APPLICATIONS
VOLUME BY INTEGRATION
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Volume by Integration, Lecture notes of Engineering Mathematics
The concept of solid of revolution and the methods to calculate its volume. It describes the Ring or Washer method, which is used when the element is perpendicular to but not touching the axis. It also explains the Shell method, which is used when the element is parallel to the axis of revolution. an example of finding the volume of a solid generated by revolving the second quadrant region bounded by the curve about R =1-x using the Ring or Washer method.
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2021/2022
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APPLICATIONS
VOLUME BY INTEGRATION
A solid of revolution is the figure formed when a plane region is revolved
about a fixed line. The fixed line is called the axis of revolution. For short,
we shall refer to the fixed line as axis.
The volume of a solid of revolution may be using the following
methods:
DISK, RING and SHELL METHOD
DISK METHOD: V = r
2
h
Find the volume of the solid generated by revolving the region bounded by the
line y = 6 – 2x and the coordinate axes about the y-axis.
r =x
( x , y )
h = dy
( 0 , 6 )
x
y
0
(3,0)
𝑉 =
0
6
𝜋 𝑥
2
𝑑𝑦
𝑉 =
0
6
𝜋
[
1
2
]
2
𝑑𝑦
𝑉 =
𝜋
4
0
6
( 6 − 𝑦 )
2
𝑑𝑦
( 6 − 𝑦 )
3
|
0
3
𝑥=
1
2
( 6 − 𝑦
)
𝑖𝑓 𝑦 = 6 − 2 x ;
𝑉 = −
𝜋
12
[
( 6 − 6
)
3
−
( 6 − 0
)
3
]
V r h
2
Ring or Washer method is used when the element (or representative strip) is perpendicular to
but not touching the axis. Since the axis is not a part of the boundary of the plane area, the strip
when revolved about the axis generates a ring or washer.
B. RING OR WASHER METHOD: V = (R
2
- r
2
)h
( x
1
, y
1
)
( x
2
, y
2
)
x = a
x = b
dx
h = dx
y
1
= g(x)
y
2
= f ( x )
❑
❑
𝑑𝑉 =
❑
❑
𝜋
[
𝑦
1
2
− 𝑦
2
2
]
𝑑𝑥
Since
𝑦
1
= 𝑓 (𝑥)
𝑦
2
=𝑔 (𝑥)
𝑉 =𝜋
𝑎
𝑏
[ 𝑔 ( 𝑥 )
2
− 𝑓 ( 𝑥 )
2
] 𝑑𝑥
r
R
C. SHELL METHOD:
The method is used when the
element (or representative strip) is
parallel to the axis of revolution.
When this strip is revolved about
the axis, the solid formed is of
hollow cylindrical form.
𝑉
𝑠 𝑒𝑙𝑙 h
= 2 𝜋 𝑟𝑡 ⋅ h
SHELL METHOD:
Find the volume of the solid generated by revolving the second quadrant region bounded
by the curve about the line.
𝒙
𝟐
= 𝟒 − 𝒚
𝑑𝑉 = 2 𝜋 ( 1 − 𝑥 ) 𝑦𝑑𝑥
𝑉 = 2 𝜋 𝑟 𝑡 h
𝑉 = 2 𝜋
− 2
0
( 1 − 𝑥 ) 𝑦 𝑑𝑥
0
− 2
0
− 2
but x
2
= 4 − 𝑦 ; y =4− x
2
0
− 2
[
( 4 − 𝑥
2
) − 𝑥
( 4 − 𝑥
2
) ] 𝑑𝑥
0
− 2
( 4 − 4 𝑥 − 𝑥
2
3
) 𝑑𝑥
𝑉 =
56 𝜋
3
𝑐𝑢. 𝑢𝑛𝑖𝑡𝑠