# Calculate Volume of Cylindrical Shells using Integration Contributed by: In this section, we will learn:
How to apply the method of cylindrical shells to find out the volume of a solid.
1. 6
APPLICATIONS OF INTEGRATION
2. APPLICATIONS OF INTEGRATION
6.3
Volumes by
Cylindrical Shells
In this section, we will learn:
How to apply the method of cylindrical shells
to find out the volume of a solid.
3. VOLUMES BY CYLINDRICAL SHELLS
Some volume problems are very
difficult to handle by the methods
discussed in Section 6.2
4. 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.
5. 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
we would have to
solve the cubic
equation y = 2x2 - x3
for x in terms of y.
 That’s not easy.
6. VOLUMES BY CYLINDRICAL SHELLS
Fortunately, there is a method—the
method of cylindrical shells—that is
easier to use in such a case.
7. CYLINDRICAL SHELLS METHOD
The figure shows a cylindrical shell
and height h.
8. 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 .
9. CYLINDRICAL SHELLS METHOD
Thus, we have:
V V2  V1
2 2
 r2 h   r h 1
2 2
 (r2  r )h 1
 (r2  r1 )(r2  r1 )h
r2  r1
2 h(r2  r1 )
2
10. CYLINDRICAL SHELLS METHOD Formula 1
Let ∆r = r2 – r1 (thickness of the shell) and
r  12  r2  r1 
Then, this formula for the volume of a
cylindrical shell becomes:
V 2 rhr
11. CYLINDRICAL SHELLS METHOD
V 2 rhr
The equation can be remembered as:
V = [circumference] [height] [thickness]
12. CYLINDRICAL SHELLS METHOD
Now, let S be the solid
obtained by rotating
region bounded by
y = f(x) [where f(x) ≥ 0],
y = 0, x = a and x = b,
where b > a ≥ 0.
13. CYLINDRICAL SHELLS METHOD
Divide the interval [a, b] into n subintervals
[xi - 1, xi ] of equal width xi and let be x
the midpoint of the i th subinterval.
14. CYLINDRICAL SHELLS METHOD
The rectangle with
base [xi - 1, xi ] and
height f ( xi ) is rotated
 The result is a
cylindrical shell with
height f ( xi ) , and
thickness ∆x.
15. CYLINDRICAL SHELLS METHOD
Thus, by Formula 1, its volume is
calculated as follows:
Vi (2 xi )[ f ( xi )]x
16. CYLINDRICAL SHELLS METHOD
So, an approximation to the volume V of S
is given by the sum of the volumes of
these shells:
n n
V  Vi  2 xi f ( xi )x
i 1 i 1
17. CYLINDRICAL SHELLS METHOD
The approximation appears to become better
as n →∞.
However, from the definition of an integral,
we know that:
n b
lim  2 xi f ( xi )x  2 x f ( x)dx
n  a
i 1
18. CYLINDRICAL SHELLS METHOD Formula 2
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:
b
V  2 xf ( x)dx
a
where 0 ≤ a < b
19. CYLINDRICAL SHELLS METHOD
The argument using cylindrical shells
makes Formula 2 seem reasonable,
but later we will be able to prove it.
20. CYLINDRICAL SHELLS METHOD
Here’s the best way to remember
the formula.
 Think of a typical shell,
cut and flattened,
circumference 2πx,
height f(x), and
thickness ∆x or dx:
b
 2 x   f( x)
a dx
circumference height thickness
21. 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.
22. CYLINDRICAL SHELLS METHOD Example 1
Find the volume of the solid obtained by
rotating about the y-axis the region
bounded by y = 2x2 - x3 and y = 0.
23. CYLINDRICAL SHELLS METHOD Example 1
We see that a typical shell has
height f(x) = 2x2 - x3.
24. CYLINDRICAL SHELLS METHOD Example 1
So, by the shell method,
the volume is: 2
V   2 x   2 x  x dx
2 3
0
2
3 4
  2 x  (2 x  x )dx
0
4 5 2
2  x  x 
1
2
1
5 0
2  8  325  165 
25. CYLINDRICAL SHELLS METHOD Example 1
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.
26. 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.
27. However, in other examples,
the methods learned in Section 6.2
may be easier.
28. CYLINDRICAL SHELLS METHOD Example 2
Find the volume of the solid obtained
by rotating about the y-axis the region
between y = x and y = x2.
29. CYLINDRICAL SHELLS METHOD Example 2
The region and a typical shell
are shown here.
 We see that the shell has radius x, circumference 2πx,
and height x - x2.
30. CYLINDRICAL SHELLS METHOD Example 2
Thus, the volume of the solid is:
1
V  2 x   x  x 2  dx
0
1
2  x  x  dx
2 3
0
3 4 1
x x  
2    
 3 4 0 6
31. CYLINDRICAL 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.
32. CYLINDRICAL SHELLS METHOD Example 3
Use cylindrical shells to find the volume of
the solid obtained by rotating about the x-axis
the region under the curve y  x from 0 to 1.
 This problem was solved using disks in Example 2
in Section 6.2
33. CYLINDRICAL SHELLS METHOD Example 3
To use shells, we relabel the curve
y x
as x = y2.
the x-axis, we see that
a typical shell has
2πy, and height 1 - y2.
34. CYLINDRICAL SHELLS METHOD Example 3
So, the volume is: 1
V  2 y   1  y 2  dy
0
1
3
2 ( y  y )dy
0
2 4 1
y y  
2    
 2 4 0 2
 In this problem, the disk method was simpler.
35. CYLINDRICAL SHELLS METHOD Example 4
Find the volume of the solid obtained by
rotating the region bounded by y = x - x2
and y = 0 about the line x = 2.
36. CYLINDRICAL SHELLS METHOD Example 4
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.
37. CYLINDRICAL SHELLS METHOD Example 4
So, the volume of the solid is:
0
V  2  2  x   x  x  dx2
1
0
2   x  3 x  2 x  dx
3 2
1
4 1
x 3 2 
2   x  x  
 4 0 2