Some applications of integration 383
When area PQRS is rotated about axis XX the volume generated is that of the pulley. The centroid of the
semicircular area removed is at a distance of
4r
3π
from its
diameter (see ‘Engineering Mathematics 6th edition’,
Chapter 58), i.e.
4(1.0)
3π
, i.e. 0.424cm from PQ. Thus
the distance of the centroid from XX is 5.0 − 0.424,
i.e. 4.576 cm.
The distance moved through in one revolution by the
centroid is 2π(4.576) cm.
Area of semicircle =
πr 2
2
=
π(1.0) 2
2
=
π
2
cm 2
By the theorem of Pappus,
volume generated = area × distance moved by
centroid =
π
2
(2π)(4.576).
i.e. volume of metal removed = 45.16 cm 3
Mass of metal removed = density × volume
= 8000 kg m −3 ×
45.16
10 6 m 3
= 0.3613 kg or 361.3 g
volume of pulley = volume of cylindrical disc
− volume of metal removed
= π(5.0) 2 (2.0) − 45.16
= 111.9 cm
3
Mass of pulley = density× volume
= 8000 kg m −3 ×
111.9
10 6 m 3
= 0.8952 kg or 895.2 g
Now try the following exercise
Exercise 151 Further problems on the
theorem of Pappus
1. A right angled isosceles triangle having a
hypotenuse of 8 cm is revolved one revolution
about one of its equal sides as axis. Determine the volume of the solid generated using
Pappus’ theorem.
[189.6 cm 3 ]
2. Using (a) the theorem of Pappus, and (b) integration, determine the position of the centroid
of a metal template in the form of a quadrant
of a circle of radius 4 cm. (The equation of a
circle, centre 0, radius r is x
2
+ y
2
= r
2 ).
⎡
⎢
⎢
⎢
⎣
On the centre line, distance
2.40 cm from the centre,
i.e. at co-ordinates
(1.70, 1.70)
⎤
⎥
⎥
⎥
⎦
3. (a) Determine the area bounded by the curve
y = 5x
2 , the x-axis and the ordinates
x = 0 and x = 3.
(b) If this area is revolved 360 ◦ about (i) the
x-axis, and (ii) the y-axis, find the volumes of the solids of revolution produced
in each case.
(c) Determine the co-ordinates of the centroid of the area using (i) integral calculus, and (ii) the theorem of Pappus.
⎡
⎢
⎢
⎢
⎣
(a) 45 square units
(b) (i) 1215π cubic units
(ii) 202.5π cubic units
(c) (2.25, 13.5)
⎤
⎥
⎥
⎥
⎦
4. A metal disc has a radius of 7.0 cm and is
of thickness 2.5 cm. A semicircular groove of
diameter 2.0 cm is machined centrally around
the rim to form a pulley. Determine the volume of metal removed using Pappus’ theorem
and express this as a percentage of the original volume of the disc. Find also the mass of
metal removed if the density of the metal is
7800 kg m −3 .
[64.90 cm 3 , 16.86%, 506.2 g]
For more on areas, mean and r.m.s. values, volumes and
centroids, see ‘Engineering Mathematics 6th edition’,
Chapters 55 to 58.
38.7 Second moments of area of
regular sections
The first moment of area about a fixed axis of a lamina
of area A, perpendicular distance y from the centroid
of the lamina is defined as Ay cubic units. The second
moment of area of the same lamina as above is given
by Ay 2 , i.e. the perpendicular distance from the centroid
of the area to the fixed axis is squared.
When area PQRS is rotated about axis XX the volume generated is that of the pulley. The centroid of the
semicircular area removed is at a distance of
4r
3π
from its
diameter (see ‘Engineering Mathematics 6th edition’,
Chapter 58), i.e.
4(1.0)
3π
, i.e. 0.424cm from PQ. Thus
the distance of the centroid from XX is 5.0 − 0.424,
i.e. 4.576 cm.
The distance moved through in one revolution by the
centroid is 2π(4.576) cm.
Area of semicircle =
πr 2
2
=
π(1.0) 2
2
=
π
2
cm 2
By the theorem of Pappus,
volume generated = area × distance moved by
centroid =
π
2
(2π)(4.576).
i.e. volume of metal removed = 45.16 cm 3
Mass of metal removed = density × volume
= 8000 kg m −3 ×
45.16
10 6 m 3
= 0.3613 kg or 361.3 g
volume of pulley = volume of cylindrical disc
− volume of metal removed
= π(5.0) 2 (2.0) − 45.16
= 111.9 cm
3
Mass of pulley = density× volume
= 8000 kg m −3 ×
111.9
10 6 m 3
= 0.8952 kg or 895.2 g
Now try the following exercise
Exercise 151 Further problems on the
theorem of Pappus
1. A right angled isosceles triangle having a
hypotenuse of 8 cm is revolved one revolution
about one of its equal sides as axis. Determine the volume of the solid generated using
Pappus’ theorem.
[189.6 cm 3 ]
2. Using (a) the theorem of Pappus, and (b) integration, determine the position of the centroid
of a metal template in the form of a quadrant
of a circle of radius 4 cm. (The equation of a
circle, centre 0, radius r is x
2
+ y
2
= r
2 ).
⎡
⎢
⎢
⎢
⎣
On the centre line, distance
2.40 cm from the centre,
i.e. at co-ordinates
(1.70, 1.70)
⎤
⎥
⎥
⎥
⎦
3. (a) Determine the area bounded by the curve
y = 5x
2 , the x-axis and the ordinates
x = 0 and x = 3.
(b) If this area is revolved 360 ◦ about (i) the
x-axis, and (ii) the y-axis, find the volumes of the solids of revolution produced
in each case.
(c) Determine the co-ordinates of the centroid of the area using (i) integral calculus, and (ii) the theorem of Pappus.
⎡
⎢
⎢
⎢
⎣
(a) 45 square units
(b) (i) 1215π cubic units
(ii) 202.5π cubic units
(c) (2.25, 13.5)
⎤
⎥
⎥
⎥
⎦
4. A metal disc has a radius of 7.0 cm and is
of thickness 2.5 cm. A semicircular groove of
diameter 2.0 cm is machined centrally around
the rim to form a pulley. Determine the volume of metal removed using Pappus’ theorem
and express this as a percentage of the original volume of the disc. Find also the mass of
metal removed if the density of the metal is
7800 kg m −3 .
[64.90 cm 3 , 16.86%, 506.2 g]
For more on areas, mean and r.m.s. values, volumes and
centroids, see ‘Engineering Mathematics 6th edition’,
Chapters 55 to 58.
38.7 Second moments of area of
regular sections
The first moment of area about a fixed axis of a lamina
of area A, perpendicular distance y from the centroid
of the lamina is defined as Ay cubic units. The second
moment of area of the same lamina as above is given
by Ay 2 , i.e. the perpendicular distance from the centroid
of the area to the fixed axis is squared.