Transcript PowerPoint Presentation - Solids of Revolution

Finding Volumes
Disk/Washer/Shell
Chapter 6.2 & 6.3
February 27, 2007
Review of Area: Measuring a length.
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 top
  bottom 
 function   function
 f (x) g(x)
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 right
  left

 function   function
 f (y) g(y)
Disk Method



Slices are perpendicular to the axis of rotation.
Radius is a function of position on that axis.
Therefore rotating about x axis gives an integral
in x; rotating about y gives an integral in y.
Find the volume of the solid generated by revolving the
3
y

x
region defined by
, y = 8, and x = 0 about the
y-axis.
Bounds?
[0,8]
Length?
x 3 y
 y
2
Area?

Volume?
 
3
8
3`
0

2
y dy
Find the volume of the solid generated by revolving the
region defined by y  2  x 2, and y = 1, about the
line y = 1
[-1,1]
Bounds?
2
(2

x
) 1
Length?
Area?
 1  x
1


2 2
Volume?   1  x
1
 dx
2 2
What if there is a “gap” between the axis of rotation and
the function?
Solids of Revolution:
We determined that a cut perpendicular to
the axis of rotation will either form a
disk (region touches axis of rotation
(AOR)) or
a washer (there is a gap between the
region and the AOR)
Revolved around the line y = 1, the region
forms a disk
However when revolved around the x-axis,
there is a “gap” between the region and
the x-axis. (when we draw the radius, the
radius intersects the region twice.)
Area of a Washer
 Area   Area 
 of
   of 

 

 Outer   Inner 
R
r
 R2   r 2 
 R 2  r 2 
Note: Both R and r are measured
from the axis of rotation.
Find the volume of the solid generated by revolving the
region defined by y  2  x 2, and y = 1, about the
x-axis using planar slices perpendicular to the AOR.
[-1,1]
Bounds?
Outside Radius?
1
Inside Radius?

Area?  2  x

1
Volume?
1
 2  x

2 2

  1 dx
2
2  x2
   1
2 2
2
Find the volume of the solid generated by revolving the
region defined by y  2  x 2, and y = 1, about the
line y=-1.
Bounds?
[-1,1]


Outside Radius? 2  x 2  1
Inside Radius?
Area?
 3  x
   2 
2 2
2
  3  x    2 dx
1
Volume?
1 1
1
2 2
2
Let R be the region in the x-y plane bounded by
4
1
y  , y  , and x  2
x
4
Set up the integral for the volume obtained by rotating R
about the x-axis using planar slices perpendicular to the
axis of rotation.
4
1
y  , y  , and x  2
x
4
Notice the gap:
Outside Radius ( R ):
4
x
Inside Radius ( r ):
1
4
2
Area:
2
 4
 1
2   x     4  dx
16
Volume:
2
 4
 1
   
 x
 4
2
Let R be the region in the x-y plane bounded by
y  2x 2 and y  3x  1
Set up the integral for the volume of the solid obtained
by rotating R about the x-axis, using planar slices
perpendicular to the axis of rotation.
Notice the gap:
y  2x 2 and y  3x  1
Outside Radius ( R ): 3x 1
Inside Radius ( r ):
2x 2
 
Area:  3x  1   2x
2
  3x  1
1
2
Volume:
1
2
  dx
  2x
2 2
2 2
Find the volume of the solid generated by revolving the
region defined by y  x , x = 3 and the x-axis about
the x-axis.
[0,3]
Bounds?
Length? (radius) y  x
Area?

 x
2
3
Volume?
  xdx
0
Note in the disk/washer methods, the focus in on the
radius (perpendicular to the axis of rotation) and the
shape it forms. We can also look at a slice that is
parallel to the axis of rotation.
Note in the disk/washer methods, the focus in on the
radius (perpendicular to the axis of rotation) and the
shape it forms. We can also look at a slice that is
parallel to the axis of rotation.
2 r
Length
of slice
 length

Area: 2 r  of


 slice 
 radius   length 
 y




d 
Volume =  2  from   of

 x
a
 AOR   slice 
b
Slice is PARALLEL to the AOR
Using y  x on the interval [0,2] revolving around
the x-axis using planar slices PARALLEL to the AOR,
we find the volume:
Length of slice?
Radius?
2  y2
y

Area? 2 y  2  y
2

Volume?
2


2
2

y
2

y
dy



0
4
1
y  , y  , and x  2
x
4
Back to example:
Find volume of the solid generated by revolving the
region about the y-axis using cylindrical slices
4 1

Length of slice ( h ):
x 4
Radius ( r ):
Area:
4
2 2 x  x 
16
Volume:
1
 dx
4
x
4
2 x  
x
1

4
Find the volume of the solid generated by revolving the
region: y  2x 2 and y  3x  1
about the y-axis, using cylindrical slices.
Length of slice ( h ):
3x  1  2x 2
Inside Radius ( r ):

x
Area: 2 x 3x  1 2x
1
Volume:


2
2

x
3x

1

2x
dx

1
2
2

Try:

Set up an integral integrating with respect
to y to find the volume of the solid of
revolution obtained when the region
bounded by the graphs of y = x2 and y = 0
and x = 2 is rotated around
a) the y-axis

b) the line y = 4
