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