Some applications of integration 381
=
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
y dx
=
1
2
5
0
(5x − x
2
)
2 dx
5
0
(5x − x
2
) dx
=
1
2
5
0
(25x
2
− 10x
3
+ x
4
) dx
125
6
=
1
2
25x 3
3
−
10x 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
about x = 2.5 and thus x could have been determined
‘on sight’.)
Now try the following exercise
Exercise 150 Further problems on centroids
In Problems 1 and 2, find the position of the centroids of the areas bounded by the given curves, the
x-axis and the given ordinates.
1. y = 3x + 2 x = 0, x = 4
[(2.5, 4.75)]
2. y = 5x 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 = 4x − x
2 which lies above the x-axis.
[(2, 1.6)]
4. Find the co-ordinates of the centroid of the area
which lies between the curve y/x = x − 2 and
the x-axis.
[(1, −0.4)]
5. Sketch the curve y 2 = 9x between the limits
x = 0 and x = 4. Determine the position of the
centroid of this area.
[(2.4, 0)]
38.6 Theorem of Pappus
A theorem of Pappus states:
‘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 curve y = f (x)
is rotated one revolution about the x-axis between
the limits x = a and x = b, the volume V generated
is given by:
volume V = (A)(2π y ), from which, y =
V
2π A
y
C
Area A
x
x 5 b
x 5 a
y 5 f(x)
y
Figure 38.11
Problem 9. (a) Calculate the area bounded by the
curve y = 2x 2 , the x-axis and ordinates x = 0 and
x = 3. (b) If this area is revolved (i) about the
x-axis and (ii) about the y-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
y dx =
3
0
2x
2 dx
=
2x 3
3
3
0
= 18 square units
=
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
y dx
=
1
2
5
0
(5x − x
2
)
2 dx
5
0
(5x − x
2
) dx
=
1
2
5
0
(25x
2
− 10x
3
+ x
4
) dx
125
6
=
1
2
25x 3
3
−
10x 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
about x = 2.5 and thus x could have been determined
‘on sight’.)
Now try the following exercise
Exercise 150 Further problems on centroids
In Problems 1 and 2, find the position of the centroids of the areas bounded by the given curves, the
x-axis and the given ordinates.
1. y = 3x + 2 x = 0, x = 4
[(2.5, 4.75)]
2. y = 5x 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 = 4x − x
2 which lies above the x-axis.
[(2, 1.6)]
4. Find the co-ordinates of the centroid of the area
which lies between the curve y/x = x − 2 and
the x-axis.
[(1, −0.4)]
5. Sketch the curve y 2 = 9x between the limits
x = 0 and x = 4. Determine the position of the
centroid of this area.
[(2.4, 0)]
38.6 Theorem of Pappus
A theorem of Pappus states:
‘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 curve y = f (x)
is rotated one revolution about the x-axis between
the limits x = a and x = b, the volume V generated
is given by:
volume V = (A)(2π y ), from which, y =
V
2π A
y
C
Area A
x
x 5 b
x 5 a
y 5 f(x)
y
Figure 38.11
Problem 9. (a) Calculate the area bounded by the
curve y = 2x 2 , the x-axis and ordinates x = 0 and
x = 3. (b) If this area is revolved (i) about the
x-axis and (ii) about the y-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
y dx =
3
0
2x
2 dx
=
2x 3
3
3
0
= 18 square units