chapter 75 centroids of simple...

© 2014, John Bird

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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

( )[ ]

[ ][ ]

333 3 3

20 0 0 0

3 3 3 3200 0 0

2d 2 d 2 d 3 18

9d 2 d 2 d

xxy x x x x x x

xxy x x x x x

= = = = =∫ ∫ ∫∫ ∫ ∫

= 2

( )( ) [ ]

3 3 22 3330 0 23 0

00

1 1d 2 d 1 4 22 2 4 d 9 09 18 18 3 9d

y x x x xy x xy x

= = = = = −

∫ ∫∫

∫= 2

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


© 2014, John Bird

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( ) [ ] [ ][ ]

4 4 4 42 3 20 0 0 0

4 4 4 420 0 0

0

d 3 2 d 3 2 d 64 16 8024 8 323d 3 2d 3 2d 2

2

xy x x x x x x x x xx

xy x x x x x x

+ + + += = = = = =

+ + + +

∫ ∫ ∫∫ ∫ ∫

= 2.50

( )( ) [ ]

4 4 224 40 0 2 3 2

4 00

0

1 1d 3 2 d 1 12 2 9 12 4 d 3 6 432 64 64d

y x x xy x x x x x x

y x

+= = = + + = + +

∫ ∫∫

∫

= [ ]1 304192 96 1664 64

+ + = = 4.75

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.

( ) [ ]

[ ]

444 4 4

4 42 31 1 1 1

4 4 4 432 2 3 31 1 1

1

5 5 54 1 (255)d 5 d 5 d 4 318.754 45 5 1055d 5 d 5 d 4 1 (63)3 33

xxy x x x x x x

xxy x x x x x

−

= = = = = = = −

∫ ∫ ∫∫ ∫ ∫

= 3.036

( )[ ]

4 4 22 2 4541 1 4 5 54 1

11

1 1d 5 d 1 1 25 5 52 2 25 d 4 1 (1023)105 210 210 5 210 210d

y x x x xy x xy x

= = = = = − =

∫ ∫∫

∫

= 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


© 2014, John Bird

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( )[ ]

252 2 2

3 40 0 0 0

2 2 2 243 30 0 0

0

2 64d 2 d 2 d 5 5

82d 2 d 2 d4

xxy x x x x x x

xxy x x x x x

= = = = =

∫ ∫ ∫∫ ∫ ∫

= 1.60

( )( ) [ ]

2 2 22 3 2720 0 62 0

00

1 1d 2 d 1 4 12 2 4 d 128 08 16 16 7 28d

y x x x xy x xy x

= = = = = −

∫ ∫∫

∫= 4.57

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.

( )04 3

0 0 02 3 2

1 1 1 10 0 0 022 2

31 1 11

3 3 10d 3 d 3 d 4 3 4 31d 3 d 3 d 0 122

x xxy x x x x x x x x

xxy x x x x x x x x

− − − −

− − −

−

+ − − + + = = = = = + + − − ++

∫ ∫ ∫∫ ∫ ∫

= 0.4166660.5

− = –0.833

© 2014, John Bird

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( )( )

0 0 22 2 05 4 301 1 4 3 20 1

11

1 1d 3 d 9 62 2 9 6 d0.5 5 4 3d

y x x x x x x xy x x x xy x

− −

−−

−

+ = = = + + = + +

∫ ∫∫

∫

= 9 3 1 9 3 105 2 3 5 2 3

− − + − = − + = 0.633

Hence, the centroid lies at (–0.833, 0.633)

© 2014, John Bird

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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, 2x =

( )

( )

( )

( )

43 4 54 4 422 2 2 3 4

0 0 0 04 4 4 432 2

20 0 00

1 16 81 1 1d 4 d 16 8 d 2 3 4 52 2 2d 4 d 4 d 2

3

x x xy x x x x x x x x

yxy x x x x x x x x

− +− − + = = = =

− − −

∫ ∫ ∫∫ ∫ ∫

= ( )1 341.333 512 204.8 (0) 17.06652

64 10.66632 (0)3

− + − =

− −

= 1.6

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 yx

= x – 2 and the x-

axis.

yx

= x – 2 i.e. y = 2 2x x− = x(x – 2) Hence, when y = 0 (i.e. the x-axis), x = 0 and x = 2

A sketch of y = 2 2x x− is shown below

© 2014, John Bird

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By symmetry, 1x =

( )

( )

( )

( )

25 32 2 2 422 2 4 3 2

0 0 0 02 2 2 232 2

20 0 00

1 41 1 1d 2 d 4 4 d 2 5 32 2 2d 2 d 2 d

3

x xxy x x x x x x x xy

xy x x x x x x x x

− +− − + = = = =

− − −

∫ ∫ ∫∫ ∫ ∫

= ( )( )

1 6.4 16 10.6667 (0) 0.5333522.6667 4 (0) 1.3333

− + − =

− − − = –0.4

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, 0x =

( )

( )

( )

( )

353 3 3 322 2 2 43 3 3 3

3 3 3 332 23 3 3

3

11 1 1 81 6d 9 d 81 18 d 2 52 2 2d 9 d 9 d 9

3

xx xy x x x x x xy

xy x x x x x x

− − − −

− −

−

− +− − + = = = =

− − −

∫ ∫ ∫∫ ∫ ∫

= ( )

( )

1 243 162 48.6 ( 243 162 48.6) 129.6227 9 ( 27 9) 36

− + − − + − =

− − − + = 3.6

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

© 2014, John Bird

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424 4

20 0 04 43/24

00

0

11 1d d 2 8 1 32 2 48 81d d 32 2 3 / 2

yyx x xx

y yx x x

= = = = =

∫ ∫∫ ∫

= 0.375

45/23/24 44

0 00 04

0

1( )d dd 2 5 / 2 3 32 3 64 122 2

8 8 8 16 5 / 2 16 5 5d3 3 3

yy yy y yxy yy

x y

= = = = = = =

∫ ∫∫∫

= 2.40

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.

© 2014, John Bird

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( )( )

( ) ( )[ ]

552

15 35 5 52 200 0 05 5 1 1 55 5

32 20 0 0 0 2

0

5 55 dd 5 d 5 d 50 02

16.6667d 5 d 5 d 5 d5 3

2

x

x x xxy x x x x x xx

y x x x x x x xx

− = = = = = =

∫∫ ∫ ∫∫ ∫ ∫ ∫

= 3.0

( )( ) ( ) ( )

25 52 5250 0

5 00

0

1 1d 5 d 1 1 5 12 2 5 d 62.5 016.6667 33.3333 33.3333 2 33.3333d

y x x x xy x xy x

= = = = = −

∫ ∫∫

∫ = 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, 0y =

( )

452

34 4 45 52

0 0 0 04 4 1 44 3 3

320 0 0 2

0

35 6 6 64 0 (32)d 3 d 3 d 2 5 5 5

2(8)2 2 4 0d 3 d 3 d3

32

x

xxy x x x x x xx

xy x x x x xx

−

= = = = = = = −

∫ ∫ ∫∫ ∫ ∫

= 38.416

= 2.4

© 2014, John Bird

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Hence, the centroid is at (2.4, 0)

7. Calculate the points of intersection of the curves x2 = 4y and 2

4y = x, and determine the position of

the centroid of the area enclosed by them.

The curves are shown in the sketch below.

Since 2 4x y= then

2

4xy = and

42

16xy =

and for the second curve, 2

4y x= i.e. 2 4y x=

Hence, equating the 2y values gives: 4

416x x= and 4 64x x= i.e. 4 64 0x x− =

from which, 3( 64) 0x x − = giving x = 0 and 3 64 0x − = i.e. 3 64x = 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)

45

42

24 3 34452

0 00 04 412 24 4

3320 30 0 2

0

25 16 4 256 1282 d 4 162 dd 24 5 16 54

4 64 32 16d 2 d 2 d 44 4 3 12 3 32

3 122

x xx xx x x x xxy x

x x xy x x x x xx x

− − − −− = = = = = = − − − −

−

∫ ∫∫∫ ∫ ∫

= 9.65.3333

= 1.80

© 2014, John Bird

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Taking moments about 0x gives: (total area) ( )y = (area of strip)(perpendicular distance of centroid of strip to 0x)

i.e. (5.3333) ( )y = 2 2 24

0

12 2 d4 2 4 4x x xx x x − − +

∫

= 2 2 5/2 5/2 44 4

0 02 d 2 d

4 8 4 4 32x x x x xx x x x x − + = + − −

∫ ∫

= 44 54

20

0

2 d 16 6.432 160x xx x x − = − = − ∫ = 9.6

from which, 9.65.3333

y = = 1.80

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 + 2x + 2x or y = 2x + 2x + 8

Equating the y-values gives: 22 5x + = 2x + 2x + 8 i.e. 2x – 2x – 3 = 0

© 2014, John Bird

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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 3 32 2 2

1 1 13 3 3

2 2 21 1 1

d 2 8 2 5 d 2 3 d

d 2 2 8 2 5 d 2 3 d

xy x x x x x x x x x xx

y x x x x x x x x− − −

− − −

+ + − + − + + = = =

+ + − + − + +

∫ ∫ ∫∫ ∫ ∫

= ( )

( )

34 3 23

3 21 1

332

1

2 3 81 27 1 2 3182 3 d 4 3 2 4 2 4 3 221 109 9 9 1 33 333

x x xx x x x

x x x

− −

−

− + + − + + − − − + − + + = = − + + − + −− + +

∫

=

80 24 2 218 104 2 3 3

2 210 103 3

− + + += = 1

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

2 2 21

12 3 2 3 2 5 d2

x x x x x x−

− + + − + + + + ∫

( )

32 2

1

34 3 2 3 2 2

1

3 132 3 d2 2

3 13 9 393 2 13 3 d2 2 2 2

x x x x x

x x x x x x x x x

−

−

= − + + + +

= − − − + + + + + +

∫

∫

= 35 4 23

4 31

1

3 39 3 2 16 392 16 d2 2 10 4 2 2

x x xx x x x x−

−

− + + + = − + + + ∫

= 81 117 3 1 3972.9 72 82 2 10 2 2

− + + + − + + − = 108.8

from which, 108.810.6666

y = = 10.20

Hence, the centroid of the enclosed area is at (1, 10.20)

© 2014, John Bird

1177


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: 2 2 28x x+ = i.e. 22 64x = and 2 32x = from which, x = 32 = 5.657 cm The triangle is shown below

Using Pappus, volume, V = ( )( )area 2 yπ

For a triangle, the centroid lies at one-third of the perpendicular height above any side as base

Hence, 1 5.657cm3

y = × = 1.8857 cm

Thus, volume, V = ( )( ) ( )1area 2 5.657 5.657 2 1.88572

yπ π = × × ×

= 189.6 3cm 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

© 2014, John Bird

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Using Pappus, volume, V = ( )( ) ( ) 6area 2 10 6 22

yπ π = × ×

= 360π = 1131 3cm

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 2 yπ

i.e. ( )( )3 21 4 1 22 3 4

r r yπ π π =

from which, ( )

3

2

24 4(4)33 32

4

r ryr

π

π π ππ= = =

= 1.70 cm

By symmetry, 1.70cmx =

Hence, the centroid of the template is at (1.70, 1.70)

(b) ( )( )( )

( )

( )

( )

( )

432 2

14 4 42 2 2

0 0 0 04 4 4 422 2 2 2

1 2 20 0 00

1 162

3d 16 d 16 d 2

4d 4 d 4 d sin 42 4 2

x

xy x x x x x x xx

x xy x x x x x x−

− − − −

= = = = − − + −

∫ ∫ ∫∫ ∫ ∫

(the numerator being an algebraic substitution – see Chapter 64 – and the

denominator being a sin θ substitution – see Chapter 65)

© 2014, John Bird

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i.e. ( )

( ) ( )

32

1 1

1 640 163 3

12.5668sin 1 2(0) 8sin 0 0(4)x

− −

− − = = + − +

= 1.70

By symmetry, 1.70cmy =

Hence, the centroid lies on the centre line OC (see diagram), at coordinates (1.70, 1.70)

The distance from 0 is given by ( )2 21.70 1.70+ = 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 = ( )333

2 30

0

5 55 d 3 03 3xx x = = − ∫ = 45 square units

(b)(i) ( ) [ ]353 3 322 2 4 5-

0 0 00

Volume d 5 d 25 d 25 5 3 05

x axisxy x x x x xπ π π π π = = = = = − ∫ ∫ ∫

= 1215π cubic units

(ii) -Volume y axis = (volume generated by x = 3) – (volume generated by y = 5x2)

= 45245 45 45

20 0 0

0

(3) d d 9 d 95 5 10y y yy y y yπ π π π − = − = − ∫ ∫ ∫

= 2459(45) (0) [405 202.5]

10π π − − = −

= 202.5π cubic units

© 2014, John Bird

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(c)(i) ( ) ( )

343 3 3

42 30 0 0 0

3

0

5 5 3 0d 5 d 5 d 4 445 45 45 45d

xxy x x x x x x

xy x

−

= = = = =∫ ∫ ∫∫

= 2.25

[ ]

353 3

2 4 50 0 03

0

1 251 1 5d 25 d 3 02 52 2 245 45 45d

xy x x x

yy x

−

= = = =∫ ∫∫

= 13.5

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) ( )2 xπ

i.e. 202.5π = (45) ( )2 xπ

from which, 202.5(45)(2 )

x ππ

= = 2.25

Similarly, volume generated when the shaded area is revolved about 0x = (area) ( )2 yπ

i.e. 1215π = (45) ( )2 yπ

from which, 1215(40)(2 )

y ππ

= = 13.5

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.

© 2014, John Bird

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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 43

rπ

from its diameter, from Problem 3 above,

i.e. 4(1.0)3π

= 0.424 cm from PQ.

Distance of centroid from XX = 7.0 – 0.424 = 6.576 cm

Distance moved in 1 revolution by the centroid = 2π(6.576) cm

Area of semicircle = 2 2

2(1.0) cm2 2 2rπ π π

= =

By Pappus, volume generated = area × distance moved by the centroid

i.e. volume of metal removed = ( )2 (6.576)2π π

= 364.90cm

Volume of disc = ( ) ( )22 37.0 2.5 384.845cmr hπ π= =

Thus, percentage of metal removed = 64.90 100%384.845

× = 16.86%

Mass of metal removed = density × volume

= 7800 3

kgm

6 364.90 10 m−× × = 0.5062 kg or 506.2 g

chapter 75 centroids of simple...

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