Ch 7.3
Volumes
Calculus Graphical, Numerical, Algebraic by
Finney, Demana, Waits, Kennedy
Volume of a Solid
The definition of a solid of known integrable cross section
area A  x  from x = 0 to x = b is the integral of A from a to b,
b
V=
 A  x  dx.
a
, the area of the cross section.
A 45o wedge is cut from a cylinder of radius 3 as shown.
Find the volume of the wedge.
You could slice this
wedge shape several
y
ways, but the simplest
x cross section is a
rectangle.
If we let h equal the height of the slice then the volume of
the slice is: V x  2 y  h  dx
 
Since the wedge is cut at a
Since
x  y 9
2
2
h
45o
angle:
45o
x
hx
y  9  x2

y
x
Even though we
started with a
cylinder, p does not
enter the
calculation!
V  x   2 9  x2  x  dx
2
V
x

2
y

h

dx


u

9

x
u  0  9
3
2
V   2 x 9  x dx du  2x dx u  3  0 h  x
0
0
1
2
V    u du
9
2
 u
3
3 9
2
2
y  9 27
 x 2  18
3
0

Cavalieri’s Theorem:
Two solids with equal altitudes and identical parallel cross
sections have the same volume.
Identical Cross Sections
p
p
V=
2
p
=
2
p
  sin x 
2
dx
0
p
1
0 2 1  cos  2x   dx
p
p 
sin 2x 
=
x+

4 
2  0
p
=
 p
4
p2
=
4
Disk Method
2
y x
Suppose I start with this curve.
1
0
1
2
3
4
My boss at the ACME Rocket
Company has assigned me to
build a nose cone in this shape.
So I put a piece of wood in a
lathe and turn it to a shape to
match the curve.

2
How could we find the
volume of the cone?
y x
One way would be to cut it into a
series of thin slices (flat cylinders)
and add their volumes.
1
0
1
2
3
4
The volume of each flat
cylinder (disk) is:
p r 2  the thickness
p
 x
2
dx
In this case:
r= the y value of the function
thickness = a small change
in x = dx

2
The volume of each flat
cylinder (disk) is:
y x
p r 2  the thickness
1
p
0
1
2
3
 x
2
dx
4
If we add the volumes, we get:
 p  x
2
4
0
dx
4
  p x dx
0

p
2
4
 8p
x2
0

This application of the method of slicing is called the
disk method. The shape of the slice is a disk, so we
use the formula for the area of a circle to find the
volume of the disk.
If the shape is rotated about the x-axis, then the formula is:
b
V  p  y dx
2
a
Since we will be using the disk method to rotate shapes
about other lines besides the x-axis, we will not have this
formula on the formula quizzes.
b
A shape rotated about the y-axis would be:
V  p  x dy
2
a

1
The region between the curve x 
, 1  y  4 and the
y
y-axis is revolved about the y-axis. Find the volume.
y
x
1
1
2
3
4
1
 .707
2
1
 .577
3
1
2
We use a horizontal disk.
The thickness is dy.
4
3
2
The radius is the x value of the
1
function 
.
dy
y
1
2
 1 
V  p
dy
 y 
1


4
 p
4
0
1
1
1
dy
y
volume of disk
0
2
 p ln y 1  p  ln 4  ln1  p ln 2  2p ln 2
4

y
500 ft
The natural draft cooling tower
shown at left is about 500 feet
high and its shape can be
approximated by the graph of
this equation revolved about
the y-axis:
x
x  .000574 y 2  .439 y  185
The volume can be calculated using the disk method with
a horizontal disk.
p
500
0
.000574 y
2
 .439 y  185 dy  24, 700, 000 ft 3
2

Disks Example
The region between the graph of f(x) = 2 + x cos x and the x
axis over the interval [-2,2] is revolved about the x-axis to
generate a solid. Find the volume of the solid
Circular Cross Sections
The region between the graph of f(x) = 2 + x cos x and the x axis
over the interval [-2,2] is revolved about the x-axis to generate a
solid. Find the volume of the solid
Area of the cross section =
A(x) = p
f  x 
2
The volume of the solid is:
V =  A(x) dx =  p  2 + x cos x  dx = 52.429 units3
2
2
-2
-2
2
End of Ch 7.3 Day 1
Ch 7.3 Day 2: Washer Method
fx = x2
gx =
3x
hx =
-3x
3
2
1
-2
2
4
3
y  2x
2
y  x2
The region bounded by
y  x 2 and y  2 x is
revolved about the y-axis.
Find the volume.
1
If we use a horizontal slice:
yx
y  2x
y
x
2
2
yx
0
1
2
The volume of the washer is:

V  p
0


4
 y
2
 y
 
2
2

 dy

1 

V   p  y  y 2  dy
0
4 

4
V p
4
0
1 2
y  y dy
4
The “disk” now has a hole in
it, making it a “washer”.
p R  p r   thickness
p  R  r  dy
2
2
2
2
outer
radius
4
1 
1
 p  y 2  y3 
12  0
2
inner
radius
 16 
 p 8  
3

8p

3

This application of the method of slicing is called the
washer method. The shape of the slice is a circle
with a hole in it, so we subtract the area of the inner
circle from the area of the outer circle.
The washer method formula is:
b
V  p  R 2  r 2 dx
a
Like the disk method, this formula will not be on the
formula quizzes. I want you to understand the formula.

y  x2
4
3
y  2x
2
1
0
1
r
y  2x
y
x
2
y  x2
yx
2
r  2 y
y2
 p  4  2 y   4  4 y  y dy
0
4
1
4
1 2
 p  3 y  y  4 y 2 dy
0
4
4
V  p  R2  r 2 dy
0
2

y

p  2   2 y
0
2

 dy
2


4
 3 2 1 3 8 
 p   y  y  y 
12
3 0
 2
16 64 
8p

 p   24    
3 3
3

3
2

y2 
 p   4  2 y    4  4 y  y dy
0
4 

4
The outer radius is:
y
R  2
2
The inner radius is:
R
4
4
If the same region is
rotated about the line x=2:
p
Washer Cross Section
The region in the first quadrant enclosed by the y-axis and the
graphs of y = cos x and y = sin x is revolved about the x-axis to form
a solid. Find its volume.
Washer Cross Section
The region in the first quadrant enclosed by the y-axis and the graphs
of y = cos x and y = sin x is revolved about the x-axis to form a solid.
Find its volume.
V=

p
4
0
=p

A(x) dx =
p
0
4

p
0
4
p
 cos x - sin x  dx
cos 2x dx
p
p
 sin 2x  4
3
=p 
=
units
2
 2  0
2
2
Volumes of Solids: End of Day 2
7.3 Day 3
The Shell Method
Grows
to over
12 feet
wide
Japanese
Spider
Crab
and lives
100 years.
Georgia
Aquarium,
Atlanta
Photo by Vickie Kelly, 2006
Greg Kelly, Hanford High School, Richland, Washington
5
4
3
y  x2  1
2
5
Find the volume
p  4   of
y 2the
1 dyregion
 4p
1 y  x 1
bounded by
, x2,
5
and y  1 revolved
about the yp  5  y dy  4p
1
axis.
5
1 

p 5 y  y 2   4p
2 1

1
0
2
1
We can use the washer method ifwe split
25  itinto1 two
  parts:
p  25     5     4p
y 1  x2
5
p 2 
2
1
outer
radius


x  y 1

2
y  1 dy  p  2 1
inner
radius
2
cylinder
thickness
Japanese Spider Crab of slice
Georgia Aquarium, Atlanta
2  
2 
 25 9 
   4p
 2 2
p
p
16
 4p
2
8p  4p
 12p

5
4
Here is another
way we could
approach this
problem:
3
y  x2  1
2
1
0
1
2
cross section
If we take a vertical slice and revolve it about the y-axis
we get a cylinder.
If we add all of the cylinders together, we can reconstruct
the original object.

5
4
3
y  x2  1
2
1
0
1
2
cross section
The volume of a thin, hollow cylinder is given by:
Lateral surface area of cylinder  thickness
 circumference  height  thickness
=2p r  h  thickness


=2p x x 2  1 dx
r
h
circumference thickness
r is the x value of the function.
h is the y value of the function.
thickness is dx.

5
4
This is called the
shell method
because we use
cylindrical shells.
3
y  x2  1
2
1
0
1
2
cross section
If we add all the cylinders from the
smallest to the largest:

2
0
=2p r  h  thickness


=2p x x 2  1 dx
r
h
circumference thickness
2p x  x 2  1 dx
2p 4  2
2
2p  x3  x dx
0
2
1 4 1 2 
2p  x  x 
2 0
4
12p

4
Find the volume generated
when this shape is revolved
about the y axis.
3
2
1
0
1
2
3
4

5
6
4 2
y   x  10 x  16
9
7
8

We can’t solve for x,
so we can’t use a
horizontal slice
directly.

4
If we take a
vertical slice
and revolve it
about the y-axis
we get a cylinder.
3
2
1
0
1
Shell method:
2
3
y
4
5

6
7
4 2
x  10 x  16
9
8

Lateral surface area of cylinder
=circumference  height
=2p r  h
Volume of thin cylinder  2p r  h  dx

4
3
2
1
0
1
Volume of thin cylinder  2p r  h  dx
 4 2

2 2p x  9 x  10 x  16  dx

8

r
circumference
h
thickness
2
3
y
4

5
6
4 2
x  10 x  16
9
7
8

 160p
 502.655 cm3
Note: When entering this into the calculator, be sure to enter
the multiplication symbol before the parenthesis.

When the strip is parallel to the axis of rotation, use the
shell method.
When the strip is perpendicular to the axis of rotation,
use the washer method.
p
Volumes Using Cylindrical Shells
The region bounded by the curve y =
x , the x-axis, and the
line x = 4 is revolved about the x-axis to generate a solid. Find the
volume of the solid using cylindrical shells.
Volumes Using Cylindrical Shells
The region bounded by the curve y =
x , the x-axis, and the
line x = 4 is revolved about the x-axis to generate a solid. Find the
volume of the solid using cylindrical shells.
V=

2
0
2p  y   4 - y
= 8p
Radius = y
x = y2
Shell height = 4 - y 2
2
 dy