7.3 Day One: Volumes by Slicing
x2
 x2
x
2
2
8
 15  2
x

4 
 3  2
 4  x
Find the volume of the pyramid:
Consider a horizontal slice through
the pyramid.
3
The volume of the slice is s2dh.
If we put zero at the top of the
pyramid and make down the
positive direction, then s=h.
3
3
Vslice  h 2 dh
0
h
s
3
3
1 3
V   h dh  h  9
0
3 0
3
dh
2
This correlates with the formula:
1
1
V  Bh   9  3  9
3
3

Method of Slicing:
1
Sketch the solid and a typical cross section.
2
Find a formula for V(x).
(Note that I used V(x) instead of A(x).)
3
Find the limits of integration.
4
Integrate V(x) to find volume.

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
x ways, but the simplest
cross section is a
rectangle.
y
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,  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

Cavalieri’s Theorem: Volume of a Sphere


V=
2

=
2

  sin x 
2
dx
0

1
0 2 1  cos  2x   dx

 
sin 2x 
=
x+

4 
2  0

=
 
4
2
=
4
7.3 Disk and Washer Methods
2
Suppose I start with this curve.
y x
My boss at the ACME Rocket
Company has assigned me to
build a nose cone in this shape.
1
0
1
2
3
4
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:
 r 2  the thickness

 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
 r 2  the thickness
1

0
1
2
3
 x
2
dx
4
If we add the volumes, we get:
   x
2
4
0
dx
4
   x dx
0


2
4
 8
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    y 2 dx
a
b
A shape rotated about the y-axis would be:
V    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  
dy
 y 
1


4
 
4
0
1
1
1
dy
y
volume of disk
0
2
  ln y 1    ln 4  ln1   ln 2  2 ln 2
4

y
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:
500 ft
x
x  .000574 y 2  .439 y  185
The volume can be calculated using the disk method with
a horizontal disk.

500
0
.000574 y
2
 .439 y  185 dy  24, 700, 000 ft 3
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  
0


4
 y
2
 y
 
2
2

 dy

1 

V     y  y 2  dy
0
4 

4
V 
4
0
1 2
y  y dy
4
The “disk” now has a hole in
it, making it a “washer”.
 R   r   thickness
  R  r  dy
2
2
2
2
outer
radius
4
1 
1
   y 2  y3 
12  0
2
inner
radius
 16 
  8  
3

8

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    R 2  r 2 dx
a

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
   4  2 y   4  4 y  y dy
0
4
1
4
1 2
   3 y  y  4 y 2 dy
0
4
4
V    R2  r 2 dy
0
2

y

  2   2 y
0
2

 dy
2


4
 3 2 1 3 8 
    y  y  y 
12
3 0
 2
16 64 
8

    24    
3 3
3

3
2

y2 
    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:

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=


4
0
=

A(x) dx =

0
4


0
4

 cos x - sin x  dx
cos 2x dx


 sin 2x  4
3
= 
=
units
2
 2  0
2
2
7.3 The Shell Method
5
Find the volume
the 4region
1 dy

1 4 y yof
2
x  1, x  2 ,
bounded by
5
and y  0  revolved
the y5  y dy  4about

1
axis.
5
4
3
y  x2  1
2
5
1 

 5 y  y 2   4
2 1

1
0
2
1
We can use the washer method ifwe split
25  itinto1 two
  parts:
  25     5     4
y 1  x2
5
 2 
2
1
outer
radius


x  y 1

2
y  1 dy    2 1
inner
radius
2
cylinder
thickness
Japanese Spider Crab of slice
Georgia Aquarium, Atlanta
2  
2 
 25 9 
   4
 2 2


16
 4
2
8  4
 12

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
=2 r  h  thickness


=2 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
=2 r  h  thickness


=2 x x 2  1 dx
r
h
circumference thickness


2 x x 2  1 dx
2  4  2
2
2  x3  x dx
0
2
1 4 1 2 
2  x  x 
2 0
4
12

Find the volume generated
when this shape is revolved
about the y axis.
4
3
2
1
0
1
2
3
y
4

5
6
4 2
x  10 x  16
9
7
8

We can’t solve for x,
so we can’t use a
horizontal slice
directly.

If we take a
vertical slice
and revolve it
about the y-axis
we get a cylinder.
4
3
2
1
0
Shell method:
1
2
3
y
4

5
6
4 2
x  10 x  16
9
7
8

Lateral surface area of cylinder
=circumference  height
=2 r  h
Volume of thin cylinder  2 r  h  dx

4
3
2
1
0
1
Volume of thin cylinder  2 r  h  dx
 4 2

2 2 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

 160
 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.

Find the volume of the solid when the region bounded by the
curve y = x , the x-axis, and the line x = 4 is revolved about
the x-axis. Find the volume of the solid using cylindrical
shells.
Find the volume of the solid when the region bounded by the
curve y = x , the x-axis, and the line x = 4 is revolved about
the x-axis. Find the volume of the solid using cylindrical
shells.
V=

2
0
2  y   4 - y
= 8
Radius = y
x = y2
Shell height = 4 - y 2
2
 dy
Find the volume of the solid of revolution formed
by revolving
2
the region bounded by the graph of x  e y and the y
axis, 0 ≤ y ≤ 1, about the x-axis. Use the Shell Method.
2
1
2
-1
Find the volume of the solid of revolution formed
by revolving
2
the region bounded by the graph of x  e y and the y
axis, 0 ≤ y ≤ 1, about the x-axis. Use the Shell Method.
1
V  2  ye
 y2
0
 1.986
2
1
   e  y 

0
 1
  1  
e

2
dy
1
2
-1
Find the volume of the solid formed by revolving the region
bounded by the graphs y = x3 + x + 1, y = 1, and x = 1 about
the line x = 2.
3
axis of
revolution
2
1
2
-1
4
Find the volume of the solid formed by revolving the region
bounded by the graphs y = x3 + x + 1, y = 1, and x = 1 about
the line x = 2.
1
0
3
axis of
revolution
2



 2  x 4  2x 3  x 2  2x dx
0
2
1
2
4
1
 x

x
x
2
=2  

 x 
4
3
 5
0
 1 1 1 
 2      1
 5 2 3 
5
-1


V  2  2  x x 3  x  1  1 dx
29

15
4
3