M53 Lec4.5 Volumes of Solids of Revolution

Volumes of Solids of Revolution

(Washers and Cylindrical Shells)

Mathematics 53

Institute of Mathematics (UP Diliman)

For today

Solids of Revolution

Solids of RevolutionA solid of revolution is a solid obtained when a plane region is revolved about a

line (in the same plane) called the axis of revolution.

Solids of Revolution

Example. The solid of revolution generated when the region bounded by x = y2

and y = x2 is revolved about the x-axis.

Solids of Revolution

Example.The solid of revolution generated when the region bounded by y = sin xand the x-axis on the interval [0, π] is revolved about the y-axis.

Solids of Revolution

Two methods to find the volume of a solid of revolution:

1 Disk or Washer Method: Use rectangles that are perpendicular to the axis of

revolution.

2 Cylindrical Shell Method: Use rectangles that are parallel to the axis of

revolution.

The Disk or Washer Method

Divide the region and orient rectangles so that they are perpendicular to the

axis of revolution.

Disks are formed if a segment of the axis of revolution is a boundary of the region.

The Disk or Washer Method

Washers are obtained when no segment of the axis of revolution is a boundary of

the region.

Disk or Washer Method

Consider the simple case where we only obtain disks.

PROBLEM: Find the volume of the solid generated when the region below is

revolved about the x-axis.

The Disk or Washer Method

DISK

V = πr2h

Disk or Washer Method

Divide [a, b] into n subintervals of

equal length ∆x.

Let x∗i be an arbitrary point in the ith

subinterval.

For the ith disk:

ri = f (x∗i )− 0 hi = ∆x

Volume of the ith disk:

Vi = πr2i hi = π [ f (x∗i )]

2 ∆x

Disk or Washer Method

Approximate volume of the solid:

V ≈n

∑i=1

Vi =n

∑i=1

π [ f (x∗i )]2 ∆x

Let n→ ∞:

V = limn→∞

n

∑i=1

π [ f (x∗i )]2 ∆x

The volume is given by

V =∫ b

aπ [ f (x)]2 dx.

Disk or Washer Method

If only disks are obtained (a boundary lies on the axis of revolution), then

V =∫ b

aπr2 dw,

where

[a, b] is the interval I covered by the region

r is given by the distance of the farther tip of an arbitrary rectangle to the axis

of revolution

dw is the width (shorter side) of an arbitrary rectangle.

Disk or Washer Method

If vertical rectangles are used (dw = dx), set up the integral in terms of x.

If horizontal rectangles are used (dw = dy), set up the integral in terms of y.

Example

Let R be the region bounded by y = x2, y = 0 and x = 2. Find the volume of the

solid generated when R is revolved about the x-axis.

2x

y

4

y = x2

(x, x2)

(x, 0)

(2, 4)

Set up in terms of x:

I = [0, 2]

r = x2 − 0 = x2

V =∫ 2

0πx4 dx

=πx5

5

∣∣∣∣∣2

0

=32π

5cubic units

ExampleFind the volume of the solid generated when R is revolved about the line x = 2.

2x

y

4

y = x2

x =√

y

(2, y)(√

y, y)

(2, 4)

x = 2

Use y:

I = [0, 4]

r = 2−√y

V =∫ 4

0π (2−√y)2 dy

=∫ 4

0π (4− 4

√y + y) dy

= π

(4y− 8y3/2

3+

y2

2

)∣∣∣∣∣4

0

=8π

3cubic units

The Disk or Washer Method

Washers are obtained when no segment of the axis of revolution is a boundary of

the region.

The Disk or Washer Method

WASHER

r2 = outer radius; r1 = inner radius

V = πr2

2h− πr2

1h = π(r2

2 − r2

1)h

Disk or Washer Method

If washers are obtained, then

V =∫ b

aπ(

r2

2 − r2

1

)dw,

where

[a, b] is the interval I covered by the region

r2 is given by the distance of the farther tip of an arbitrary rectangle to the axis

of revolution

r1 is given by the distance of the nearer tip of an arbitrary rectangle to the axis

of revolution

dw is the width (shorter side) of an arbitrary rectangle.

ExampleFind the volume of the solid generated when R is revolved about the line y = 4.

2x

y

y = x2

(x, x2)

(x, 0)

(2, 4)

y = 4

Use x:

I = [0, 2]

r2 = 4− 0 = 4

r1 = 4− x2

V =∫ 2

0π[42 − (4− x2)2

]dx

=∫ 2

0π(8x2 − x4)dx

= π

(8x3

3− x5

5

)∣∣∣∣20

=224π

15cubic units

ExampleFind the volume of the solid generated when R is revolved about the y-axis.

2x

y

4

x =√

y

(2, y)(√

y, y)

(2, 4)

Use y:

V =∫ 4

0π([

2− 0]2 − [√y− 0

]2)dy

=∫ 4

0π(4− y)dy

= π

(4y− y2

2

)∣∣∣∣40

= 8π cubic units

Solids of Revolution

Two methods to find the volume of a solid of revolution:

1 Disk or Washer Method: Divide the region into rectangles that are

perpendicular to the axis of revolution

2 Cylindrical Shell Method: Divide the region into rectangles that are parallel to

the axis of revolution

The Cylindrical Shell Method

Cylindrical Shell Method: Orient the rectangles so that they are parallel to the

axis of revolution.

Problem: Find the volume of the solid generated when the region is revolved

about the y-axis.

The Cylindrical Shell Method

The solid formed by revolving each rectangle is called a cylindrical shell.

The Cylindrical Shell Method

CYLINDRICAL SHELL

V = π(r2

2 − r2

1)h

r2 = outer radius; r1 = inner radius

The Cylindrical Shell Method

Volume of a Cylindrical Shell:

V = π(r2

2 − r2

1)h

= π(r2 + r1)(r2 − r1)h

= 2π

(r2 + r1

2

)h(r2 − r1)

= 2πrh∆r

where r =r2 + r1

2, ∆r = r2 − r1

V = 2π(average radius)(height)(thickness)

The Cylindrical Shell Method

Problem: Find the volume of the solid generated when the region is revolved

about the y-axis.

The Cylindrical Shell Method

Divide [a, b] into n subintervals of equal

length ∆x.

Let x∗i be the midpoint of the ith subinterval.

For the ith cylindrical shell:

average radius = x∗i − 0 = x∗i

height = f (x∗i )− 0 = f (x∗i )

thickness = ∆x

Volume of the ith cylindrical shell:

Vi = 2πx∗i f (x∗i )∆x

The Cylindrical Shell Method

Approximate the volume of the solid:

V ≈n

∑i=1

Vi =n

∑i=1

2πx∗i f (x∗i )∆x

Let n→ ∞:

V = limn→∞

n

∑i=1

2πx∗i f (x∗i )∆x

The volume is given by

V =∫ b

a2πx f (x) dx.

The Cylindrical Shell Method

We can think of the formula for the volume using cylindrical shells as

V =∫ b

a2πrh dw,

where

[a, b] is the interval I covered by the region

r is the distance of an arbitrary rectangle to the axis of revolution (radius of

cylindrical shell generated)

h is the height of an arbitrary rectangle (height of a cylindrical shell generated)

dw is the width of an arbitrary rectangle.

The Cylindrical Shell Method

V =∫ b

a2πrh dw

The Cylindrical Shell Method

In using the cylindrical shell method:

1 Orient an arbitrary rectangle so that it is parallel to the axis of revolution.

2 Determine if the integral will be set up in terms of x or y.

If vertical rectangles are used, dw = dx, so set up the integral in terms of x.

If horizontal rectangles are used, dw = dy, so set up the integral in terms of y.

Example

Let R be the region bounded by y = x2, x = 0 and x = 2. Find the volume of the

solid generated when R is revolved about the y-axis.

2x

y

4

(x, x2)

(x, 0)

(2, 4)

Set up in terms of x:

I = [0, 2]

r = x− 0 = xh = x2 − 0 = x2

V =∫ 2

02πx(x2) dx

=∫ 2

02πx3 dx

=2πx4

4

∣∣∣∣20

= 8π cubic units

Example

Find the volume of the solid generated when the region R, bounded by y = x2,

x = 2 and the x-axis is revolved about x = 2.

2x

y

4

y = x2

(x, x2)

(x, 0)

(2, 4)

x = 2

I = [0, 2]

r = 2− x

h = x2 − 0 = x2

V =∫ 2

02π(2− x)(x2) dx

=∫ 2

02π(2x2 − x3) dx

= 2π

(2x3

3− x4

4

)∣∣∣∣20

=8π

3cubic units

Example

Find the volume of the solid generated when the region R, bounded by y = x2,

x = 2 and the x-axis is revolved about the x-axis.

2x

y

4

y = x2

x =√

y

(2, y)(√

y, y)

(2, 4)

Use y:

I = [0, 4]

r = y− 0 = y

h = 2−√y

V =∫ 4

02πy (2−√y) dy

=∫ 4

02π(

2y− y3/2)

dy

= 2π

(y2 − 2y5/2

5

)∣∣∣∣∣4

0

=32π

5cubic units

Example

Find the volume of the solid generated when the region R, bounded by y = x2,

x = 2 and the x-axis is revolved about y = 4.

2x

y

4

y = x2

x =√

y

(2, y)(√

y, y)

(2, 4)y = 4

V =∫ 4

02π(4− y)[2−√y]dy

=∫ 4

02π(

8− 2y− 4y1/2 + y3/2)

dy

= 2π

(8y− y2 − 8y3/2

3+

2y5/2

5

)∣∣∣∣∣4

0

=224π

15cubic units

Volumes of Solids of Revolution

Remarks1 If the formula for the height of the rectangle is not the same throughout the

region, divide the region into the appropriate number of subregions.

2 The better method is usually that for which the formula for the height of the

rectangle changes least throughout the region.

Volumes of Solids of Revolution

To find the volume of a solid of revolution:

1 Determine if it is better to use vertical or horizontal rectangles.

This depends on the region revolved and not on the axis of revolution.

2 Determine if you will use x or y.3 Determine the method.

Disk or Washer Method if the rectangles are perpendicular to the axis of revolution.

Cylindrical Shell if the rectangles are parallel to the axis of revolution.

Example

Let R be the region enclosed by x = 3y2 and x = 4− y2. Set up the integral that

gives the volume of the solid generated when R is revolved about the line y = 1.

Find the y-coordinates of the points of intersection:

3y2 = 4− y2

4y2 = 4

y = ±1

Find the x-coordinates of the points of intersection:

y = ±1⇒ x = 3

Points of intersection:

(3, 1) and (3,−1)

ExampleSet up the integral that gives the volume of the solid generated when R is revolved

about the line y = 1.

x

y

(3, 1)

(3,−1) x = 3y2

x = 4− y2x = 4− y2

(0, 0) (4, 0)

y = 1Use y.

Use Cylindrical Shells.

V =∫ 1

−12π(1− y)[(4− y2)− 3y2]dy

ExampleWhat if the region is to be revolved about x = −1?

x

y

x = 3y2

x = 4− y2x = 4− y2

(3, 1)

(3,−1)

x = −1

(0, 0) (4, 0)Use Washers.

V = π([(4− y2) + 1]2 − [(3y2) + 1]2)dy

ExampleLet R be the region below, with boundaries by y = −3x + 8, y = 3x + 2 and

y = x2 − 2. Set up the integral equal to the volume of the solid generated when Ris revolved about the line x = 2.

x

y

y = 3x + 2

y = −3x + 8

y = x2 − 2

Points of intersection:

y = 3x + 2 and y = x2 − 2:

3x + 2 = x2 − 2

x2 − 3x− 4 = 0

x = 4 or x = −1

y = 14 y = −1

y = −3x + 8 and y = x2 − 2:

x2 − 2 = −3x + 8

x2 + 3x− 10 = 0

(x + 5)(x− 2) = 0

x = −5 or x = 2

y = 23 y = 2

y = 3x + 2 and y = −3x + 8:

3x + 2 = −3x + 8

x = 1

y = 5

ExampleLet R be the region below, with boundaries by y = −3x + 8, y = 3x + 2 and

y = x2 − 2.

x

y

y = 3x + 2

y = −3x + 8

y = x2 − 2

(−1,−1)

(1, 5)

(2, 2)

Points of intersection:

y = 3x + 2 and y = x2 − 2:

(4, 14), (−1,−1)

y = 3x + 2 and y = −3x + 8:

(1, 5)

y = −3x + 8 and y = x2 − 2:

(−5, 23), (2, 2)

Use x.

ExampleSet up the integral that gives the volume of the solid generated when R is revolved

about the line x = 2.

x

y

y = 3x + 2

y = −3x + 8

y = x2 − 2

(−1,−1)

(1, 5)

(2, 2)

x = 2

Use Cylindrical Shells.

V =∫ 1

−12π(2− x)[(3x + 2)− (x2 + 2)] dx +∫ 2

12π(2− x)[(−3x + 8)− (x2 + 2)] dx

ExampleWhat if R is to be revolved about the line y = −3?

x

y

y = 3x + 2

y = −3x + 8

y = x2 − 2

(−1,−1)

(1, 5)

(2, 2)

y = −3

Use Washers.

V =∫ 1

−1π([(3x + 2) + 3]2 − [(x2 − 2) + 3]2)dx +∫ 2

1π([(−3x + 8) + 3]2 − [(x2 − 2) + 3]2)dx

End-of-class Exercise

Find the volume of the solid of revolution generated when the plane region below

bounded by x = 2− y2, y = x and the x-axis is revolved about the line x = 2.

(No need to simplify the integrand.)

Use y:

I = [0, 1]

r2 = 2− y r1 = 2− (2− y2) = y2

V =∫ 1

0π

[(2− y)2 −

(y2)2]

dy

End-of-class Exercise

Find the volume of the solid of revolution generated when the plane region below

bounded by x = 2− y2, y = x and the x-axis is revolved about the line y = 2.

(No need to simplify the integrand.)