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Section 5.2: Volume: The Disk Method
Practice HW from Larson Textbook (not to hand in)
p. 322 # 1-27 odd
Solids of Revolution
In this section, we want to examine how to find the volume of a solid of revolution,
which is formed by rotating a region in a plane about a line.
Examples: Figures 5.14 p. 315
Disc Method
Consider a rectangle of length R and width w.
If we rotate this rectangle about this axis, we sweep out a circular disk.
Volume of Disk = (Area of Disk)(Width of Disk) = R 2 w .
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Now, suppose we have a function R(x) and rotate it about the x-axis.
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V  Volume of Disk i  ( R 2 )x   [ R( xi )] 2 x
To generate the volume of the entire solid, we divide up the interval from x = a to x = b
into n equal subintervals, for n rectangles, and add up the volumes of the disks formed by
rotating each rectangle around the x axis.
Volume of a single Disk i   [ R( xi )] 2 x
n
n
i 1
i 1
Volume of the Solid    [ R( xi )] 2 x   [ R( xi )] 2 x
n
2
b
Exact Volume of the Solid = lim   R( xi ) x    R( x)2 dx
n 
We can also rotate around the y-axis.
i 1
a
4
d
Exact Volume of the Solid =   R( y )2 dy
c
Disc Method
To find the volume of a solid of revolution, we use
b
Horizontal Axis of Revolution: Volume    R( x)2 dx
a
d
Vertical Axis of Revolution: Volume    R( y )2 dy
c
Notes
1. The representative rectangle in the disk method is always perpendicular to the axis of
rotation.
2. It is important to realize that when finding the function R representing the radius of the
solid, measure from the axis of revolution.
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6
Example 1: Find the volume of the solid form by rotating the region generated by the
graphs bounded by x = 0, y = 0 and y  4  x 2 , where x  0 about the x-axis.
Solution:
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Example 2: Find the volume of the solid generated by revolving the region bounded by
y  2x 2 , x = 0, and y = 8 where x  0 about the
a. y axis.
b. the line y = 8.
Solution:
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The Washer Method
Involves extending the disk method to cover solids of revolution with a hole. Consider
the following rectangle, where w is the width, R is the distance from the axis of
revolution to the outer edge of the rectangle, and r is the distance from the axis of
revolution to the inner edge of the rectangle.
If we rotate this rectangle about this axis, we sweep out a circular washer.
Look at Figures 5.18-5.20 p. 318
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Now, consider the region bound by an outer radius R(x) and inner radius r(x) and suppose
we rotate this region around the x axis.
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Then
b
b
Volume of Washer    R( x)2 dx    r ( x)2 dx
a
 

a
 

Volume of entire solid
Volume of entire hole
b


   R( x)2  r ( x)2 dx
a
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Example 3: Find the volume of the solid obtained by rotating the region bounded by
y  x 2 , y  4 , and x  0 about the
a. The x axis
b. x = 3.
Solution (Part a): We consider the following region given by the following graphs:
If we rotate about the x-axis, we must find the radius of the entire solid (outer radius) and
the radius of the hole (inner radius). Here, the outer radius is R( x)  4 and inner radius is
r ( x)  x 2 . Hence, the volume of the washer revolved about the x-axis (a horizontal axis) is
Volume  


2
2
 R( x)  r ( x)  dx
x2
x 0
x 2
 (4)
2

 ( x 2 ) 2 dx
x 0
x 2
 (16  x
4
 dx
Substitute R( x)  4 and r ( x)  x 2
Square and simplify
x 0
x2
1
  (16 x  x 5 )
5
x 0
1
  (16(2)  (2) 5 )   (0)
5
32
  (32  )
5
32 (5) 32
160 32
128
(
 )  (
 ) 
1 (5) 5
5
5
5
Integrate
Substitute in limits of integratio n
Simplify
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(Part b): We consider the following region given by the following graphs:
If we rotate about the x = 3 (a vertical axis), the radius of the entire solid (outer radius) is
R ( y )  3  0  3 and the radius of the hole (inner radius) is r ( y)  3  y . Hence, the
volume is
y 4
2
2
 R( y)  r ( y)  dy
Volume  
y 0
y 4

 (3)
2

 (3  y ) 2 dy
Substitute R( y )  3 and r ( x)  3  y
y 0
y 4
 (9  (9  6


y  y ) dy
FOIL (3  y ) 2  9  6 y  y
x 0
y 4

 (9  9  6

y  y ) dy
Distribute  sign
y 0
y 4

 (6 y
1/ 2

 y ) dy
Simplify - cancel 9' s
y 0
continued on next page
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y 4

 (6 y
1/ 2

 y ) dy
Simplify - cancel 9' s
y 0
y 4


 y 3/ 2 1 
  6
 y2 
3
2 



2

 y 0
Integrate
y 4
1 

   4 y3/ 2  y 2 
2  y 0

Simplify - note
1
  ( 4 ( 4 ) 3 / 2  ( 4 ) 2 )   ( 0)
2
Substitute in limits of integratio n
1
  (4(8)  (16)) - 0
2
6
2
 6  4
3
3
2
Simplify - note (4) 3/2  (41 / 2 ) 3  ( 4 ) 3  (2) 3  8
  (32  8)
 24
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