This document discusses calculating the volume of solids of revolution formed by rotating an area bounded by graphs around an axis. It provides the formula for finding the volume of a cylindrical shell as well as the formula for finding the total volume of a solid of revolution by summing the volumes of infinitely thin cylindrical shells. It includes two example problems demonstrating how to set up and solve the integrals to find the volume of solids of revolution.

1. Volumes by Cylindrical Shells

2. The volume of a right circular cylindrical shell with radius r, height h,
and infinitesimal thickness dx, is given by:
Vshell = 2πrh dx
If one slits the cylinder down a side and unrolls it into a rectangle,
the height of the rectangle is the height of the cylinder, h, and the
length of the rectangle is the circumference of a circular end of
the cylinder, 2πr. So the area of the rectangle (and the surface of
the cylinder) is 2πrh. Multiply this by a (slight) thickness dx to get
the volume.

3. In the diagram, the yellow region is revolved about the
y-axis. Two of the shells are shown. For each value of x
between 0 and a (in the graph), a cylindrical shell is
obtained, with radius x and height f(x). Thus, the
volume of one of these shells (with thickness dx) is given
by
Vshell = 2π x f(x) dx.

4. Summing up the volumes of all these
infinitely thin shells, we get the total volume
of the solid of revolution:
V=
a
a
0
0
ò 2p xf (x)dx = 2p ò xf (x)dx

5. Example 1: Find the volume of the solid of revolution formed
by rotating the region bounded by the x-axis and the graph of
y = x from x=0 to x=1, about the y-axis.
ù
V = ò 2p x x dx = 2p ò x dx = 2p × 2x ú =
1
0
5 ú
ú0
1
û
5
5 öù
4p
4p æ 2
ç1 - 0 2 ÷ú =
5
5 è
øú0
û
2
1
3
2
5 1
2

6. Example 2: Find the volume of the solid of revolution formed by
rotating the finite region bounded by the graphs of
y = x -1
2
and y = ( x -1) about the y-axis.

7. 2
V = ò 2p x
1
(
x -1 - ( x -1)
2
)
2
æ2
ö
2
dx = 2p ç ò x x -1 dx - ò x ( x -1) dx ÷ =
è1
ø
1
æ
ö
2
2p ç ò ( u +1) u du - ò ( u +1) u du÷ =
è
ø
æ æ 3 1ö
ö
3
2
2p ç ò ç u 2 + u 2 ÷ du - ò ( u + u ) du÷ =
ç
÷
è
ø
è
ø
u = x – 1 so x = u + 1
du = dx
3
é 5
ù
4
3
2
2
ê 2u 2u u u ú
2p ê
+
- - ú=
5
3
4 3
ê
ú
ë
û
5
3
é
4
3ù
2
2
ê 2 ( x -1) + 2 ( x -1) - ( x -1) - ( x -1) ú =
2p
ê
5
3
4
3 ú
ê
ú1
ë
û
2
é 2 2 1 1ù
29 29p
=
2p ê + - - ú = 2p
60 30
ë 5 3 4 3û

8. Time to Practice !!!
EXAMPLE Find the volume of the solid
obtained by rotating the region bounded by
y = x - x 2 and y = 0 about the line x = 2 .

9. One More Example