Volume of Revolution, Shell Method
A flat sheet of plastic is of length L, width W and thickness dx
What is the volume of this sheet?
The volume dV of the sheet is:
dV = L W dxL
This sheet is rolled into a hollow cylinder of height L and radius r
L
r
The width W of the sheet has now become the circumference of a circle of radius r
rThe circumference C of a circle of radius r is C = 2r
This is called a shell.W
W
The volume dV of the sheet is:
dV = L W dx
L
Since W has become C, the volume dV of the shell can be written as:
L
rC = 2r
dV = L 2r dx
If L and r are functions of x then dV is: dV = 2 L(x)r(x) dx
If L and r are functions of y then dV is: dV = 2 L(y)r(y) dy
Consider the graph of a function f(x)
L
The area enclosed by this graph with the x-axis from x = a to x = b is shaded.
x = a x = b
f(x)
We will use the shell method to find the volume generated when this area is revolved about the x-axis.
Take a strip of area dA of width dx parallel to the y-axis
Take note: For shell method, dA is taken parallel to the axis of revolution.
What is the length of this area strip?
Length of the strip = upper function – lower function.
dA
dx
= f(x) – 0 = f(x)
L(x) = f(x)
L(x) = f(x)
Revolution on the y-axis
If this area is revolved about the y-axis, a shell is generated.
L
x = a x = b
f(x)
What is the radius of this shell?
Radius of the shell is the distance of dA from the axis of revolution.
The distance of dA from the y-axis is x
The radius of the shell is:
dA
dxr(x) = x
The length of the shell is: L(x) = f(x)
L(x) = f(x)
The volume dV of the shell is:
dV = 2 L(x)r(x) dx
The volume V generated by revolving the area from x = a to x = b is:
2 ( ) ( )b
a
V r x L x dx
2 ( ) ( )b
a
V r x L x dx
Revolution on the x-axis
L
f(y)
In order to revolve this area about the x-axis, dA of width dy is taken parallel to the x-axis
To find the length of dA, the functions must of y
dy
f(y) is on the right of dA and g(y) on its left.
L(y) = f(y) – g(y)
The length of dA is
When dA is revolved about the x-axis, a shell is generated
g(y)
L(y) = f(y) – g(y)
The radius r(y) of the shell is the distance of dA from the axis of revolution.
The distance of dA from the x-axis is y
yy
r(y) = y
dA
L
f(y)
dAdy
L(y) = f(y) – g(y)
r(y) = y
L(y) = f(y) – g(y)
The volume dV of the shell generated is:
ydV = 2 r(y)L(y) dy
The volume V generated by revolving the area from y = c to y = d is:
y = c
y = d
2 ( ) ( )d
c
V r y L y dy
2 ( ) ( )d
c
V r y L y dy
L
To use shell method to obtain the volume of revolution, we use the following steps
1. Take dA parallel to the axis
2. Obtain the length of dA
2 ( ) ( )d
c
V r y L y dy 2 ( ) ( )b
a
V r x L x dx
If dA is vertical, L(x) = g(x) – f(x)
If dA is horizontal, L(y) = g(y) – f(y)
3. Obtain the radius of revolution
Radius r(x) or r(y) is the distance of dA from the axis of revolution.
3. Use the appropriate formula to obtain V
= Function above – function below
= right function – left function
Draw the graphs of y = x2, y = 0, x = 4 and shade the area enclosed.
y = 0
To revolve this area about the y-axis, take dA parallel to it.
x = 4
dx
The length of dA is: L(x) = x2 – 0
Use the shell method to find the volume of revolution of the area enclosed by y = x2, y = 0 and x = 4 about the y axis.
dA
y = x2
= x2
x2
When dA is revolved about the y-axis, the radius of the shell generated is the distance of dA from y-axis.
xr(x) = x
L(x) = x2
xdxThe volume dV of the shell is:
dV = 2 r(x)L(x) dx
= 2 x · x2 dx
= 2 x3 dx
y = 0
x = 4
dx
dA
y = x2
x2
x
L(x) = x2
xdx
dV= 2 x3 dx
The position of dA can change from x = 0 to x = 4
The volume V generated by revolving the area enclosed about the y-axis is:
2 ( ) ( )b
a
V r x L x dx 32
b
a
x dx 44
0
24
x
442
4
128
y = 0
x = 4
dy
Use the shell method to find the area of revolution of the area enclosed by y = x2, y = 0 and x = 4 about the x axis.
dA
y = x2
y
Take dA of width dy parallel to the x-axis.
Express the functions to the left and right of dA as functions of y
The function on the left is: y = x2
Solving for x gives: x y( )f y y
( )f y y
The function on the right is: x = 4
g(y) = 4
The length of dA is:
L(y) = g(y) – f(y) 4 y The radius of revolution is the distance of dA from the x-axis
4 y
r(y) = y
y = 0
x = 4
dydA
y = x2
y
To find the limits, we find the points of intersection of x = y and x = 4
y = 4 gives: y = 16
The position of dA varies from y = 0 to y = 16
( )f y y
The volume of revolution is: 4 y
y = 16
2 ( ) ( )d
c
V r y L y dy
16
0
2 4y y dy
( ) 4L y y r(y) = y
y = 0
x = 4
dydA
y = x2
y
( )f y y
4 y
y = 16
16
0
2 4V y y dy
16
3/ 2
0
2 4y y dy 162 5 / 2
0
42
2 5/ 2
y y
2 5/24 16 162
2 5 / 2
1024
5
x = 0
Use the shell method to find the volume of revolution of the area enclosed by y = 2x, x = 0 and y = 4 about the y axis.
Take dA of width dx parallel to the y-axisdx
The length of dA is:
y = 4
Draw the graphs of y = 2x, x = 0, y = 4 and shade the area enclosed.
y = 2x
L(x) = 4 – 2x 4 – 2x
r(x) = xx
The volume of revolution V is:
2
0
2 (4 2 )x x dx
Obtain limits by solving 2x = 4 x = 2
2 ( ) ( )b
a
V r x L x dx x = 2
x = 0
dx y = 4
y = 2x
4 – 2x
x
2
0
2 (4 2 )x x dx 2
2
0
2 (4 2 )x x dx 22 3
0
2 4 22 3
x x
2 32 22 4 2
2 3
16
3
x = 2
Find the volume generated by revolving the area enclosed by y = x2 and y = 4x – x2 about the line x = 2
Draw the graphs of y = x2, y = 4x – x2 and x = 2 and shade the area enclosed. y = x2
y = 4x – x2
x = 2
Take dA of width dx parallel to x = 2
Find the points of intersection of y = x2 and y = 4x – x2.
4x – x2 = x2
4x – x2 – x2 = 0
2x(2 – x) = 0 x = 0, 2
x = 0 x = 2
dx
The length of dA is:
L(x) = 4x – x2 – x2
The distance of dA from x = 2 is:
x
2
2 – x
2 – x
r(x) = 2 – x
= 4x –2x2
y = x2
y = 4x – x2
x = 2
x = 0 x = 2
dx
L(x) = 4x – 2x2
The volume of revolution V is:
x
2
2 – x
r(x) = 2 – x
2
0
2 ( ) ( )V r x L x dx 2
2
0
2 (2 )(4 2 )x x x dx 2
2
0
2 (2 )(4 2 )x x x dx 2
2 3
0
2 (8 8 2 )x x x dx
y = x2
y = 4x – x2
x = 2
x = 0 x = 2
dx
x
2
2 – x
22 3
0
2 (8 8 2 )x x x dx 22 3 4
0
2 8 8 22 3 4
x x x
2 3 42 2 22 8 8 2
2 3 4
64
2 16 83
16
3
Use the shell method to find the volume of the solid generated by revolving the region bounded by y = 2 – x, y = 0 and x = 4 about the x-axis. y = 2 – x
Draw the graphs of y = 2 – x , y = 0 and x = 4 and shade the area enclosed.
y = 0
x = 4
Draw dA of width dy parallel to the x-axis.dA dy
In order to find the length of dA, we need to write the functions on the left and right of dA as functions of y
y = 2 – x Solving for x gives:
x = 2 – y
x = 2 – y
L(y) = 4 – (2 – y)
4
dA is below the x-axis.
The distance of dA from the x-axis is: – y
– y
r(y) = – y
L(y) = 2 + y
y = 2x
y = 0
x = 4
dA dy2 – y
4
– y
r(y) = – y L(y) = 2 + y
The volume of revolution V is:0
2
2 ( ) ( )V r y L y dy
To find the limits, solve 2 – y = 4
y = -2
y = -2
0
2
2 ( )(2 )y y dy
0
2
2
2 ( 2 )y y dy
y = 2x
y = 0
x = 4
dA dy2 – y
4
– y
y = -2
02
2
2 ( 2 )V y y dy
02 3
2
2 22 3
y y
3
2 ( 2)2 0 ( 2)
3
82 0 4
3
8
3