Higher Engineering Mathematics, Sixth Edition

380 Higher Engineering Mathematics

38.5 Centroids

Alaminais a thin flat sheet having uniform thickness.

Thecentre of gravityof a lamina is the point where

it balances perfectly, i.e. the lamina’scentre of mass.

When dealing with an area (i.e. a lamina of negligible

thickness and mass) thetermcentre of areaorcentroid

is used for the point where the centre of gravity of a

lamina of that shape would lie.

Ifxandydenote the co-ordinates of the centroidC

of areaAof Fig. 38.9, then:

x=

∫b

a

xydx

∫b

a

ydx

and y=

1

2

∫b

a

y^2 dx

∫b

a

ydx

0

Area A

x 5 ax 5 b

y

x

y 5 f(x)

y

x

C

Figure 38.9

Problem 7. Find the position of the centroid of

the area bounded by the curvey= 3 x^2 ,thex-axis

and the ordinatesx=0andx=2.

If(x,y)are co-ordinates of the centroid of the given

area then:

x=

∫ 2

0

xydx

∫ 2

0

ydx

=

∫ 2

0

x( 3 x^2 )dx

∫ 2

0

3 x^2 dx

=

∫ 2

0

3 x^3 dx

∫ 2

0

3 x^2 dx

=

[

3 x^4

4

] 2

0

[x^3 ]^20

=

12

8

=1.5

y=

1

2

∫ 2

0

y^2 dx

∫ 2

0

ydx

=

1

2

∫ 2

0

( 3 x^2 )^2 dx

8

=

1

2

∫ 2

0

9 x^4 dx

8

=

9

2

[

x^5

5

] 2

0

8

=

9

2

(

32

5

)

8

=

18

5

=3.6

Hence the centroid lies at (1.5, 3.6)

Problem 8. Determine the co-ordinates of

the centroid of the area lying between the curve

y= 5 x−x^2 and thex-axis.

y= 5 x−x^2 =x( 5 −x).Wheny=0,x=0orx=5.

Hence the curve cuts thex-axis at 0 and 5 as shown

in Fig. 38.10. Let the co-ordinates of the centroid be

(x,y)then, by integration,

x=

∫ 5

0

xydx

∫ 5

0

ydx

=

∫ 5

0

x( 5 x−x^2 )dx

∫ 5

0

( 5 x−x^2 )dx

=

∫ 5

0

( 5 x^2 −x^3 )dx

∫ 5

0

( 5 x−x^2 )dx

=

[

5 x^3

3 −

x^4

4

] 5

0

[

5 x^2

2 −

x^3

3

] 5

0

8

C

6

4

2

12345 x

y

y 55 x 2 x^2

y

x

0

Figure 38.10