Higher Engineering Mathematics

SOME APPLICATIONS OF INTEGRATION 379

H

where it balances perfectly, i.e. the lamina’scen-
tre of mass. When dealing with an area (i.e. a
lamina of negligible thickness and mass) the term
centre of areaorcentroidis used for the point
where the centre of gravity of a lamina of that shape
would lie.
Ifxandydenote the co-ordinates of the centroid
Cof 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 x^ =^ ax^ =^ bx

y

x

y

C

Area A

y = f(x)

Figure 38.9

Problem 7. Find the position of the centroid

of the area bounded by the curvey= 3 x^2 , the

x-axis and the ordinatesx=0 andx=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(3x^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

(3x^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(5x−x^2 )dx

∫ 5

0

(5x−x^2 )dx

=

∫ 5

0

(5x^2 −x^3 )dx

∫ 5

0

(5x−x^2 )dx

=

[

5 x^3

3 −

x^4

4

] 5

0

[

5 x^2

2 −

x^3

3

] 5

0

Figure 38.10