Higher Engineering Mathematics, Sixth Edition

382 Higher Engineering Mathematics

y

0

6

12

18

123 x

y 52 x^2

x

y

Figure 38.12

(b) (i) When the shaded area of Fig. 38.12 is

revolved 360◦about thex-axis, the volume

generated

=

∫ 3

0

πy^2 dx=

∫ 3

0

π( 2 x^2 )^2 dx

=

∫ 3

0

4 πx^4 dx= 4 π

[

x^5

5

] 3

0

= 4 π

(

243

5

)

=194.4πcubic units

(ii) When the shaded area of Fig. 38.12 is

revolved 360◦about they-axis, the volume

generated

=(volume generated byx= 3 )

−(volume generated byy= 2 x^2 )

=

∫ 18

0

π( 3 )^2 dy−

∫ 18

0

π

(y

2

)

dy

=π

∫ 18

0

(

9 −

y

2

)

dy=π

[

9 y−

y^2

4

] 18

0

= 81 πcubic units

(c) If the co-ordinates of the centroid of the shaded

area in Fig. 38.12 are(x,y)then:

(i) by integration,

x=

∫ 3

0

xydx

∫ 3

0

ydx

=

∫ 3

0

x( 2 x^2 )dx

18

=

∫ 3

0

2 x^3 dx

18

=

[

2 x^4

4

] 3

0

18

=

81

36

=2.25

y=

1

2

∫ 3

0

y^2 dx

∫ 3

0

ydx

=

1

2

∫ 3

0

( 2 x^2 )^2 dx

18

=

1

2

∫ 3

0

4 x^4 dx

18

=

1

2

[

4 x^5

5

] 3

0

18

=5.4

(ii) using the theorem of Pappus:

Volume generated when shaded area is

revolved aboutOY=(area)( 2 πx).

i.e. 81 π=( 18 )( 2 πx),

from which, x=

81 π

36 π

=2.25

Volume generated when shaded area is

revolved aboutOX=(area)( 2 πy).

i.e. 194. 4 π=( 18 )( 2 πy),

from which, y=

194. 4 π

36 π

=5.4

Hence the centroid of the shaded area in

Fig. 38.12 is at (2.25, 5.4).

Problem 10. A metal disc has a radius of 5.0cm

and is of thickness 2.0cm. A semicircular groove of

diameter 2.0cm is machined centrally around the

rim to form a pulley. Determine, using Pappus’

theorem, the volume and mass of metal removed

and the volume and mass of the pulley if the density

of the metal is 8000kgm−^3.

A side view of the rim of the disc is shown in Fig. 38.13.

5.0 cm

2.0 cm

X X

SR

QP

Figure 38.13