6.3 volumes by cylindrical shells applications of integration in this section, we will learn: how to...
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6.3
Volumes by
Cylindrical Shells
APPLICATIONS OF INTEGRATION
In this section, we will learn:
How to apply the method of cylindrical shells
to find out the volume of a solid.
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Some volume problems are very difficult to handle
by the methods discussed in Section 5.2.
The next example will illustrate this claim.
VOLUMES BY CYLINDRICAL SHELLS
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Let us consider the problem of finding the volume
of the solid obtained by rotating about the y-axis
the region bounded by y = 2x2 - x3 and y = 0.
VOLUMES BY CYLINDRICAL SHELLS
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If we slice perpendicular to the y-axis, we get a
washer. However, to compute the inner radius and the outer
radius of the washer, we would have to solve the cubic equation y = 2x2 - x3 for x in terms of y.
That is not easy.
VOLUMES BY CYLINDRICAL SHELLS
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The figure on the right shows
a cylindrical shell with inner
radius r1, outer radius r2, and
height h.
CYLINDRICAL SHELLS METHOD
Fortunately, there is a method, the method of
cylindrical shells, that is easier to use in such
a case.
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Its volume V is calculated by subtracting the volume
V1 of the inner cylinder from the volume of the outer
cylinder V2 .
CYLINDRICAL SHELLS METHOD
2 1
2 22 1
2 22 1
2 1 2 1
2 12 1
( )
( )( )
2 ( )2
V V V
r h r h
r r h
r r r r h
r rh r r
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Let ∆r = r2 – r1 (thickness of the shell) and
r = (r2 + r1)/2 (average radius of the shell).
Then, this formula for the volume of a cylindrical shell becomes: 2V rh r
Formula 1CYLINDRICAL SHELLS METHOD
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The equation can be remembered as:
V = [circumference] [height] [thickness]
CYLINDRICAL SHELLS METHOD
2V rh r
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Now, let S be the solid obtained by rotating about
the y-axis the region bounded by y = f (x), where
f(x) ≥0, y = 0, x = a and x = b, where b > a ≥ 0.
CYLINDRICAL SHELLS METHOD
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Divide the interval [a, b] into n subintervals [xi - 1, xi ]
of equal width and let be the midpoint of the
ith subinterval.
x ix
CYLINDRICAL SHELLS METHOD
( )iy f x
ix
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The rectangle with base [xi - 1, xi ] and height is rotated about the y-axis.
The result is a cylindrical shell with average radius , height , and thickness ∆x.
( )if x
( )if xix
CYLINDRICAL SHELLS METHOD
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Thus, by Formula 1, its volume is calculated as
follows:
So, an approximation to the volume V of S is given
by the sum of the volumes of these shells:
(2 )[ ( )]i i iV x f x x
CYLINDRICAL SHELLS METHOD
1 1
2 ( )n n
i i ii i
V V x f x x
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The approximation appears to become better as
However, from the definition of an integral, we
know that:
1
lim 2 ( ) 2 ( )n b
i i ani
x f x x x f x dx
CYLINDRICAL SHELLS METHOD
.n
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Thus, the following appears plausible.
The volume of the solid obtained by rotating about the y-axis the region under the curve y = f(x) from a to b, is:
where 0 ≤ a < b
2 ( )b
aV xf x dx
Formula 2CYLINDRICAL SHELLS METHOD
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The argument using cylindrical shells makes
Formula 2 seem reasonable, but later we will
be able to prove it.
CYLINDRICAL SHELLS METHOD
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Here is the best way to remember the formula.
Think of a typical shell, cut and flattened, with radius x, circumference 2πx, height f(x), and thickness ∆x or dx:
2 ( )b
athicknesscircumference height
x f x dx
CYLINDRICAL SHELLS METHOD
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This type of reasoning will be helpful in other
situations, such as when we rotate about lines
other than the y-axis.
CYLINDRICAL SHELLS METHOD
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Find the volume of the solid obtained by rotating
about the y-axis the region bounded by y = 2x2 - x3
and y = 0.
Example 1CYLINDRICAL SHELLS METHOD
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We see that a typical shell has radius x,
circumference 2πx, and height given by
f(x) = 2x2 – x3.
Example 1CYLINDRICAL SHELLS METHOD
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So, by the shell method, the volume is:
2 2 3
0
2 3 4
0
24 51 12 5 0
32 165 5
2 2
2 (2 )
2
2 8
V x x x dx
x x x dx
x x
Example 1CYLINDRICAL SHELLS METHOD
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It can be verified that the shell method gives the
same answer as slicing.
The figure shows a
computer-generated
picture of the solid
whose volume we
computed in the
example.
Example 1CYLINDRICAL SHELLS METHOD
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Comparing the solution of Example 1 with the
remarks at the beginning of the section, we see
that the cylindrical shells method is much easier
than the washer method for the problem.
We did not have to find the coordinates of the local maximum. We did not have to solve the equation of the curve for x in terms
of y.
However, in other examples, the methods learned
in Section 5.2 may be easier.
NOTE
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Find the volume of the solid obtained by rotating
about the y-axis the region between y = x and y = x2.
Example 2CYLINDRICAL SHELLS METHOD
The region and a typical shell are shown here. We see that the shell has radius x, circumference 2πx,
and height x - x2.
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Thus, the volume of the solid is:
1 2
0
1 2 3
0
13 4
0
2
2
23 4 6
V x x x dx
x x dx
x x
Example 2CYLINDRICAL SHELLS METHOD
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As the following example shows, the shell method
works just as well if we rotate about the x-axis.
We simply have to draw a diagram to identify the radius and height of a shell.
CYLINDRICAL SHELLS METHOD
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Use cylindrical shells to find the volume of the
solid obtained by rotating about the x-axis the
region under the curve from 0 to 1.
This problem was solved using disks in Example 2 in Section 6.2
y x
Example 3CYLINDRICAL SHELLS METHOD
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To use shells, we relabel the curve
as x = y2.
For a rotation about the x-axis, we see that a typical shell has radius y, circumference 2πy, and height 1 - y2.
y x
Example 3CYLINDRICAL SHELLS METHOD
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So, the volume is:
In this problem, the disk method was simpler.
1 2
0
1 3
0
12 4
0
2 1
2 ( )
22 4 2
V y y dy
y y dy
y y
Example 3CYLINDRICAL SHELLS METHOD
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Find the volume of the solid obtained by rotating
the region bounded by y = x - x2 and y = 0 about
the line x = 2.
Example 4CYLINDRICAL SHELLS METHOD
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The figures show the region and a cylindrical shell
formed by rotation about the line x = 2, which has
radius 2 - x, circumference 2π(2 - x), and height
x - x2.
Example 4CYLINDRICAL SHELLS METHOD
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So, the volume of the solid is:
0 2
1
0 3 2
1
143 2
0
2 2
2 3 2
24 2
V x x x dx
x x x dx
xx x
Example 4CYLINDRICAL SHELLS METHOD