...
0 + 10
...
0
+ 20
...
25 m
Simpson’s rule (see (d) above)
The time base is divided into 6 strips each of width
1 s and the length of the ordinates measured
...
0) + 4(2
...
75
3

Speed v (m/s) 0 2
...
5 8
...
5 17
...
0
Determine the distance travelled in 6 seconds (i
...

the area under the v/t graph), using (a) the
trapezoidal rule (b) the mid-ordinate rule
(c) Simpson’s rule

4
...
25
2
...
75

Area = (width of interval)(sum of
mid-ordinates)
(d) Simpson’s rule
To determine the area PQRS of Figure 28
...

(ii) Erect ordinates in the middle of each interval
(shown by broken lines in Figure 28
...

(iii) Accurately measure ordinates y1 , y2, y3, etc
...


+ 17
...
5 + 12
...
33 m
Problem 2
...
Soundings of
the depth are made at equal intervals of 3 m across
the river and are as shown below
...
2 3
...
5 4
...
4 0

A graph of speed/time is shown in Figure 28
...

(a)

Trapezoidal rule (see (b) above)
The time base is divided into 6 strips, each of
width 1 s, and the length of the ordinates measured
...
0
area = (1)
+ 2
...
5
2

+ 8
...
5 + 17
...
75 m

Calculate the cross-sectional area of the flow of
water at this point using Simpson’s rule
From (d) above,
1
Area = (3)[(0+0) + 4(2
...
5+2
...
3+4
...
4 + 15] = 51
...
2

Practice Exercise 110 Areas of irregular
figures (answers on page 352)
1
...


3
...

Determine the area enclosed by the curve,
the x-axis and ordinates x = 0 and x = 3 by
(a) the trapezoidal rule (b) the mid-ordinate
rule (c) Simpson’s rule
...
0 4
...
0 14
...
0 29
...
e
...

The shape of a piece of land is shown in
Figure 28
...
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)
...


140 160 200 190 180 130

50 50 50 50 50 50

Figure 28
...


If the cross-sectional areas A1 , A2 , A3 ,
...
5), by
Simpson’s rule
Volume, V =

A1

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


Volumes of irregular solids

Plot the graph of y = 2x 2 + 3 between x = 0
and x = 4
...


Time t (s) 0 1

The deck of a ship is 35 m long
...

Width (m) 0 2
...
2 6
...
8 4
...
0 2
...


259

d

d
[(A1 + A7 ) + 4(A2 + A4 + A6 )
3
+ 2(A3 + A5 )]

A2

A3

d

d

A4

d

A5 A6

d

A7

d

Figure 28
...
A tree trunk is 12 m in length and has
a varying cross-section
...
52, 0
...
59, 0
...
72, 0
...
97 m2
...
5 above, where d = 2 m, A1 = 0
...
55 m2 , and so on
...
52 + 0
...
55 + 0
...
84)
3
+ 2(0
...
72)]
2
= [1
...
08 + 2
...
13 m3
3
Problem 4
...

Calculate the capacity of the reservoir in litres
Using Simpson’s rule for volumes gives
Volume =

=

10
[(210 + 170) + 4(250 + 350 + 230)
3
+ 2(320 + 290)]
10
[380 + 3320 + 1220] = 16 400m3
3

260 Basic Engineering Mathematics
16 400 m3 = 16 400 × 106 cm3
...
6
=
7
y=

= 16 400 000 = 16
...
The areas of equidistantly spaced sections of
the underwater form of a small boat are as follows:
1
...
78, 3
...
12, 2
...
24 and 0
...

Determine the underwater volume if the
sections are 3 m apart
...
To estimate the amount of earth to be
removed when constructing a cutting, the
cross-sectional area at intervals of 8 m were
estimated as follows:
0, 2
...
7, 4
...
1, 2
...

Estimate the volume of earth to be excavated
...
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 as follows
...

Distance from
one end (m)

0

2

4

6

y

y1 y2 y3 y4 y5 y6 y7
d

d

d

d

d

d

d

b

Figure 28
...
7(a)),

cycle

is

zero

(see

(b) over half a cycle is 0
...
7(b)) is 0
...
7(c)) is 0
...

π

Circumference (m) 2
...
25 3
...
32
Distance from
one end (m)

8

Circumference (m)

5
...
82

10

12
6
...
3 Mean or average values of
waveforms

area under curve
length of base, b

(b)

V
Vm
0

The mean or average value, y, of the waveform shown
in Figure 28
...
7

Irregular areas and volumes, and mean values

20

0

1 2 3 4 t (ms)

220

3
2
1
0

1 2 3

21
22
23

4 5 6 t (s)

Problem 6
...
9
Current (mA)

Current (A)

Voltage (V)

Problem 5
...
8

261

5

0

4

8 12 16 20 24 28 t (ms)

2

4

(b)

Current (A)

Voltage (V)

(a)
10

0

2 4 6 8 t (ms)

210

2

0

6

8 10 12 t (ms)

Figure 28
...
8

(a) One cycle of the trapezoidal waveform (a) is completed in 10 ms (i
...
the periodic time is 10 ms)
...
33 A
3s

Average value of waveform =

(c) A half cycle of the voltage waveform (c) is
completed in 4 ms
...
5 V
4 × 10−3s

=

area under curve
length of base
30 × 10−6 As
10 × 10−3 s

= 3 mA
(b) One cycle of the saw-tooth waveform (b) is completed in 5 ms
...
6 A

262 Basic Engineering Mathematics
Problem 7
...

0

1

2

3

4

5 6

Power (kW) 0 14 29 51 45 23 0
Plot a graph of power against time and, by using the
mid-ordinate rule, determine (a) the area under the
curve and (b) the average value of the power
The graph of power/time is shown in Figure 28
...

Graph of power/time

10

0 308608908
␲
2

1808
␲

2708
3␲
2

3608
2␲

␪

Figure 28
...
The base is divided into 6 intervals, each of
width 30◦
...


50
40
Power (kW)

Voltage (V)

Time (h)

Problem 8
...
11 shows a sinusoidal
output voltage of a full-wave rectifier
...
588 V,

30

At 45◦ the height of the mid-ordinate is
10 sin 45◦ = 7
...

The results are tabulated below
...
0 21
...
0 49
...
0 10
...
588 V

Figure 28
...
071 V

(a)

75◦

10 sin 75◦ = 9
...
659 V

135◦

10 sin 135◦ = 7
...
588 V

The time base is divided into 6 equal intervals, each
of width 1 hour
...
10) and measured
...
10
...
636 V

= (1)[7
...
5 + 42
...
5 + 37
...
0]
= 167 kWh (i
...
a measure of electrical energy)
(b) Average value of waveform =

area under curve
length of base

167 kWh
6h
= 27
...
636
sum of mid-ordinates
=
= 6
...
)
For a sine wave the actual mean value is
0
...
37 V
...
An indicator diagram for a steam
engine is shown in Figure 28
...
The base line has

2

0
22

20 t (ms)

10

3
...
0

3
...
9

2
...
0 cm

(c)

Figure 28
...
0
cm
...
0)[(3
...
6) + 4(4
...
9 + 1
...
5 + 2
...
2 + 34
...
4] = 34 cm2
3
area of diagram
(b) Mean height of ordinates =
length of base
34
= 2
...
83 cm × 100 kPa/cm = 283kPa

Now try the following Practice Exercise
Practice Exercise 112 Mean or average
values of waveforms (answers on page 352)
1
...
13 over a half
cycle
...
Find the average value of the periodic waveforms shown in Figure 28
...


Voltage (mV)

Figure 28
...
7 1
...
14

3
...


Voltage (V)

been divided into 6 equally spaced intervals and the
lengths of the 7 ordinates measured with the results
shown in centimetres
...
9 2
...
9 5
...
5 0
Plot a graph of current against time and estimate the area under the curve over the 30 ms
period, using the mid-ordinate rule, and determine its mean value
...
Determine, using an approximate method, the
average value of a sine wave of maximum
value 50 V for (a) a half cycle (b) a complete
cycle
...
An indicator diagram of a steam engine is
12 cm long
...
90, 5
...
22, 3
...
32, 3
...
16 cm
...


Revision Test 11 : Volumes, irregular areas and volumes, and mean values
This assignment covers the material contained in Chapters 27 and 28
...

1
...
Calculate the volume
of the alloy in cubic centimetres
...
Determine how many cubic metres of concrete are
required for a 120 m long path, 400 mm wide and
10 cm deep
...
Find the volume of a cylinder of radius 5
...
5 cm
...

(3)
4
...
35 m wide and has a diameter of 0
...
What area will it roll in making 40
revolutions?
(4)
5
...
5 cm and
base diameter 6
...

(3)
6
...
1
...
70 cm

4
...
Find (a) the volume and (b) the surface area of a
sphere of diameter 25 mm
...
A piece of alloy with dimensions 25 mm by
60 mm by 1
...
Assuming no wastage, calculate the height of the
cylinder in centimetres, correct to 1 decimal
place
...
Determine the volume (in cubic metres) and the
total surface area (in square metres) of a solid
metal cone of base radius 0
...
20 m
...

(6)
12
...
2 m by 90 cm by 60 cm
...

(4)
13
...

(8)
14
...
0 m long and diameter 1
...
2
...
(1 litre = 1000 cm3
...
6 cm

3
...
0 m

Figure RT11
...
1

7
...
4 cm3
...
(4)
8
...
It has a bore of 80 mm
and an outside diameter of 100 mm
...
(4)

(5)
15
...
0 cm and end diameters 60
...
0 cm
...
A boat has a mass of 20 000 kg
...
If the model is
made of the same material as the boat, determine
the mass of the model (in grams)
...
Plot a graph of y = 3x 2 + 5 from x = 1 to x = 4
...

(16)
18
...
The inside
diameter of the tower at different heights is given
in the following table
...
0 10
...
0 20
...
0 13
...
7

8
...
0

Determine the area corresponding to each diameter and hence estimate the capacity of the tower in
cubic metres
...
A vehicle starts from rest and its velocity is
measured every second for 6 seconds, with the
following results
...
2 2
...
7 5
...
0 9
...
e
...

(6)

List of formulae
Laws of indices:

Areas of plane figures:

a m × a n = a m+n
a m/n =

√
n m
a

am
an

= a m−n (a m )n = a mn

a −n =

Quadratic formula:
If ax 2 + bx + c = 0

1
an

Area = l × b

(i) Rectangle

a0 = 1
b

√
−b ± b2 − 4ac
x=
2a

then

Equation of a straight line:

l

(ii) Parallelogram Area = b × h

y = mx + c

Definition of a logarithm:
If y = a x

then

h

x = loga y

Laws of logarithms:
log(A × B) = log A + log B
 
A
= log A − log B
log
B

b

(iii) Trapezium

log An = n × log A

1
Area = (a + b)h
2
a

Exponential series:
ex = 1 + x +

x2 x3
+
+···
2! 3!

h

(valid for all values of x)
b

Theorem of Pythagoras:
b 2 = a 2 + c2

(iv) Triangle

Area =

1
×b×h
2

A

c

B

b
h

a

C
b

List of formulae
Area = πr 2 Circumference = 2πr

(v) Circle

(iii) Pyramid
If area of base = A and
perpendicular height = h then:

r

s

␪

Volume =

1
× A×h
3

r

2π radians = 360 degrees

Radian measure:

h

For a sector of circle:
θ◦
(2πr) = rθ
360

(θ in rad)

1
θ◦
(πr 2 ) = r 2 θ
360
2

(θ in rad)

s=

arc length,

shaded area =

Equation of a circle, centre at origin, radius r:
x 2 + y2 = r 2
Equation of a circle, centre at (a, b), radius r:

Total surface area = sum of areas of triangles
forming sides + area of base
(iv) Cone
1
Volume = πr2 h
3
Curved Surface area = πrl

(x − a)2 + (y − b)2 = r 2

Total Surface area = πrl + πr2

Volumes and surface areas of regular
solids:
l

(i) Rectangular prism (or cuboid)

h

Volume = l × b × h
Surface area = 2(bh + hl + lb)

r

(v) Sphere

l

h

4
Volume = πr3
3
Surface area = 4πr2

b

(ii) Cylinder
Volume = πr2 h
Total surface area = 2πrh + 2πr2
r

r

h

337

338 Basic Engineering Mathematics
Areas of irregular figures by approximate
methods:
Trapezoidal rule


 
width of 1 first + last
Area ≈
interval
2 ordinate

+ sum of remaining ordinates
Mid-ordinate rule
Area ≈ (width of interval)(sum of mid-ordinates)
Simpson’s rule



1 width of
first + last
Area ≈
ordinate
3 interval




sum of even
sum of remaining
+4
+2
ordinates
odd ordinates

For a general sinusoidal function y = A sin (ωt ± α),
then
A = amplitude
ω = angular velocity = 2π f rad/s
ω
= frequency, f hertz
2π
2π
= periodic time T seconds
ω
α = angle of lead or lag (compared with
y = A sin ωt )

Cartesian and polar co-ordinates:
If co-ordinate (x, y) = (r, θ) then

y
r = x 2 + y 2 and θ = tan−1
x
If co-ordinate (r, θ) = (x, y) then
x = r cosθ and y = r sin θ

Mean or average value of a waveform:
area under curve
length of base
sum of mid-ordinates
=
number of mid-ordinates

mean value, y =

Triangle formulae:
Sine rule:
Cosine rule:

b
c
a
=
=
sin A sin B
sin C
a 2 = b2 + c2 − 2bc cos A

B

b

a

If a = first term and d = common difference, then the
arithmetic progression is: a, a + d, a + 2d,
...

The n’th term is: arn−1
Sum of n terms, Sn =

A
c

Arithmetic progression:

a (1 − r n )
a (r n − 1)
or
(1 − r )
(r − 1)

If − 1 < r < 1, S∞ =

a
(1 − r )

C

Area of any triangle
1
= × base × perpendicular height
2
1
1
1
= ab sin C or
ac sin B or
bc sin A
2
2
2

a +b+c
= [s (s − a) (s − b) (s − c)] where s =
2

Statistics:
Discrete data:

#
mean, x¯ =

x

n
)
(
 #

(x − x¯ )2

standard deviation, σ =
n

List of formulae
Grouped data:

#
fx
mean, x¯ = #
f

)
 #$
%(

f (x − x)
¯ 2

#
standard deviation, σ =
f

Standard integrals
y
axn
cos ax

Standard derivatives
sin ax

y or f(x)

dy
= or f (x)
dx

axn

anxn−1

sin ax

a cos ax

cos ax

−a sin ax

eax

aeax

ln ax

1
x

eax
1
x

&

a

y dx
x n+1
+ c (except when n = −1)
n +1

1
sin ax + c
a
1
− cos ax + c
a
1 ax
e +c
a
ln x + c

339

Answers

Answers to practice exercises
Chapter 1

Chapter 2

Exercise 1 (page 2)
1
...

7
...

13
...

17
...
16 m
3
...
£565
6
...
−36 121
9
...
1487
12
...
−70872
15
...
25 cm
d = 64 mm, A = 136 mm, B = 10 mm

1
...

4
...

7
...


5
...

13
...


Exercise 2 (page 5)
(a) 468 (b) 868
2
...

(a) 259 (b) 56
8
...


2
...

6
...

10
...


(a) 12 (b) 360
(a) 90 (b) 2700
(a) 3 (b) 180
(a) 15 (b) 6300
(a) 14 (b) 53 900

Exercise 4 (page 8)
2
...
68

3
...
5

DOI: 10
...
00040-5

3
...

10
1
14
...
1
21
6
...


16
...


2
6
8
2
...

35
9
11
3
1
7
...

5
13
5
3
2
12
...
3
2
5
12
4
3
1
17
...
13
4
9
(a) £60, P£36, Q£16

7
...

15
...


22
9
8
25
3
16
4
27
17
60

4
...

12
...

20
...
5
9
...
−33
10
...
1
2
12
14
...


71
8
11
15
3
16
51
8
52
17
20

19
...

15
5
...
4
20
...
2880 litres

Exercise 7 (page 14)
1
18
7
6
...
2
1
...
22
11
...

(a) 4 (b) 24
(a) 10 (b) 350
(a) 2 (b) 210
(a) 5 (b) 210
(a) 14 (b) 420 420

2
...

3
...

7
...


1
7
4
11
17
30
43
77
9
1
40

1
...
4

1
9
19
7
...
2
20
2
...
1
8
...
7
3

4
...

15
5
...
0
...
14
...
1
...


13
20

7
40
21
1
141
(b)
(c)
(d)
(e)
25
80
500
11
3
7
...
10
25
200
11
1
7
(b) 4
(c) 14
(d) 15
40
8
20
2
...
4
40
41
10
...
(a)

11
...
625

9
250

3
...


6
125

9
...

4
...

10
...


2
...
11
...
185
...
8307
5
...
1581
6
...
571
5
...
1
...
0
...
068
11
...
5 ×10
12
...
5 ×103
−6
−3
4
...
202
...
18
...
6
...
0
...
11
...
0
...
(a) 300 V (b) 100 Hz (c) 0
...
412 rad (or 23
...
(a) v = 120 sin100πt volts
(b) v = 120 sin (100πt + 0
...
i = 20 sin 80πt −
6
i = 20 sin(80πt − 0
...
3
...
488) m
7
...
75◦ lagging
(b) −2
...
363 A (d) 6
...
423 ms

Chapter 23
Exercise 90 (page 207)
1
...
1 mm, c = 28
...
A = 52◦2 , c = 7
...
152 cm,
area = 25
...
D = 19◦48 , E = 134◦12 , e = 36
...
E = 49◦ 0 , F = 26◦ 38 , f = 15
...
6 mm2
5
...
420 cm,
area = 6
...
811 cm, area = 0
...
K = 47◦ 8 , J = 97◦ 52 , j = 62
...
2 mm2 or K = 132◦52 , J = 12◦8 ,
j = 13
...
0 mm2

Exercise 93 (page 212)
1
...
42◦, 59
...
20◦ 2
...
23 m (b) 38
...
40
...
05◦
4
...
8 cm 5
...
2 m
6
...
3 mm, y = 142 mm 7
...
13
...

2
...

4
...

6
...

8
...
83, 59
...
83, 1
...
61, 20
...
61, 0
...
47, 116
...
47, 2
...
55, 145
...
55, 2
...
62, 203
...
62, 3
...
33, 236
...
33, 4
...
83, 329
...
83, 5
...
68, 307
...
68, 5
...
294, 4
...
(1
...
960)
(−5
...
500)
4
...
884, 2
...
353, −5
...
(−2
...
207)
(0
...
299)
8
...
252, −4
...
04, 12
...
04, 12
...
51, −32
...
51, −32
...
47
...

3
...

7
...


Answers to practice exercises
Exercise 103 (page 234)

Chapter 25
Exercise 96 (page 221)
1
...
t = 146◦

351

2
...
(i) rhombus (a) 14 cm2 (b) 16 cm (ii) parallelogram
(a) 180 mm2 (b) 80 mm (iii) rectangle (a) 3600 mm2
(b) 300 mm (iv) trapezium (a) 190 cm2 (b) 62
...
35
...
(a) 80 m (b) 170 m 4
...
2 cm2
5
...
1200 mm
7
...
560 m2
2
9
...
4 cm
10
...
43
...
32

1
...

7
...

11
...

16
...

20
...
2376 mm2
3
...
1709 mm
6
...
(a) 106
...
9 cm2
2
21
...
17
...
07 cm2
(a) 59
...
8 mm
12
...
2 cm
8
...
48 cm 14
...
5◦ 15
...
698 rad (b) 804
...
10
...
24%
19
...
8 mm
7
...
(a) 2 (b) (3, −4)
2
...
Circle, centre (0, 1), radius 5
4
...

2
...

5
...


482 m2
(a) 50
...
9 mm2 (c) 3183 mm2
2513 mm2
4
...
19 mm (b) 63
...
01 cm2 (b) 129
...
5773 mm2
2
1
...
1932 mm2 2
...
(a) 0
...

4
...

8
...

12
...

15
...

19
...
2 m3
2
...
8 cm3
3
2
(a) 3840 mm (b) 1792 mm
972 litres
6
...
500 litres
3
9
...
3 cm3 (b) 61
...
44 m
(a) 2400 cm3 (b) 2460 cm2 11
...
04 m
1
...
8796 cm3
4
...
9 cm2
2
...
28060 cm3 , 1
...
22 m by 8
...
62
...
4
...
80 ha

2
...
3
...
45
...
259
...
2
...
47
...
38
...
12730 km 7
...
13 mm

Exercise 106 (page 246)
1
...
1 cm3 , 159
...
7
...
81 cm2
3
...
1 cm3 , 113
...
5
...
3 cm
6
...
(a) 268 083 mm3 or 268
...
06 cm2
8
...
53 cm
9
...
09 × 1012 km3 10
...
(a) 0
...
481 (c) 4
...
(a) 210◦ (b) 80◦ (c) 105◦
4
...

2
...

6
...
(a)

5890 mm2 or 58
...
55 cm3 (b) 84
...
13
...
393

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