Volume of Solids of Revolution
from section 13.3
Consider a region R in the xy-plane. Take any point (x, y) of the region. If we rotate this point about
the x-axis , it generates a circle whose radius is |y| and therefore the perimeter of the circle is 2π |y|. By
rotating the region R we get a solid which is the union of the circle generated. Therefore the volume of
the solid is the continuous sum of the circles. This continuous sum is the double integral. So the
volume of the solid of revolution is
∫∫
2π |y|d(x, y)
R
If the region R is rotated about the y-axis, then the volume will be:
∫∫
2π |x|d(x, y)
R
The general formula is:
∫∫
{
}
2π distance between (x, y) and the axis of revolution d(x, y)
R
Note Some modifications might be necessary depending on the problem. The next two examples show
how to apply the modifications.
Section 13.3 exercise 17 Use double integral
 to find the volume of the solid of revolution obtained by
 y = 4x2 − 4x
about the line y = −2.
rotating the region bounded by the curves
 y = x3
Solution The first step is the plotting of the region and finding the intersection of the two curves.

 y = 4x2 − 4x
 y = x3
⇒
4x2 − 4x = x3
⇒
x3 − 4x2 + 4x = 0
1
⇒
x(x2 − 4x + 4) = 0
⇒

 0
x=
 2
Take an arbitrary point (x, y) in the region. The distance between this point and the line y = −2 is
y + 2. So according to the formula, we have
∫∫
Volume =
[
2π (y + 2)d(x, y) = 2π
R
]y=x3
∫ x=2 [
1 2
= 2π
y + 2y
x=0
=π
∫ 2
0
[
2
]
∫ x=2 ∫ y=x3
x=0
y=4x2 −4x
(y + 2)dy dx
dx = after substitution and simplifying =
y=4x2 −4x
(x6 − 16x4 + 36x3 − 32x2 + 16x)dx
1
16
32
= π x7 − x5 + 9x4 − x3 + 8x2
7
5
3
]2
=
0
668π
105
2