Higher Engineering Mathematics

380 INTEGRAL CALCULUS

=

625

3

−

625

4

125

2

−

125

3

=

625

12

125

6

=

(

625

12

)(

6

125

)

=

5

2

=2.5

y=

1

2

∫ 5

0

y^2 dx

∫ 5

0

ydx

=

1

2

∫ 5

0

(5x−x^2 )^2 dx

∫ 5

0

(5x−x^2 )dx

=

1

2

∫ 5

0

(25x^2 − 10 x^3 +x^4 )dx

125

6

=

1

2

[

25 x^3

3

−

10 x^4

4

+

x^5

5

] 5

0

125

6

=

1

2

(

25(125)

3

−

6250

4

+ 625

)

125

6

=2.5

Hence the centroid of the area lies at (2.5, 2.5).

(Note from Fig. 38.10 that the curve is symmetrical
aboutx= 2 .5 and thusxcould have been determined
‘on sight’.)

Now try the following exercise.

Exercise 151 Further problems on

centroids

In Problems 1 and 2, find the position of the cen-

troids of the areas bounded by the given curves,

thex-axis and the given ordinates.

1. y= 3 x+ 2 x=0,x= 4 [(2.5, 4.75)]


2. y= 5 x^2 x=1,x= 4 [(3.036, 24.36)]


3. Determine the position of the centroid of a
   sheet of metal formed by the curve
   y= 4 x−x^2 which lies above thex-axis.
   [(2, 1.6)]
   4. Find the co-ordinates of the centroid of the
   area which lies between the curvey/x=x− 2
   and thex-axis. [(1,−0.4)]
   5. Sketch the curvey^2 = 9 xbetween the limits
   x=0 andx=4. Determine the position of
   the centroid of this area.
   [(2.4, 0)]

38.6 Theorem of Pappus

A theorem of Pappusstates:

‘If a plane area is rotated about an axis in its own

plane but not intersecting it, the volume of the solid

formed is given by the product of the area and the

distance moved by the centroid of the area’.

With reference to Fig. 38.11, when the curvey=f(x)

is rotated one revolution about thex-axis between

the limitsx=aandx=b, the volumeVgenerated

is given by:

volume V=(A)(2πy), from which,y=

V

2 πA

Figure 38.11

Problem 9. (a) Calculate the area bounded by

the curvey= 2 x^2 , thex-axis and ordinatesx= 0

andx=3. (b) If this area is revolved (i) about the

x-axis and (ii) about they-axis, find the volumes

of the solids produced. (c) Locate the position

of the centroid using (i) integration, and (ii) the

theorem of Pappus.

(a) The required area is shown shaded in

Fig. 38.12.

Area=

∫ 3

0

ydx=

∫ 3

0

2 x^2 dx

=

[

2 x^3

3

] 3

0

=18 square units