Let
be continuous and nonnegative for
and let
denote the region bounded by the graph of
the
-axis,
and the lines
and
The volume of the solid obtained by rotating
about the
-axis
is

Find the volume of the cone obtained by revolving about the -axis the region bounded above by the graph of and below by the -axis for

**Solution.** Using formula

with
we obtain

Since this cone has height
and base of radius
this result agrees with the formula from geometry for the volume of this cone:

Find the volume of the solids obtained by rotating the region bounded by the graphs of and about the -axis.

**Solution**

The two graphs cross at
and
since the
equation
implies
or
Since
for
the region is bounded above by the graph of
and below by the graph of
,
we can view the solid
as the ``difference'' of two other solids. Let
be the solid obtained by rotating the region bounded by the graph of
about the x-axis for
and let
be the solid obtained by rotating the region bounded by the graph of
about the
-axis
for
The the original solid
is the solid that results when
is removed from
The desired volume
is, therefore, the difference between the volume of
and the volume of
Using equation

twice, we obtain

As you may have observed, the volume in Example 2 could have been calculated
by the single integral

In general, if the region
is bounded above by the graph of
and below by the graph of
and
,
then the formula for volume is

The region in the first quadrant bounded by the graph of and the coordinate axes is rotated about the -axis. Find the volume of the resulting solid.

**Solution**

This problem is similar to that of Example 1 except that the role of
and
are reversed. Solving the given equation for
as a function of
gives
Since the rotation is about the
-axis,
the integration will be with
and the limits of integration are from
to
We obtain

The result of Example 3 generalizes as follows:

If the region
is bounded by the graph of the continuous function
and the
-axis
from
to
then the volume of the solid obtained by rotating
about the
-axis
is

If we ``slice'' a solid of revolution
in the direction perpendicular to the axis of rotation, we obtain a circular
disc. For example, if
is obtained by revolving the region
bounded by the graph of a nonnegative function
about the x-axis
for
the slice for which
is a disc with radius
Consequently, its area
is
.
In fact, we can interpret formula

for the volume of
as the integral of the cross-sectional area function
from
to
Our next objective is to show that the volume of any solid cross-sectional
area can be calculated in this way.

In particular, suppose that
is the solid for which the area
of each cross-section perpendicular to the
-axis
is known
(
need not be a solid of revolution). Moreover, suppose that
extends from
to
.
We partition the interval
into subintervals of equal length
with endpoints
Using the partition, we subdivide solids
into solids
of thickness
where each
corresponds to the interval
.
By selecting one number
in each subinterval
,
we approximate the volume of
by the volume
of the cylinder with face area
;
that is,

Summing the
for
yields an approximation to the volume
of

We assume that, as
, the Riemann sum on the right-hand side approaches the volume
Now, if
is continuous for
the sum also approaches the definite integral

Thus, we obtain the formula for the volume of solids with known
cross-sectional area.

Let
denote the area of the cross section of
for
.
If function
is continuous on
the volume
of
is

Notice this equation generalizes the familiar formula for the cylinder with base area and height It also contains formula for the volume of a solid of revolution as a special case, since for such solids,

The base of a solid is a circle of radius cm. All cross sections perpendicular to a particular axis are squares. Find the volume of this solid.

**Solution**

If we impose an
-coordinate
system on the circular base so that the x-axis corresponds to the given axis,
then the equation for the boundary of the base is

The equation for the upper semicircle is
,
and the equation for the lower semicircle is
.
A cross section perpendicular to the x-axis will therefore intersect this
circular base in a chord of length
Since this chord is one side of the square cross section, the area of the
cross section is

The smallest and largest values of
are, respectively,
and
The volume is therefore

Find the volume of the solid generated by rotating the region bounded by the graph of and the -axis for about the line

**Solution**

A cross section taken at location
consists of a circle of radius
from which a smaller circle of radius
has been removed. The cross-sectional area is therefore

We
can now apply equation

to
find that