2. 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? L
The volume dV of the sheet is: L
dV = L W dx
This sheet is rolled into a hollow cylinder of height L and radius r W
This is called a shell.
The width W of the sheet has now become the circumference of a circle of radius r
The circumference C of a circle of radius r is r
C = 2r

3. The volume dV of the sheet is:r
dV = L W dx
C = 2r
Since W has become C, the volume dV of the shell can be written as: L L
dV = L 2r dx W
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

4. Revolution on the y-axis
Consider the graph of a function f(x)
The area enclosed by this graph with the x-axis from x = a to x = b is shaded.
f(x)
We will use the shell method to find the volume generated when this area is revolved about the x-axis. L dA
L(x) = f(x) dx
Take a strip of area dA of width dx parallel to the y-axis
x = a
x = b
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.
= f(x) – 0
= f(x)
L(x) = f(x)

5. If this area is revolved about the y-axis, a shell is generated.
What is the radius of this shell?
Radius of the shell is the distance of dA from the axis of revolution.
f(x) L dA
The distance of dA from the y-axis is x
L(x) = f(x)
The radius of the shell is:
r(x) = x dx
x = a
x = b
The length of the shell is:
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:

6. Revolution on the x-axis
In order to revolve this area about the x-axis, dA of width dy is taken parallel to the x-axis
g(y)
f(y) L
To find the length of dA, the functions must of y dy dA y y
L(y) = f(y) – g(y)
f(y) is on the right of dA and g(y) on its left.
The length of dA is
L(y) = f(y) – g(y)
When dA is revolved about the x-axis, a shell is generated
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
r(y) = y

7. The volume dV of the shell generated is: f(y)
L(y) = f(y) – g(y)
r(y) = y
y = d
The volume dV of the shell generated is:
f(y) L dy
dV = 2 r(y)L(y) dy dA y
L(y) = f(y) – g(y)
y = c
The volume V generated by revolving the area from y = c to y = d is:

8. 1. Take dA parallel to the axis
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 L
If dA is vertical, L(x) = g(x) – f(x)
= Function above – function below
If dA is horizontal, L(y) = g(y) – f(y)
= right function – left function
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

9. Draw the graphs of y = x2, y = 0, x = 4 and shade the area enclosed.
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.
Draw the graphs of y = x2, y = 0, x = 4 and shade the area enclosed.
y = x2
To revolve this area about the y-axis, take dA parallel to it.
The length of dA is:
L(x) = x2 – 0
= x2
x = 4
When dA is revolved about the y-axis, the radius of the shell generated is the distance of dA from y-axis. dA x2
y = 0 dx
r(x) = x x
The volume dV of the shell is: dx x
dV = 2 r(x)L(x) dx
L(x) = x2
= 2 x · x2 dx
= 2 x3 dx

10. The position of dA can change from x = 0 to x = 4
dV= 2 x3 dx
The position of dA can change from x = 0 to x = 4
y = x2
The volume V generated by revolving the area enclosed about the y-axis is:
x = 4 dA x2
y = 0 dx x dx x
L(x) = x2

11. Take dA of width dy parallel to the x-axis.
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.
Take dA of width dy parallel to the x-axis.
Express the functions to the left and right of dA as functions of y
y = x2
The function on the left is:
y = x2
x = 4
Solving for x gives: dy dA y
The function on the right is:
x = 4
y = 0
g(y) = 4
The length of dA is:
L(y) = g(y) – f(y)
The radius of revolution is the distance of dA from the x-axis
r(y) = y

12. The position of dA varies from y = 0 to y = 16
r(y) = y
To find the limits, we find the points of intersection of x = y and x = 4
y = 16
y = x2
y = 4 gives:
y = 16
The position of dA varies from y = 0 to y = 16
x = 4 dy
The volume of revolution is: dA y
y = 0

13. y = 16
y = x2
x = 4 dy dA y
y = 0

14. 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.
x = 0
y = 2x
Draw the graphs of y = 2x, x = 0, y = 4 and shade the area enclosed.
y = 4 dx
Take dA of width dx parallel to the y-axis
The length of dA is:
L(x) = 4 – 2x
4 – 2x
r(x) = x
Obtain limits by solving 2x = 4
x = 2 x
The volume of revolution V is:
x = 2

15. x = 0
y = 2x
y = 4 dx
4 – 2x x
x = 2

16. Find the volume generated by revolving the area enclosed by y = x2 and y = 4x – x2 about the line x = 2
x = 2
Draw the graphs of y = x2, y = 4x – x2 and x = 2 and shade the area enclosed. 2
y = x2
Find the points of intersection of y = x2 and y = 4x – x2. x
4x – x2 = x2
y = 4x – x2
2 – x
4x – x2 – x2 = 0
2x(2 – x) = 0
x = 0, 2 dx
Take dA of width dx parallel to x = 2
x = 0
x = 2
The length of dA is:
L(x) = 4x – x2 – x2
= 4x –2x2
The distance of dA from x = 2 is:
2 – x
r(x) = 2 – x

17. L(x) = 4x – 2x2
r(x) = 2 – x
x = 2
The volume of revolution V is: 2
y = x2 x
y = 4x – x2
2 – x dx
x = 0
x = 2

18. x = 2 2
y = x2 x
y = 4x – x2
2 – x dx
x = 0
x = 2

19. Draw dA of width dy parallel to the x-axis. x = 2 – y
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
– y dA dy
Draw dA of width dy parallel to the x-axis.
x = 2 – y
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 4
x = 4
y = 2 – x
Solving for x gives:
x = 2 – y
L(y) = 4 – (2 – y)
L(y) = 2 + y
dA is below the x-axis.
The distance of dA from the x-axis is:
– y
r(y) = – y

20. To find the limits, solve 2 – y = 4
L(y) = 2 + y
r(y) = – y
y = 2x
To find the limits, solve 2 – y = 4
y = -2
y = 0
– y
The volume of revolution V is: dA dy
2 – y
y = -2 4
x = 4

21. y = 2x
y = 0
– y dA dy
2 – y
y = -2 4
x = 4