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Finding Volumes
Chapter 6.2February 22, 2007
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In General:
Vertical Cut: Horizontal Cut:
A =topfunction
⎛⎝⎜
⎞⎠⎟−
bottomfunction
⎛⎝⎜
⎞⎠⎟dxa
b
∫ a≤b
A =rightfunction
⎛⎝⎜
⎞⎠⎟−
leftfunction
⎛⎝⎜
⎞⎠⎟dyc
d
∫ c≤d
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In-class Integrate:
Set up an integral to find the area of the region bounded by:
y =5e−x, y=2e3x, and x ≥0
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Find the area of the region bounded by
Bounds? In terms of y: [-2,1]
Points: (0,-2), (3,1)Right Function?
Left Function?
Area?
x =2 + yx =4 −y2
[(4 −y2−2
1
∫ )−(2 + y)]dy
x + y2 =4 and x−y=2
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Volume & Definite Integrals
We used definite integrals to find areas by slicing the region and adding up the areas of the slices.
We will use definite integrals to compute volume in a similar way, by slicing the solid and adding up the volumes of the slices.
For Example………………
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Blobs in SpaceVolume of a blob:
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Blobs in SpaceVolume of a blob:
Cross sectional area at height h: A(h)
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Blobs in SpaceVolume of a blob:
Cross sectional area at height h: A(h)
Volume =
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ExampleSolid with cross sectional area A(h) = 2h at height h.Stretches from h = 2 to h = 4. Find the volume.
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ExampleSolid with cross sectional area A(h) = 2h at height h.Stretches from h = 2 to h = 4. Find the volume.
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ExampleSolid with cross sectional area A(h) = 2h at height h.Stretches from h = 2 to h = 4. Find the volume.
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ExampleSolid with cross sectional area A(h) = 2h at height h.Stretches from h = 2 to h = 4. Find the volume.
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Volumes:
We will be given a “boundary” for the base of the shape which will be used to find a length.
We will use that length to find the area of a figure generated from the slice .
The dy or dx will be used to represent the thickness.
The volumes from the slices will be added together to get the total volume of the figure.
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Find the volume of the solid whose bottom face is the circle and every cross section perpendicular to the x-axis is a square.
x2 + y2 ≤1
Bounds?
Top Function?
Bottom Function?
[-1,1]
y = 1−x2
y =− 1−x2
Length? 1−x2 − − 1−x2( )
=2 1− x2
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Find the volume of the solid whose bottom face is the circle and every cross section perpendicular to the x-axis is a square.
x2 + y2 ≤1
We use this length to find the area of the square.
Length? =2 1− x2
Area? 2 1−x2( )2
4 1−x2( )
4 1−x2( )dx−1
1
∫
Volume?
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Find the volume of the solid whose bottom face is the circle and every cross section perpendicular to the x-axis is a square.
x2 + y2 ≤1
What does this shape look like?
4 1−x2( )dx−1
1
∫Volume?
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Find the volume of the solid whose bottom face is the circle and every cross section perpendicular to the x-axis is a circle with diameter in the plane.
x2 + y2 ≤1
Length? =2 1− x2
Area?
p 1− x2( )
2
p 1 − x2( )
p 1− x2( )dx−1
1
∫Volume?
2 1−x2
2Radius:
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Using the half circle [0,1] as the base slices perpendicular to the x-axis are isosceles right triangles.
x2 + y2 ≤1
Length? =2 1− x2
Area?12
2 1−x2( ) 2 1−x2( )
Volume? 2 1−x2( )dx0
1
∫
Bounds? [0,1]
Visual?
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The base of the solid is the region between the curve and the interval [0,π] on the x-axis. The cross sections perpendicular to the x-axis are equilateral triangles with bases running from the x-axis to the curve.
y =2 sinx
Bounds?
Top Function?
Bottom Function?
[0,π]
y =2 sinx
y =0
Length? 2 sin x
Area of an equilateral triangle?
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Area of an Equilateral Triangle?
S
S
S
S/2
S
S
Sqrt(3)*S/2
S/2
Area = (1/2)b*h
=12
⎛⎝⎜
⎞⎠⎟ S( )
32S
⎛⎝⎜
⎞⎠⎟=
34S2
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The base of the solid is the region between the curve and the interval [0,π] on the x-axis. The cross sections perpendicular to the x-axis are equilateral triangles with bases running from the x-axis to the curve.
y =2 sinx
Bounds?
Top Function?
Bottom Function?
[0,π]
y =2 sinx
y =0
Length? 2 sin x
Area of an equilateral triangle?3
4(2 sin x )2
34
(2 sin x )2dx0
p
∫Volume?
34
(S)2⎛⎝⎜
⎞⎠⎟
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Find the volume of the solid whose bottom face is the circle and every cross section perpendicular to the x-axis is a square with diagonal in the plane.
x2 + y2 ≤1
We used this length to find the area of the square whose side was in the plane….
Length? =2 1− x2
Area with the length representing the diagonal?
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Area of Square whose diagonal is in the plane?
D
SS
S2 + S2 =D 2
2S2 =D 2
S2 =D 2
2⇒ S =
D2
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Find the volume of the solid whose bottom face is the circle and every cross section perpendicular to the x-axis is a square with diagonal in the plane.
x2 + y2 ≤1
Length of Diagonal? =2 1− x2
Length of Side?
2 1−x2
2= 2 1− x2
2(1−x2 )Area?
Volume? 2 1−x2( )dx−1
1
∫
(S =D2)
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Solids of Revolution
We start with a known planar shape and rotate that shape about a line resulting in a three dimensional shape known as a solid of revolution. When this solid of revolution takes on a non-regular shape, we can use integration to compute the volume.
For example……
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The Bell!
Volume:
Area:
p f (x)( )2
p f (x)( )2 dxa
b
∫
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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
p x( )2
p xdx0
3
∫
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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
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Solids of RevolutionRotate a region about an axis.
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Solids of RevolutionRotate a region about an axis.
Such as: Region between y = x2 and the y-axis, for 0 ≤ x ≤ 2, about the y-axis.
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Solids of RevolutionRotate a region about an axis.
Such as: Region between y = x2 and the y-axis, for 0 ≤ x ≤ 2, about the y-axis.
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Solids of RevolutionFor solids of revolution, cross sections are circles, so we can use the formula
Volume = p r(h)( )2 dha
b
∫
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Solids of RevolutionFor solids of revolution, cross sections are circles, so we can use the formula
Usually, the only difficult part is determining r(h).A good sketch is a big help.
Volume = p r(h)( )2 dha
b
∫
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ExampleRotate region between x = sin(y), 0 ≤ y ≤ π, and the y axis about the y axis.
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ExampleRotate region between x = sin(y), 0 ≤ y ≤ π, and the y axis about the y axis.
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ExampleRotate region between x = sin(y), 0 ≤ y ≤ π, and the y axis about the y axis.
We need to figure out whatthe resulting solid looks like.
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Aside: Sketching Revolutions1. Sketch the curve; determine the region.
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Aside: Sketching Revolutions1. Sketch the curve; determine the region.
2. Sketch the reflection over the axis.
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Aside: Sketching Revolutions1. Sketch the curve; determine the region.
2. Sketch the reflection over the axis.
3. Sketch in a few “revolution” lines.
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ExampleRotate region between x = sin(y), 0 ≤ y ≤ π, and the y axis about the y axis.
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ExampleRotate region between x = sin(y), 0 ≤ y ≤ π, and the y axis about the y axis.
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ExampleRotate region between x = sin(y), 0 ≤ y ≤ π, and the y axis about the y axis.
Now find r(y), the radius at height y.
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ExampleRotate region between x = sin(y), 0 ≤ y ≤ π, and the y axis about the y axis.
Now find r(y), the radius at height y.
(Each slice is a circle.)y
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ExampleRotate region between x = sin(y), 0 ≤ y ≤ π, and the y axis about the y axis.
y
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ExampleRotate region between x = sin(y), 0 ≤ y ≤ π, and the y axis about the y axis.
x = sin(y)
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ExampleRotate region between x = sin(y), 0 ≤ y ≤ π, and the y axis about the y axis.
So r(y) = sin(y).
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ExampleRotate region between x = sin(y), 0 ≤ y ≤ π, and the y axis about the y axis.
So r(y) = sin(y).
Therefore, volume is
p r y( )( )2dy
a
b
∫
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ExampleRotate region between x = sin(y), 0 ≤ y ≤ π, and the y axis about the y axis.
So r(y) = sin(y).
Therefore, volume is
p r y( )( )2dy
a
b
∫ = p sin y( )( )2dy
a
b
∫
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ExampleRotate region between x = sin(y), 0 ≤ y ≤ π, and the y axis about the y axis.
So r(y) = sin(y).
Therefore, volume is
What are the limits?
p r y( )( )2dy
a
b
∫ = p sin y( )( )2dy
a
b
∫
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ExampleRotate region between x = sin(y), 0 ≤ y ≤ π, and the y axis about the y axis.
So r(y) = sin(y).
Therefore, volume is
What are the limits?
The variable is y, so the limits are in terms of y…
= p sin y( )( )2dy
a
b
∫p r y( )( )2dy
a
b
∫
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ExampleRotate region between x = sin(y), 0 ≤ y ≤ π, and the y axis about the y axis.
Upper limit: y = π
Lower limit: y = 0
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ExampleRotate region between x = sin(y), 0 ≤ y ≤ π, and the y axis about the y axis.
Volume is
= p sin y( )( )2dy
0
π
∫= p sin y( )( )2dy
a
b
∫p r y( )( )2dy
a
b
∫
=p sin y( )( )2dy
0
π
∫ =p12
−12
cos 2y( )⎛⎝⎜
⎞⎠⎟dy
0
π
∫
=py2
−14
sin 2y( )⎛⎝⎜
⎞⎠⎟
0
π
=p 2
2
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ExampleRevolve the region under the curve y = 3e–x, for 0 ≤ x ≤ 1, about the x axis.
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ExampleRevolve the region under the curve y = 3e–x, for 0 ≤ x ≤ 1, about the x axis.
(Since we rotate about x, the slices are perpendicular to x.So x is our variable.)
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ExampleRevolve the region under the curve y = 3e–x, for 0 ≤ x ≤ 1, about the x axis.
First, get the region:
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ExampleRevolve the region under the curve y = 3e–x, for 0 ≤ x ≤ 1, about the x axis.
First, get the region:
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ExampleRevolve the region under the curve y = 3e–x, for 0 ≤ x ≤ 1, about the x axis.
First, get the region:When we rotate, this will become a radius.
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ExampleRevolve the region under the curve y = 3e–x, for 0 ≤ x ≤ 1, about the x axis.
Next, revolve the region:
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ExampleRevolve the region under the curve y = 3e–x, for 0 ≤ x ≤ 1, about the x axis.
Next, revolve the region:
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ExampleRevolve the region under the curve y = 3e–x, for 0 ≤ x ≤ 1, about the x axis.
Radius at point x is 3e–x.
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ExampleRevolve the region under the curve y = 3e–x, for 0 ≤ x ≤ 1, about the x axis.
Radius at point x is 3e–x.
Limits are from x = to x =
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ExampleRevolve the region under the curve y = 3e–x, for 0 ≤ x ≤ 1, about the x axis.
Radius at point x is 3e–x.
Limits are from x = 0 to x = 1
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ExampleRevolve the region under the curve y = 3e–x, for 0 ≤ x ≤ 1, about the x axis.
Radius at point x is 3e–x.
Limits are from x = 0 to x = 1
Volume:
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ExampleRevolve the region under the curve y = 3e–x, for 0 ≤ x ≤ 1, about the x axis.
Radius at point x is 3e–x.
Limits are from x = 0 to x = 1
Volume:
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ExampleRevolve the region under the curve y = 3e–x, for 0 ≤ x ≤ 1, about the x axis.
Radius at point x is 3e–x.
Limits are from x = 0 to x = 1
Volume:
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ExampleRevolve the region under the curve y = 3e–x, for 0 ≤ x ≤ 1, about the x axis.
Radius at point x is 3e–x.
Limits are from x = 0 to x = 1
Volume:
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Remember for this method: Slices are perpendicular to the
axis of rotation. Radius is then 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.
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ExampleRotate y = x2, from x = 0 to x = 4, about the x-axis.
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ExampleRotate y = x2, from x = 0 to x = 4, about the x-axis.
First, sketch:
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ExampleRotate y = x2, from x = 0 to x = 4, about the x-axis.
Next, rotate about the x axis:
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ExampleRotate y = x2, from x = 0 to x = 4, about the x-axis.
Next, rotate about the x axis:
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ExampleRotate y = x2, from x = 0 to x = 4, about the x-axis.
Next, find r(x): r(x) =
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ExampleRotate y = x2, from x = 0 to x = 4, about the x-axis.
Next, find r(x): r(x) =
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ExampleRotate y = x2, from x = 0 to x = 4, about the x-axis.
Next, find r(x): r(x) = x2
Limits:
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ExampleRotate y = x2, from x = 0 to x = 4, about the x-axis.
Next, find r(x): r(x) = x2
Limits: x = 0 to x = 4
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ExampleRotate y = x2, from x = 0 to x = 4, about the x-axis.
Next, find r(x): r(x) = x2
Limits: x = 0 to x = 4
Volume:
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ExampleRotate y = x2, from x = 0 to x = 4, about the x-axis.
Next, find r(x): r(x) = x2
Limits: x = 0 to x = 4
Volume:
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ExampleRotate y = x2, from x = 0 to x = 4, about the x-axis.
Next, find r(x): r(x) = x2
Limits: x = 0 to x = 4
Volume:
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ExampleRotate y = x2, from x = 0 to x = 4, about the x-axis.
Next, find r(x): r(x) = x2
Limits: x = 0 to x = 4
Volume:
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Find the volume of the solid generated by revolving the region defined by , on the interval [1,2] about the x-axis.
y =x2
Bounds?
Length? (radius)
Area?
Volume?
[1,2]
y =x2
p x2( )2
p x4dx1
2
∫
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Find the volume of the solid generated by revolving the region defined by , on the interval [1,2] about the x-axis.
y =x2
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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
p y3( )2
p y3`( )2dy
0
8
∫
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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
p 1− x2( )2
p 1− x2( )2dx
−1
1
∫
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What if there is a “gap” between the axis of rotation and the function?
(Temporarily end powerpointLook at Washer2.gifWasher5.gifWasher8.gifWasher10.gif)
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*Find the volume of the solid generated by revolving the region defined by , and y = 1, about the x-axis.
y =2−x2
Bounds?
Outside Radius?
Inside Radius?
Area?
[-1,1]
2 −x2
1
p 2 − x2( )2
− π 1( )2
Volume? p 2 − x2( )2
− π 1( )2( )
−1
1
∫ dx
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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( )
p 3 − x2( )2
− π 2( )2
Volume? p 3 − x2( )2
− π 2( )2( )
−1
1
∫ dx
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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 x = 4
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Summary Described how a plane region rotated
about an axis describes a solid.
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Summary Described how a plane region rotated
about an axis describes a solid. Found volumes of solids of revolution
using the area of a circle
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Summary Described how a plane region rotated
about an axis describes a solid. Found volumes of solids of revolution
using the area of a circle Determined the variable as given by the
axis of rotation.
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Solids of RevolutionIf there is a gap between the function and the axis of rotation, we have a washer and use:
If there is NO gap, we have a disk and use:
Volume = p R(h)( )2 − (r(h))2( )dha
b
∫
p r(h)( )2 dha
b
∫Volume =