CHAPTER 75 CENTROIDS OF SIMPLE SHAPES
EXERCISE 290 Page 789
1. Find the position of the centroid of the area bounded by the curve, the x-axis and the given
ordinates: y = 2x ; x = 0, x = 3
A sketch of the area is shown below
3
 2 x3 
xy d x ∫ x ( 2 x ) d x ∫ 2 x 2 d x  3 
∫
0
0
0
0
=
x
=
=
= =
3
3
3
3
2
∫ y d x ∫ 2x d x ∫ 2x d x [ x ]0
3
3
0
3
0
[18]
[9]
=2
0
1 3 2
1 3
2
3
y dx
2x) d x
(
∫
∫
1 3
4  x3 
2
0
0
2
2
2
=
y
=
=
=
dx
(4x )=
[9 − 0] = 2
3
∫


0
9
18
18  3  0 9
y
d
x
∫
0
Hence, the centroid is at (2, 2)
2. Find the position of the centroid of the area bounded by the curve, the x-axis and the given
ordinates: y = 3x + 2 ; x = 0, x = 4
A sketch of the area is shown below
1166
© 2014, John Bird
4
xy d x
∫
0
=
x =
4
∫ yd x
0
+ 2) d x
∫ 0 x ( 3 x=
4
∫ 3x + 2 d x
4
∫
0
4
4
3x 2 + 2 x d x
[ x3 + x 2 ] 0
=
=
4
4
3x 2


+
x
x
3
2
d
∫0
+
x
2
 2

0
0
+ 16]
[64
=
[ 24 + 8]
80
= 2.50
32
1 4 2
1 4
2
y dx
3x + 2 ) d x
(
∫
∫
1 4
1
4
0
0
=
y 2 4
= 2
=
x
( 9 x 2 + 12 x + 4 ) d=
[3 x 3 + 6 x 2 + 4 x ] 0
∫
0
32
64
64
∫ yd x
0
=
1
304
[192 + 96 + 16] = = 4.75
64
64
Hence, the centroid is at (2.50, 4.75)
3. Find the position of the centroid of the area bounded by the curve, the x-axis and the given
ordinates: y = 5x2 ; x = 1, x = 4
A sketch of the area is shown below.
4
 5x4 
2
3
xy d x ∫ x ( 5 x ) d x ∫ 5 x d x  4 
∫
1
1
1
1
=
=
= 4= =
x
4
4
4
3
∫1 y d x ∫1 5 x 2 d x ∫1 5x 2 d x  5 x 
 3 1
4
4
4
5 4 4
5
4 −1 ]
(255)
[
4 = 4= 318.75 = 3.036
5 3 3
5
105
4 −1 ]
(63)
[
3
3
1 4 2
1 4
2
4
5x2 ) d x
y dx
(
∫
∫
1 4
1  25 x5 
5
5
1
1
2
2
4
25 x =
dx
15 ]
(1023)
y
=
=
=
=
[ 45 −=
4
∫


1
105
210
210  5  1 210
210
d
y
x
∫
1
= 24.36
Hence, the centroid lies at (3.036, 24.36)
4. Find the position of the centroid of the area bounded by the curve, the x-axis and the given
ordinates: y = 2x3 ; x = 0, x = 2
A sketch of the area is shown below
1167
© 2014, John Bird
2
 2 x5 
 64 
3
4


xy
x
x
x
x
x
x
d
2
d
2
d
(
)
5  0  5 
∫0= ∫0 = ∫0 = =
= 1.60
=
x
2
2
2
2
8]
3dx
3dx
[
2 x4 

y
x
x
x
d
2
2
∫0
∫0
∫0
 4 
0
2
2
2
1 2 2
1 2
2
2
y dx
( 2 x3 ) d x 1 2
∫
∫
4  x7 
1
0
0
2
2
6
=
=
=
=
y
x )d x
( 4=
[128 − 0] = 4.57
2
∫


0
8
16
16  7  0 28
y
x
d
∫
0
Hence, the centroid is at (1.60, 4.57)
5. Find the position of the centroid of the area bounded by the curve, the x-axis and the given
ordinates: y = x(3x + 1) ; x = –1, x = 0
A sketch of the area is shown below.
0
 3x 4 x3 
  3 1 
+ 
0− −
2
3
2


+
+
xy
d
x
x
3
x
x
d
x
3
x
x
d
x
(
)
3  −1   4 3  
∫
∫
∫
 4
−1
−1
−1
=
=
=
=
x =
0
0
0
0
 
1 
2 + xd x
2 + xd x
x2 

y
x
x
x
d
3
3
3
0 −  −1 +  
∫ −1
∫ −1
∫ −1
+
x


2 
2  −1
 
0
0
0
=
1168
−0.416666
= –0.833
0.5
© 2014, John Bird
1 0 2
1 0
2
y dx
3x 2 + x ) d x
(
∫
∫
−1
−1
y= 2 0
= 2
=
0.5
∫ yd x
−1
0
 9 x5 6 x 4 x3 
∫ −1 ( 9 x 4 + 6 x3 + x 2 ) d x =  5 + 4 + 3  −1
0
  9 3 1  9 3 1
= 0 −  − + −   = − + = 0.633
  5 2 3  5 2 3
Hence, the centroid lies at (–0.833, 0.633)
1169
© 2014, John Bird
EXERCISE 291 Page 791
1. Determine the position of the centroid of a sheet of metal formed by the curve y = 4x – x2 which
lies above the x-axis.
y = 4x – x2 = x(4 – x)
i.e. when y = 0, x = 0 and x = 4
The area of the sheet of metal is shown sketched below
By symmetry,
x=2
4
1 16 x 3 8 x 4 x 5 
1 4 2
1 4
1 4
2
2
2
3
4
y dx
( 4x − x ) d x
(16 x − 8 x + x ) d x 2  3 − 4 + 5  0
∫
∫
∫
0
0
0
2
=
= 2
=
y 2=
4
4
4
4
2 dx
2 dx
 2 x3 
d
4
−
4
−
y
x
x
x
x
x
(
)
(
)
∫0
∫0
∫0
 2 x − 3 
0
1
( 341.333 − 512 + 204.8 ) − (0) 
17.0665
2
=
= 1.6
=
64 
10.666

 32 −  − (0)
3 

Hence, the coordinates of the centroid are (2, 1.6)
2. Find the coordinates of the centroid of the area that lies between the curve
y
= x – 2 and the xx
axis.
y
= x – 2 i.e. y = x 2 − 2 x = x(x – 2)
x
Hence, when y = 0 (i.e. the x-axis), x = 0 and x = 2
A sketch of y = x 2 − 2 x is shown below
1170
© 2014, John Bird
By symmetry, x = 1
2
1  x5
4 x3 
4 +
1 2 2
1 2 2
1 2 4
2
−
x
3
2
∫0 y d x 2 ∫0 ( x − 2 x ) d x 2 ∫0 ( x − 4 x + 4 x ) d x 2  5
3  0
=
y 2=
=
=
2
2
2
2
2 − 2x d x
2 − 2x d x
x3


y
d
x
x
x
(
)
(
)
2
∫0
∫0
∫0
 3 − x 
0
1
( 6.4 − 16 + 10.6667 ) − (0) 
0.53335
2
=
= –0.4
=
−1.3333
( 2.6667 − 4 ) − (0)
Hence, the coordinates of the centroid are (1, –0.4)
3. Determine the coordinates of the centroid of the area formed between the curve y = 9 – x2 and the
x-axis.
A sketch of y = 9 – x2 is shown below. By symmetry, x = 0
3
1
x5 
3 +
1 3 2
1 3
1 3
2
81
6
−
x
x
2
2 + x4 d x
d
9
d
81
18
−
−
y
x
x
x
x
(
)
(
)
∫ −3
2 
5  −3
2 ∫ −3 = 2 ∫ −3
=
=
y 2=
3
3
3
3
x3 

∫ −3 y d x
∫3 ( 9 − x 2 ) d x
∫ −3 ( 9 − x 2 ) d x
9 x − 3 
−3
1
( 243 − 162 + 48.6 ) − (−243 + 162 − 48.6) 
129.6
= 2
= 3.6
=
36
( 27 − 9 ) − (−27 + 9)
Hence, the coordinates of the centroid are (0, 3.6)
4. Determine the centroid of the area lying between y = 4x2, the y-axis and the ordinates y = 0 and
y=4
The curve y = 4x2 is shown in the sketch below
1171
© 2014, John Bird
4
1  y2 
1 4 2
1 4y
∫0 x d x 2 ∫0 4 d x 2  8  0 1 3
= 0.375
=
=
= =
x= 2 4
4
8 8
4
y
1  y 3/2 
x
d
x
∫0
∫0 2 d x 2  3 / 2  0 3
4
=
y
4
4
0
0
xy d y ∫
∫=
∫ xd y
4
0
y
( y) d y
2=
8
3
∫
4
0
1  y 5/2 
y 3/2
d y 2 5 / 2

 0 3  32  3  64  12
2
= 2.40
=
=
= =
 
8
8
16  5 / 2  16  5  5
3
3
Hence, the coordinates of the centroid are (0.375, 2.40)
5. Find the position of the centroid of the area enclosed by the curve y =
5x , the x-axis and the
ordinate x = 5
The curve y =
5x is shown in the sketch below.
1172
© 2014, John Bird
5
1


2 dx
x
5
x

xy d x ∫ x 5 x d x ∫ 0 
∫
0
0
 =

x =
=
=
5
5
1
5
∫0 y d x ∫0 5x d x
∫ 5 x2 d x
5
5
( )
( )
(
5
5
 5
 x2 
5 
5
 
 2  0 ( 50 ) − ( 0 ) 
=
= 3.0
5
[16.6667]
 3
 x2 
5 
3
 
 2 0
3
5∫ x2 d x
=
5 1
5∫ x2 d x
0
0
)
2
1 5 2
1 5
5
y dx
5x d x
∫
∫
5
1
1
1
0
0
 5x2 
2
2
( 62.5 ) − ( 0 ) 
y =
=
=
=
=
(5x ) d x
5
∫


0
16.6667
33.3333
33.3333  2  0 33.3333 
d
y
x
∫
0
= 1.875
Hence, the centroid is at (3.0, 1.875)
6. Sketch the curve y2 = 9x between the limits x = 0 and x = 4. Determine the position of the centroid
of this area.
The curve y2 = 9x is shown sketched below.
By symmetry, y = 0
4
 5
 3x 2 
 5 
3
4
4
4
6 5 6 5
6
x 
4 − 0 
(32)
xy d x ∫ x 3 x d x ∫ 3 x 2 d x  2 
∫


0
0
0
0
5= 5 = 5
=
=
=
=
=
x
4
4
1
4
4
∫0 y d x ∫0 3 x d x ∫0 3x 2 d x  3x 32  2  x3  2  43 − 0 2(8)
 3 


 2 0
(
)
=
1173
38.4
= 2.4
16
© 2014, John Bird
Hence, the centroid is at (2.4, 0)
7. Calculate the points of intersection of the curves x2 = 4y and
y2
= x, and determine the position of
4
the centroid of the area enclosed by them.
The curves are shown in the sketch below.
Since x 2 = 4 y then y =
and for the second curve,
x2
4
and y 2 =
y2
= x i.e. y 2 = 4 x
4
Hence, equating the y 2 values gives:
from which,
x4
16
x4
= 4x
16
and x 4 = 64 x
i.e.
x 4 − 64 x =
0
x( x 3 − 64) =
0 giving x = 0 and x 3 − 64 =
0 i.e. x3 = 64 and x =
3
64 = 4
When x = 0, y = 0 and when x = 4, y = 4
Hence, the curves intersect at (0, 0) and (4, 4)
4
4
∫0 xy d x
=
x =
4
∫ yd x
0
∫
4
0

x2

∫
4
0
2
 5

 2x 2 x4 
 5 − 
3
3
x2 
4
16 
4 5 256 128
x

x − d x
2 −
4 −
− 16
2
d
x
x
∫0
2


4 
0
5 =
16
5
4
=
=
=
4
1
4
4 3 64
32 16
x2
x2
 3

2 −
dx
2
d
4 −
−
x−
x
x
∫
3
0
 2x 2 x 
4
4
3
12
3 3
 3 − 
12 

 2
0
9.6
=
= 1.80
5.3333
1174
© 2014, John Bird
Taking moments about 0x gives:
( )
(total area) y = (area of strip)(perpendicular distance of centroid of strip to 0x)
i.e.
( ) ∫
(5.3333) y =
=
x2   1 
x2  x2 

−
−
x
x
2
2
dx
 
+
0 
4   2 
4  4 

4
x 2 
x2 

2
x
−
x
+

 d x=
∫0 
4 
8 
x5/2 x5/2 x 4 

2
x
+
−
− d x
∫0 
4
4 32 
4
4
4
x4 

 2 x5 
= ∫  2x −  d x =
x −
=
16 − 6.4 = 9.6
0
32 
160  0


4
from which,
y=
9.6
= 1.80
5.3333
Hence, the centroid of the enclosed area is at (1.80, 1.80)
8. Sketch the curves y = 2x2 + 5 and y – 8 = x(x + 2) on the same axes and determine their points of
intersection. Calculate the coordinates of the centroid of the area enclosed by the two curves.
The curves are shown in the sketch below.
y – 8 = x(x + 2) is equivalent to y = 8 + x 2 + 2x
or
y = x 2 + 2x + 8
Equating the y-values gives: 2 x 2 + 5 = x 2 + 2x + 8 i.e. x 2 – 2x – 3 = 0
1175
© 2014, John Bird
i.e.
(x – 3)(x + 1) = 0
from which,
x = 3 and x = –1
When x = –1, y = 7 and when x = 3, y = 23
Hence, the coordinates of the points of intersection of the two curves occurs at (–1, 7) and
(3, 23)
3
=
x
x ( x 2 + 2 x + 8 ) − ( 2 x 2 + 5 )  d x
−1 
=
3
2 ( x 2 + 2 x + 8 ) − ( 2 x 2 + 5 )  d x
−1
xy d x ∫
∫=
∫ yd x ∫
−1
3
−1
3
∫ x ( − x + 2 x + 3) d x
∫ ( − x + 2 x + 3) d x
3
2
−1
3
2
−1
3
 x 4 2 x3 3x 2 
 81
27   1 2 3  
− +
+
− + 18 +  −  − − +  
3
2




−
+
+
2
3
d
x
x
x
x
(
)
3
2  −1
∫1
2   4 3 2 
 4
 4
= −
=
= 
3
2

1

 x3
10
2 + 3x 
−9 + 9 + 9 ) −  + 1 − 3  
(
−
+
x

 3

3
3


−1
3
−
=
80
24 2
2
10
+ 18 +
+
4
2 3 = 3 =1
2
2
10
10
3
3
Taking moments about 0x gives:
( )
(total area) y = (area of strip)(perpendicular distance of centroid of strip to 0x)
( ) ∫
i.e. (10.6666) y =
3
−1
( − x 2 + 2 x + 3)  ( − x 2 + 2 x + 3) + 2 x 2 + 5 d x
1
2

13 
3
+ 2 x + 3)  x 2 + x +  d x
−1
2
2
3
3
13
9
39 
= ∫  − x 4 − x3 − x 2 + 3 x 3 + 2 x 2 + 13 x + x 2 + 3 x +  d x
−1
2
2
2 
 2
∫ (−x
3
=
2
3
39 
 3
 3 x5 2 x 4 16 x 2 39 
= ∫  − x 4 + 2 x3 + 16 x +  d x =
+
+
+ x
−
−1
2 
4
2
2  −1
 2
 10
3

81
117   3 1
39  
=  −72.9 + + 72 +
 −  + + 8 −   = 108.8
2
2   10 2
2 

from which,
y=
108.8
= 10.20
10.6666
Hence, the centroid of the enclosed area is at (1, 10.20)
1176
© 2014, John Bird
EXERCISE 292 Page 794
1. A right-angled isosceles triangle having a hypotenuse of 8 cm is revolved one revolution about one
of its equal sides as axis. Determine the volume of the solid generated using Pappus’s theorem.
If the two equal sides of the isosceles triangle are each x, then by Pythagoras:
x2 + x2 =
82 i.e.
2 x 2 = 64
and x 2 = 32
from which, x =
32 = 5.657 cm
The triangle is shown below
(
Using Pappus, volume, V = ( area ) 2π y
)
For a triangle, the centroid lies at one-third of the perpendicular height above any side as base
Hence, y=
Thus,
1
× 5.657 cm = 1.8857 cm
3
(
)
1

volume, V = ( area ) 2π y =
 × 5.657 × 5.657  ( 2π ×1.8857 )
2

= 189.6 cm3
2. A rectangle measuring 10.0 cm by 6.0 cm rotates one revolution about one of its longest sides as
axis. Determine the volume of the resulting cylinder by using the theorem of Pappus.
The rectangle is shown below
1177
© 2014, John Bird
(
)
6
Using Pappus, volume, V = ( area ) 2π y =
(10 × 6 )  2π × 
2

= 360π = 1131 cm3
3. Using (a) the theorem of Pappus and (b) integration, determine the position of the centroid of a
metal template in the form of a quadrant of a circle of radius 4 cm (the equation of a circle, centre
0, radius r is x2 + y2 = r2).
(a) A sketch of the template is shown below.
(
Using Pappus, volume, V = ( area ) 2π y
)
(
14 3 1
2
 π r  = (π r ) 2π y
23
 4
i.e.
)
2 3
πr
4r 4(4)
3
= 1.70 cm
=
=
=
y
3π
3π
 π r2 

 ( 2π )
 4 
from which,
By symmetry, x = 1.70 cm
Hence, the centroid of the template is at (1.70, 1.70)
4
4
∫0 xy d x
(b) x =
=
4
∫ yd x
0
x 2 ) ) d x ∫ x (16 − x 2 ) 2 d x
∫0 (16 − =
0
=
4
4
2 − x2 d x
2 − x2 d x
4
4
(
)
(
)
∫0
∫0
4
x
(
1
4
3
 1
2 2
−
−
16
x
) 
 2(


3


2

0
x x
 42
−1
 2 sin 4 + 2
4
( 42 − x 2 ) 
0
(the numerator being an algebraic substitution – see Chapter 64 – and the
denominator being a sin θ substitution – see Chapter 65)
1178
© 2014, John Bird
i.e. x


1
 3 
64
− ( 0 ) − 16 2  


3




3
=

= 1.70
 ( 8sin −1 1 + 2(0) ) − ( 8sin −1 0 + 0(4) )   12.566




By symmetry, y = 1.70 cm
Hence, the centroid lies on the centre line OC (see diagram), at coordinates (1.70, 1.70)
(1.702 + 1.702 )
The distance from 0 is given by
= 2.40 cm
4. (a) Determine the area bounded by the curve y = 5x2, the x-axis and the ordinates x = 0 and x = 3
(b) If this area is revolved 360° about (i) the x-axis, and (ii) the y-axis, find the volumes of the
solids of revolution produced in each case.
(c) Determine the coordinates of the centroid of the area using (i) integral calculus, and (ii) the
theorem of Pappus.
(a) The area is shown in the sketch below
Shaded area =
∫
3
0
3
5 3
 5 x3 
5x =
dx  =
( 3 − 0 ) = 45 square units

 3 0 3
2
∫
(b)(i) Volume=
x -axis
3
0
dx
π y=
2
dx
∫ π ( 5 x )=
3
2
0
2
25π ∫
3
0
3
 x5 
x=
d x 25π  =
5π [35 − 0]

5
 0
4
= 1215π cubic units
(ii) Volume y -axis = (volume generated by x = 3) – (volume generated by y = 5x2)
=
∫
45
0
π (3) d y − ∫
2
45
0
45
45 
y
y2 
 y

π   d y =π ∫  9 −  d y =π 9 y − 
0
5
10  0
5




452 
= π  9(45) −
 − (0) = π [405 − 202.5] = 202.5π cubic units
10 


1179
© 2014, John Bird
3
(c)(i)
=
x
3
3
0
0
xy d x ∫
∫=
∫ yd x
3
 5x4 
5 4
2
3
(3 − 0)
x ( 5 x ) d x ∫ 5 x d x  4 
0
0
= 2.25
=
= = 4
45
45
45
45
3
0
3
1  25 x5 
1 3 2
1 3
5 5
4
y dx
25 x d x 2  5 
[3 − 0]
∫
∫


0
0
0
2 =
2
= 13.5
=
y 2=
=
3
45
45
45
∫ yd x
0
Hence, (2.25, 13.5) are the coordinates of the centroid
(
(ii) Using Pappus, volume generated when the shaded area is revolved about 0y = (area) 2π x
i.e.
from which,
(
202.5π = (45) 2π x
x=
)
202.5π
= 2.25
(45)(2π )
(
Similarly, volume generated when the shaded area is revolved about 0x = (area) 2π y
i.e.
from which,
(
1215π = (45) 2π y
y=
)
)
)
1215π
= 13.5
(40)(2π )
Hence, (2.25, 13.5) are the coordinates of the centroid
5. A metal disc has a radius of 7.0 cm and is of thickness 2.5 cm. A semicircular groove of diameter
2.0 cm is machined centrally around the rim to form a pulley. Determine the volume of metal
removed using Pappus’ theorem and express this as a percentage of the original volume of the disc.
Find also the mass of metal removed if the density of the metal is 7800 kg m–3
A side view of the rim of the disc is shown below.
1180
© 2014, John Bird
When area PQRS is rotated about axis XX the volume generated is that of the pulley. The centroid
of the semi-circular area removed is at a distance of
i.e.
4r
from its diameter, from Problem 3 above,
3π
4(1.0)
= 0.424 cm from PQ.
3π
Distance of centroid from XX = 7.0 – 0.424 = 6.576 cm
Distance moved in 1 revolution by the centroid = 2π(6.576) cm
π r 2 π (1.0) 2 π
Area of semicircle =
=
=
cm 2
2
2
2
By Pappus,
volume generated = area × distance moved by the centroid
π 
i.e. volume of metal removed =   ( 2π (6.576) ) = 64.90 cm3
2
Volume of =
disc = π r 2 h π=
( 7.0 ) ( 2.5) 384.845cm3
2
Thus, percentage of metal removed =
64.90
×100% = 16.86%
384.845
Mass of metal removed = density × volume
= 7800
kg
× 64.90 ×10−6 m3 = 0.5062 kg or 506.2 g
m3
1181
© 2014, John Bird