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.)