Section 7.2
Solids of Revolution
1st Day
Solids with Known Cross
Sections
Method of Slicing
1
Sketch the solid and a typical cross section.
2
Find a formula for A(x) = the area of a cross section.
(When you multiply A(x) by dx, you will have the third
length necessary to find the volume)
3
Find the limits of integration.
4
Integrate A(x) to find volume.

Examples
1.
Find the volume of the solid whose base is
bounded by the circle x2 + y2 = 9 and each
cross section perpendicular to the x-axis is a
square.
y




x












side of each square = 2y
y  9  x2
Area = (2y)2 = 4y2

V  2 4 
A x   4
3
0

9  x  dx
9 x
2
2
2
2
y
V  8  9  x 2  dx
3
0
3

x 
V  8 9 x  
3 0

3
V  144 u
3
x
2.
Find the volume of the solid whose base is
bounded by the graphs f(x) = x + 1 and
g(x) = x2 – 1 whose cross sections
perpendicular to the x-axis are equilateral
triangles. Area of an equilateral triangle is
2
given by
s 3
A
4
Side of equilateral triangle =
f(x) – g(x)
side   x  1   x  1
2
 2 x x
2
3
2 2
A x  
2 x x 

4
3 4
3
2

x

2
x

3
x
 4x  4

4
3 2 4
3
2
V
x

2
x

3
x
 4 x  4  dx


4 1
81 3 3

u
40
3.
The region bounded by the graph of
y = 2x – x2 and the x-axis is the base of a
solid. For this solid each cross section
perpendicular to the x-axis is a semicircle.
Diameter of semicircle = 2x – x2
1
2
radius of semicircle   2 x  x 
2
1 1
2 
A  x      2 x  x 
2 2

2
1 1
2 
A  x      2 x  x 
2 2

2
1
2 2
A x     2x  x 
8
1 2
V     4 x 2  4 x 3  x 4  dx
8 0
2
  u3
15
2nd Day
Disk Method
2
Suppose I start with this curve.
y x
1
0
1
2
3
4
How did we first find the approximation for area under
this curve?
Drawing rectangles and finding the area of each
rectangle and adding them together.
If I take the graph and rotate it about the x-axis, I will get a
3-dimensional solid that is cone shaped.
2
y x
1
0
1
2
3
4
How could we find the volume of the cone?
One way would be to cut it into a series of thin slices (flat
cylinders) and add their volumes.
Where do these flat cylinders (disks) come from?
They are created by rotating an infinitely many rectangles
around the axis of revolution.
2
y x
1
0
1
2
3
4
Volume of a Cylinder = r2h
What is the radius of each cylinder(disk) in the cone?
What is the height of each cylinder(disk) in the cone?
The radius of each cylinder(disk) is y coordinate of its
rectangle.
The height of each cylinder(disk) is Δx.
Volume of each cylinder(disk) = (y2)Δ x
2
y x
1
0
1
2
3
4
How do we find the volume of the solid?
Add the volume of all the cylinders(disks).
4
V  π
0
4
 x  dx
 π  x dx
2
4
π 2
V  x  8π cubic units
2 0
0
2
y x
1
0
1
2
3
4
Examples
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
3
1
 .707
2
1
 .577
3
4
1
2
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
0
1
2
4
4  1 
1
V   π
dy   π dy
1
 y 
1
y


volume of disk
 π ln y 1
4
 π  ln 4  ln1
 π ln 22
 2π ln 2 u 3
Find the volume of the solid formed by rotating the
region bounded by the x-axis (0 ≤ x ≤ ) and the
graph of
f  x   sin x
about the x-axis.
radius of a disk  sin x
Area of a disk  

sin x

2
  sin x

Volume of the solid   0 sin x dx
    cos x 0

    cos   cos 0
 2 u 3
Find the volume of the solid by revolving by
f(x) = 2 – x2 and g(x) = 1 about the line y = 1.
2
2
radius of a disk   2  x   1  1  x
Area of a disk   1  x

2 2
Volume of the solid   1 1  x
1

2 2
dx
   1  2x  x  dx
1
2
4
1
1

2 3 x 
 x  x  
3
5  1

16 3

u
15
5
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.
Solid of Revolution Formula with a horizontal axis:
b
V  π  y dx
2
a
Solid of Revolution Formula with a vertical axis:
b
V  π  x dy
a
End of 2nd Day
2
3rd Day
Washer Method
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:
0
1
2
The volume of the washer is:
The “disk” now has a hole in
it, making it a “washer”.
 R   r   thickness
  R  r  dy
2
2
outer
radius
2
2
inner
radius
The volume of the washer is:
4
3
y  2x
2
 R
y  x2
1
yx
2
yx
y  2x
y
x
2
0
1
2
4 
V   

0

2
  r 2   thickness
  R 2  r 2  dy
outer
radius
 y
2
inner
radius
 y
 
2
2

 dy


V   

0

4
 y
2
 y
 
2
2

 dy

1 2

V     y  y  dy
0
4 

4
1 2

V     y  y  dy
0
4 

4
1 2 1 3
  y  y 
12  0
2
4
 16  8 3
  8   0  
u
3

 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: V
b
   R  r dx
a
2
2
y  x2
4
3
y  2x
2
1
0
y  x2
yx
y  2x
y
x
2
If the same region is
rotated about the line
x = 2:
The outer radius is:
1
2
r
R
y
R  2
2
The inner radius is:
r  2 y
4
V    R 2  r 2 dy
0
2

y

   2    2  y
0
2

4
 dy
2

y 
    4  2 y    4  4 y  y dy
0
4 

2
4

2

y
  4  2 y 
 4  4 y  y dy
0
4
4
1
2
1 2
   3 y  y  4 y dy
0
4
4
1
2
1 2
V    3 y  y  4 y dy
0
4
3 4
 3 2 1 3 8 2
    y  y  y 
12
3 0
 2
4
16 64 

    24    0 
3 3


8 3

u
3
Examples
Find the volume of the solid by revolving the
region bounded by the graphs of
y  x and y  x
about the x-axis.
2
outer radius  x
inner radius  x 2


Volume of the solid  0 

1
 
2
x
x
1
4

    x  x  dx
0
1
x
x 
   
 2 5 0
2
3 3

u
10
5

2 2
 dx

Find the volume of the solid formed by the
equations:
y  1  x , y  1, and x  4
rotated about the line x = 6.
y  1 x
x  y 1
x   y  1
2
outer radius  6   y  1
inner radius  2
x   y  1
2
2


Volume of the solid     6   y  1
1 
3

2 2

 2  dy

2
3
4
3
2

    y  4 y  6 y  20 y  21 dy
1
y

4
3
2
    y  2 y  10 y  21 y 
5
1
192 3

u
5
5
3