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