finding volumes disk/washer/shell
DESCRIPTION
Finding Volumes Disk/Washer/Shell. Chapter 6.2 & 6.3 February 27, 2007. Review of Area: Measuring a length. Vertical Cut:. Horizontal Cut:. Disk Method. Slices are perpendicular to the axis of rotation. Radius is a function of position on that axis. - PowerPoint PPT PresentationTRANSCRIPT
Finding VolumesDisk/Washer/Shell
Chapter 6.2 & 6.3February 27, 2007
Review of Area: Measuring a length.
Vertical Cut: Horizontal Cut:
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f (x)( )− g(x)( )
rightfunction
⎛⎝⎜
⎞⎠⎟−
leftfunction
⎛⎝⎜
⎞⎠⎟
f (y)( )− g(y)( )
topfunction
⎛⎝⎜
⎞⎠⎟−
bottomfunction
⎛⎝⎜
⎞⎠⎟
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 region defined by , y = 8, and x = 0 about the y-axis.
y =x3
Bounds?
Length?
Area?
Volume?
[0,8]
x = y3
π y3( )2
π y3`( )2dy
0
8
∫
Find the volume of the solid generated by revolving the region defined by , and y = 1, about the line y = 1
y =2−x2
Bounds?
Length?
Area?
Volume?
[-1,1]
(2−x2 )−1
π 1− x2( )2
π 1− x2( )2dx
−1
1
∫
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)) ora 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
R
R
r
AreaofOuter
⎛
⎝⎜⎜
⎞
⎠⎟⎟−
AreaofInner
⎛
⎝⎜⎜
⎞
⎠⎟⎟
π R2( ) − π r2( )
π R2( ) − r2( )⎡⎣ ⎤⎦
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 , and y = 1, about the x-axis using planar slices perpendicular to the AOR.
y =2−x2
Bounds?
Outside Radius?
Inside Radius?
Area?
[-1,1]
2−x2
1
π 2 − x2( )2
− π 1( )2
Volume? π 2 − x2( )2
− π 1( )2( )
−1
1
∫ dx
Find the volume of the solid generated by revolving the region defined by , and y = 1, about the line y=-1.
y =2−x2
Bounds?
Outside Radius?
Inside Radius?
Area?
[-1,1]
2−x2( )− −1( )
1− −1( )
π 3 − x2( )2
− π 2( )2
Volume? π 3 − x2( )2
− π 2( )2( )
−1
1
∫ dx
Let R be the region in the x-y plane bounded by
Set up the integral for the volume obtained by rotating R about the x-axis using planar slices perpendicular to the axis of rotation.
y =4x, y=
14, and x =2
Notice the gap: y =4x, y=
14, and x =2
Outside Radius ( R ):
Inside Radius ( r ):
4x
14
Area: π4x
⎛⎝⎜
⎞⎠⎟
2
− π14
⎛⎝⎜
⎞⎠⎟
2
Volume: π4x
⎛⎝⎜
⎞⎠⎟
2
− π14
⎛⎝⎜
⎞⎠⎟
2
dx2
16
∫
Let R be the region in the x-y plane bounded by
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.
y =2x2 and y=3x−1
Notice the gap:
Outside Radius ( R ):
Inside Radius ( r ):
3x−1
2x2
Area: π 3x − 1( )2 − π 2x2( )2
Volume: π 3x −1( )2 − π 2x2( )2
( )dx12
1
∫
y =2x2 and y=3x−1
Find the volume of the solid generated by revolving the region defined by , x = 3 and the x-axis about the x-axis.
y = x
Bounds?
Length? (radius)
Area?
Volume?
[0,3]
y = x
π x( )2
π xdx0
3
∫
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.
2πrlengthofslice
⎛
⎝⎜⎜
⎞
⎠⎟⎟Area:
2πr
Length of slice
Volume =
2πradiusfromAOR
⎛
⎝⎜⎜
⎞
⎠⎟⎟
lengthofslice
⎛
⎝⎜⎜
⎞
⎠⎟⎟
a
b
∫ dyx
⎛⎝⎜
⎞⎠⎟
Slice is PARALLEL to the AOR
Using on the interval [0,2] revolving around the x-axis using planar slices PARALLEL to the AOR, we find the volume:
y = x
Radius?
Length of slice?
Area?
Volume?
y
2−y2
2π y( ) 2 −y2( )
2π y( ) 2 −y2( )dy0
2
∫
Back to example:Find volume of the solid generated by revolving the region about the y-axis using cylindrical slices
y =4x, y=
14, and x =2
Length of slice ( h ):
Radius ( r ):
4x−14
x
Area: 2πx4x−14
⎛⎝⎜
⎞⎠⎟
Volume: 2πx4x−14
⎛⎝⎜
⎞⎠⎟dx
2
16
∫
Find the volume of the solid generated by revolving the region:about the y-axis, using cylindrical slices.
Length of slice ( h ):
Inside Radius ( r ):
3x−1−2x2
x
Area: 2πx 3x−1−2x2( )
Volume: 2πx 3x−1−2x2( )dx12
1
∫
y =2x2 and y=3x−1
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