APPLICATIONS OF INTEGRATION
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Transcript of APPLICATIONS OF INTEGRATION
APPLICATIONS OF INTEGRATIONAPPLICATIONS OF INTEGRATION
6
6.3Volumes 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.
Some volume problems are very
difficult to handle by the methods
discussed in Section 6.2
VOLUMES BY CYLINDRICAL SHELLS
Let’s 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
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’s not easy.
VOLUMES BY CYLINDRICAL SHELLS
Fortunately, there is a method—the
method of cylindrical shells—that is
easier to use in such a case.
VOLUMES BY CYLINDRICAL SHELLS
The figure shows a cylindrical shell
with inner radius r1, outer radius r2,
and height h.
CYLINDRICAL SHELLS METHOD
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
Thus, we have:
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
CYLINDRICAL SHELLS METHOD
Let ∆r = r2 – r1 (thickness of the shell) and
(average radius of the shell).
Then, this formula for the volume of a cylindrical shell becomes:
2V rh r
Formula 1
12 12r r r
CYLINDRICAL SHELLS METHOD
The equation can be remembered as:
V = [circumference] [height] [thickness]
CYLINDRICAL SHELLS METHOD
2V rh r
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
Divide the interval [a, b] into n subintervals
[xi - 1, xi ] of equal width and let be
the midpoint of the i th subinterval.
xix
CYLINDRICAL SHELLS METHOD
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
Thus, by Formula 1, its volume is
calculated as follows:
(2 )[ ( )]i i iV x f x x
CYLINDRICAL SHELLS METHOD
So, an approximation to the volume V of S
is given by the sum of the volumes of
these shells:
1 1
2 ( )n n
i i ii i
V V x f x x
CYLINDRICAL SHELLS METHOD
The approximation appears to become better
as n →∞.
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
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
The argument using cylindrical shells
makes Formula 2 seem reasonable,
but later we will be able to prove it.
CYLINDRICAL SHELLS METHOD
Here’s 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
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
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
We see that a typical shell has
radius x, circumference 2πx, and height f(x) = 2x2 - x3.
Example 1CYLINDRICAL SHELLS METHOD
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
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
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.
NOTE
However, in other examples,
the methods learned in Section 6.2
may be easier.
NOTE
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.
Example 2CYLINDRICAL SHELLS METHOD
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
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
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
To use shells, we relabel the curve
as x = y2.
For 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
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
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
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
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