LESSON 15 VOLUMES BY DISKS
Examples Find the volume of the solid of revolution using disks if the region,
which is bounded by the graphs of the given equations, is revolved about the given
axis.
1.
y 
2
, y  0 , x  3 , x  8 ; revolve about the x-axis
x
NOTE: The graph of the equation y  0 is the x-axis. The graph of the
equation x  3 is a vertical line which crosses the x-axis at 3. The graph of
the equation x  8 is a vertical line which crosses the x-axis at 8.
y
2 

 x, 
x 

Radius =
y 
3
x 3
x 8
2
; Width = dx
x
2
x
x
x 8
Picture of the solid of revolution.
The volume of the i th disk, which is obtained by rotating the i th rectangle
2
2


2
about the x-axis, is given by the product  r dx =    dx . In order to
x
find the volume of the solid of revolution, we will need to sum the volume of
all the disks, which are obtained by rotating all the vertical rectangles about
the x-axis, using integration.
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~anderson/1860
The first vertical rectangle in the region corresponds to when x  1 and the
last vertical rectangle in the region corresponds to when x  4 . Thus, we
sum from x  3 to x  8 . Thus, we obtain the following.

8
3
2
2
   dx = 
x
8

8
3
4
dx = 4 
x2

8
3
x1 
1 
2
 4   =
x dx = 4  
=

1 3
 x 3
8
8
1
1
5 
5
1 
 4    =  4     =  4   
 =
3
6
 x 3
8
 24 
5
Answer:
6
2.
y 
5
, y  0 , x   9 , x   4 ; revolve about the x-axis
x
NOTE: The graph of the equation y  0 is the x-axis. The graph of the
equation x   9 is a vertical line which crosses the x-axis at  9 . The
graph of the equation x   4 is a vertical line which crosses the x-axis at
 4.
5
y 
y
x

 x,


5
x




Radius =
9
x  9
4
x
x
x  4
Picture of the solid of revolution.
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~anderson/1860
5
; Width = dx
x
The volume of the i th disk, which is obtained by rotating the i th rectangle
2
 5 
2
 dx . In order


r
dx
about the x-axis, is given by the product
= 

 x 
to find the volume of the solid of revolution, we will need to sum the volume
of all the disks, which are obtained by rotating all the vertical rectangles
about the x-axis, using integration.
The first vertical rectangle in the region corresponds to when x   9 and the
last vertical rectangle in the region corresponds to when x   4 . Thus, we
sum from x   9 to x   4 . Thus, we obtain the following.

4
9

 

2
5
x

 dx = 



4
9
25
dx =  25
x

4
9
1
dx =  25 ln x
x
 25 ( ln 4  ln 9 ) = 25 ( ln 9  ln 4 ) = 25 ln
Answer: 25 ln
3.

4
9
=
9
4
9
4
y   3x 2 , y   6 , x  0 ; revolve about the y-axis
2
NOTE: The graph of the equation y   3x is a parabola whose vertex is at
the origin and opens downward. The graph of the equation y   6 is a
horizontal line which crosses the y-axis at  6 . The graph of the equation
x  0 is the y-axis.
2
y   3x 2  x  
y

3
x 

y
y
 x  
3
3
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~anderson/1860
y
x




y


y
, y 
3

6
y   3x 2
y  6
Radius =

y
; Width = dy
3
Picture of the solid of revolution.
The volume of the i th disk, which is obtained by rotating the i th rectangle
2

y 
2
about the y-axis, is given by the product  r dy =    3  dy . In order


to find the volume of the solid of revolution, we will need to sum the volume
of all the disks, which are obtained by rotating all the horizontal rectangles
about the y-axis, using integration.
The first horizontal rectangle in the region corresponds to when y   6 and
the last horizontal rectangle in the region corresponds to when y  0 . Thus,
we sum from y   6 to y  0 . Thus, we obtain the following.
2

0
6


y 
    dy = 
3 


0
6

 y
   dy = 
3
 3
0

0
6
 y2 
y dy =  
3 2   6 =
0



y 2   6 =  ( 0  36 ) = 6 
6
6
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~anderson/1860
Answer: 6 
4.
x  8  2 y , y  0 , y  3 , x  0 ; revolve about the y-axis
NOTE: The graph of the equation x  8  2 y is a line whose y-intercept is
the point ( 0 , 4 ) and whose x-intercept is the point ( 8 , 0 ) . The graph of the
equation y  0 is the x-axis. The graph of the equation y  3 is a
horizontal line which crosses the y-axis at 3. The graph of the equation
x  0 is the y-axis.
x  8  2y
y
4
y 3
3
 8  2 y, y 
y
8
x
Radius = 8  2 y ; Width = dy
Picture of the solid of revolution.
NOTE: The solid is a circular cone with its top chopped off.
The volume of the i th disk, which is obtained by rotating the i th rectangle
2
2
about the y-axis, is given by the product  r dy =  ( 8  2 y ) dy . In order
to find the volume of the solid of revolution, we will need to sum the volume
of all the disks, which are obtained by rotating all the horizontal rectangles
about the y-axis, using integration.
The first horizontal rectangle in the region corresponds to when y  0 and
the last horizontal rectangle in the region corresponds to when y  3 . Thus,
we sum from y  0 to y  3 . Thus, we obtain the following.
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~anderson/1860

3
0
4
 ( 8  2 y ) 2 dy = 

3
0

3
0
[ 2 ( 4  y ) ] 2 dy = 

3
0
4 ( 4  y ) 2 dy =
( 4  y ) 2 dy
Let u  4  y
Then du   dy
4

3
0
( 4  y ) 2 dy =  4 
4
u3 
4
u3
= 4  3  =
3
1
 

3
0
( 4  y ) 2 (  1 ) dy =  4 
4
1
=

1
4
u 2 du = 4 

4
1
u 2 du
4
4
4
( 64  1 ) =
( 63 ) =
( 21 ) =
3
3
1
84 
Answer: 84 
5.
y  x 2  6 x , y  0 ; revolve about the x-axis
2
NOTE: The graph of the equation y  x  6 x is a parabola that opens
upward from its vertex. We need to find the coordinates of this vertex. Since
y  x 2  6 x  y  x ( x  6 ) , then the x-coordinates of the x-intercepts of
the graph of the parabola are x  0 and x  6 . Thus, the axis of symmetry
for the parabola is the vertical line x  3 . Since the vertex of the parabola
lies on this axis of symmetry, then the x-coordinate of the vertex is x  3 .
Since y  x ( x  6 ) , then we can find the x-coordinate for the vertex by
y  3 (  3 )   9 . The graph of the equation y  0 is the x-axis.
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~anderson/1860
y
y  x2  6x
x
6
x
2
Radius = x  6 x ; Width = dx
( x, x 2  6 x)
Picture of the solid of revolution. Picture of half of the solid of revolution
from x  0 to x  3 .
The volume of the i th disk, which is obtained by rotating the i th rectangle
2
2
2
about the x-axis, is given by the product  r dx =  ( x  6 x ) dx . In
order to find the volume of the solid of revolution, we will need to sum the
volume of all the disks, which are obtained by rotating all the vertical
rectangles about the x-axis, using integration.
The first vertical rectangle in the region corresponds to when x  0 and the
last vertical rectangle in the region corresponds to when x  6 . Thus, we
sum from x  0 to x  6 . Thus, we obtain the following.

6
0
 ( x 2  6 x ) 2 dx = 
 x 5 12 x 4
36 x 3
 


5
4
3


6
0
( x 4  12 x 3  36 x 2 ) dx =
6

 x5
  =  
 3 x 4  12 x 3
0
 5
6

  =
0
6
 3 2

x  x  15 x  60  0 =
[ 216 ( 36  90  60 )  0 ] =
5
5

1296
[ 216 ( 6 ) ] =
5
5
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~anderson/1860
Answer:
6.
1296
5
y  2 sin 3 x , y  0 , x 

; revolve about the x-axis
2
NOTE: The graph of the sine function y  2 sin 3 x has an amplitude of 2
2
and a period of
. The graph of the equation y  0 is the x-axis. The
3


graph of the equation x 
is a vertical line which crosses the x-axis at .
2
2
y
( x , 2 sin 3 x )
2
Radius = 2 sin 3 x ; Width = dx
x
2

3
x

2
2
3
x
Radius =  2 sin 3 x
( x , 2 sin 3 x )
Width = dx
y  2 sin 3 x
x 

2
Picture of the solid of revolution.
NOTE: The radius of the each disk, which is obtained by rotating a

rectangle, for values of x between 0 and
is 2 sin 3 x . The radius of the
3
each disk, which is obtained by rotating a rectangle, for values of x between


2
and
is  2 sin 3 x . Thus, the square of radius is 4 sin 3 x for all
3
2

values of x between 0 and .
2
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~anderson/1860
The volume of the i th disk, which is obtained by rotating the i th rectangle
2
2
about the x-axis, is given by the product  r dx =  ( 4 sin 3 x ) dx =
4 sin 2 3 x dx . In order to find the volume of the solid of revolution, we will
need to sum the volume of all the disks, which are obtained by rotating all the
vertical rectangles about the x-axis, using integration.
The first vertical rectangle in the region corresponds to when x  0 and the

last vertical rectangle in the region corresponds to when x  . Thus, we
2

sum from x  0 to x  . Thus, we obtain the following.
2

 /2
0
4  sin 2 3 x dx = 4 

 /2
0
sin 2 3 x dx
Let u  3 x
Then du  3 dx
4

 /2
0
4
3
sin 2 3 x dx =
3 / 2
4   u sin 2 u  
 

3  2
4   0
3 / 2


( 2 u  sin 2 u ) 
3
0
=

 /2
0
sin 2 3 x 3 dx =
4
3

3 / 2
0
sin 2 u du =
3 / 2
4 1

 ( 2 u  sin 2 u ) 
=
3 4
0
=

[ 3  sin 3  ( 0  sin 0 ) ] =
3


[ 3  0  ( 0  0 ) ] =
( 3 )   2
3
3
Answer: 
7.
2
x  cos y , x  0 , y    , y   ; revolve about the y-axis
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~anderson/1860
NOTE: The graph of the cosine function x  cos y has an amplitude of 1
and a period of 2  . The graph of the equation x  0 is the y-axis. The
graph of the equation y    is a horizontal line which crosses the y-axis
at   . The graph of the equation y   is a horizontal line which crosses
the y-axis at  .
y
x  cos y

( cos y , y )
y 
y

2
( cos y , y )
x
y
Radius =  cos y
Width = dy
 2
( cos y , y )

Radius = cos y ; Width = dy
y
y  
Picture of the solid of revolution.
NOTE: The radius of the each disk, which is obtained by rotating a


rectangle, for values of y between 
and
is cos y . The radius of the
2
2
each disk, which is obtained by rotating a rectangle, for values of y between

  and  is  cos y . The radius of the each disk, which is obtained by
2

rotating a rectangle, for values of y between
and  is also  cos y .
2
2
Thus, the square of radius is co s y for all values of y between   and
.
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~anderson/1860
The volume of the i th disk, which is obtained by rotating the i th rectangle
2
about the y-axis, is given by the product  r dy =  cos y dy . In order to
find the volume of the solid of revolution, we will need to sum the volume of
all the disks, which are obtained by rotating all the horizontal rectangles
about the y-axis, using integration.
The first horizontal rectangle in the region corresponds to when y    and
the last horizontal rectangle in the region corresponds to when y   . Thus,
we sum from y    to y   . Thus, we obtain the following.


 cos 2 y dy = 




cos 2 y dy = 2 


0
cos 2 y dy since the integrand is
even
2


0


2
 u sin 2 u  

( 2 u  sin 2 u )  =
cos 2 y dy = 2   
 =
4 0
4
 2
0




( 2 u  sin 2 u )  =
[ 2   sin 2   ( 0  sin 0 ) ] =
2
2
0


[ 2  0  ( 0  0 ) ] =
( 2 )   2
2
2
Answer: 
8.
2
y  ln 4 x and y  ln 20 ; revolve about the y-axis
1
. The graph of the
4
equation y  ln 20 is a horizontal line which crosses the y-axis at ln 20 .
The graph of y  ln 4 x crosses the x-axis at x 
Since we are rotating the region about the y-axis, our rectangles will be
horizontal. In order to measure horizontal lengths, we use the x-coordinates
of points. Thus, we will need to solve for x in the equation y  ln 4 x .
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~anderson/1860
y  ln 4 x  4 x  e y  x 
1 y
e
4
y  ln 4 x
y
y  ln 20
ln 20
 1 y

 e , y 
4


y
1
4
5
x
1 y
e ; Width = dy
4
Radius =
Picture of the solid of revolution.
NOTE: The region is not bounded below. Thus, the lower limit of
integration will be negative infinity.
The volume of the i th disk, which is obtained by rotating the i th rectangle
2
1 y
2
about the y-axis, is given by the product  r dy =   e  dy . In order
4 
to find the volume of the solid of revolution, we will need to sum the volume
of all the disks, which are obtained by rotating all the horizontal rectangles
about the y-axis, using integration.
The first horizontal rectangle in the region corresponds to when y    and
the last horizontal rectangle in the region corresponds to when y  ln 20 .
Thus, we sum from y    to y  ln 20 . Thus, we obtain the following.

ln 20

2
1 
  e y  dy =
4 

ln 20

 ln 20 2 y
 1

  e 2 y  dy =
e dy =
16   
 16

Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~anderson/1860
ln 20


1

lim  e 2 y 
lim 
e 2 y dy =
16 t     2
16 t    t
t
ln 20
=

lim ( e 2 ln 20  e 2 t ) =
32 t   


400 

lim ( e ln 400  e 2 t ) =
lim ( 400  e 2 t ) =
( 400  0 ) =
=
32 t   
32 t   
32
32
100 
25 
=
2
8
2
NOTE: e 2 ln 20 = exp( 2 ln 20 ) = exp( ln 20 ) = exp( ln 400 )  400
lim e 2 t  0
t  
Answer:
25 
2
Copyrighted by James D. Anderson, The University of Toledo
www.math.utoledo.edu/~anderson/1860