Section 7.3: Volume by Cylindrical Shells
Let f be a continuous function such that f (x) ≥ 0 for all x in [a, b]. What is the volume
of the solid obtained by revolving the region under the curve y = f (x) for a ≤ x ≤ b about
the y-axis?
Consider a vertical strip of width dx. If this strip is revolved about the x-axis, we obtain a
cylindrical shell with height f (x) and thickness dx. The shell has volume
dV = 2πxf (x)dx.
Summing the volumes of these shells for a ≤ x ≤ b, we obtain the volume of the solid.
Theorem: (The Shell Method)
If R is the region under the curve y = f (x) on the interval [a, b], then the volume of the solid
obtained by revolving R about the y-axis is
Z b
xf (x)dx.
V = 2π
a
1
Example: Find the volume of the solid obtained by revolving the region bounded by y = ,
x
y = 0, x = 1, and x = 4 about the y-axis.
Using the Shell Method,
Z
V
4
= 2π
1
Z
= 2π
4
1dx
1
= 6π.
1
x
dx
x
Example: Find the volume of the solid obtained by revolving the region bounded by y = x−x2
and y = 0 about the y-axis.
Using the Shell Method,
1
Z
V
x(x − x2 )dx
= 2π
Z0 1
(x2 − x3 )dx
1
1 3 1 4 = 2π
x − x 3
4
0
π
=
.
6
= 2π
0
2
Example: Find the volume of the solid obtained by revolving the region bounded by y = e−x ,
y = 0, x = 0, and x = 1 about the y-axis.
Using the Shell Method,
1
Z
2
xe−x dx.
V = 2π
0
2
Let u = x , du = 2xdx. Then
Z
V
1
e−u du
0
1
= −πe−u 0
1
.
= π 1−
e
= π
Example: Find the volume of the solid obtained by revolving the region bounded by x = y 2 ,
x = 0, y = 2, and y = 5 about the x-axis.
Using the Shell Method,
Z
V
5
= 2π
y(y 2 )dy
Z2 5
y 3 dy
2
4 5
y = 2π
4 2
π
(625 − 16)
=
2
609π
=
.
2
= 2π
Example: Find the volume of the solid obtained by revolving the region bounded by y = x2 ,
y = 0, x = 1, and x = 2 about the line x = 3.
Using the Shell Method,
Z
V
2
= 2π
(3 − x)x2 dx
Z1 2
(3x2 − x3 )dx
1
2
1 4 3
= 2π x − x 4
1
13π
.
=
2
= 2π
Example: Find the volume of the solid obtained by revolving the region bounded by y =
and y = x2 about the line x = −1.
Using the Shell Method,
Z
V
1
= 2π
Z0 1
= 2π
√
(1 + x)( x − x2 )dx
√
( x − x2 + x3/2 − x3 )dx
0
= 2π
=
29π
.
30
1
2 3/2 1 3 2 5/2 1 4 x − x + x − x 3
3
5
4
0
√
x
Note: For many problems, more than one method can be used to compute the volume of a
solid of revolution.
Example: Consider the region R bounded by y = x2 and y = 2x. Set up (but do not evaluate)
integrals to find the volume of the solid obtained by revolving R about the x-axis using the
Washer Method and Shell Method.
Using the Washer Method,
2
Z
2
2
Z
2 2
(4x2 − x4 )dx.
[(2x) − (x ) ]dx = π
V =π
0
0
Using the Shell Method,
Z
4
y
V = 2π
0
√
y
y−
dy = 2π
2
Z
0
4
1 2
3/2
y − y dy.
2
π
. Set up
2
(but do not evaluate) integrals to find the volume of the solid obtained by revolving R about
the y-axis using the Disk Method and Shell Method.
Example: Consider the region R bounded by y = cos x, y = 0, x = 0, and x =
Using the Shell Method,
π/2
Z
x cos xdx.
V = 2π
0
Using the Disk Method,
Z
V =π
0
1
[cos−1 (y)]2 dy.