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 .
