# Calculating Volumes of Slicing

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

## Example 1

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:

## Example 2

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

## Example 3

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

# Solids of Known Cross-Sectional Area

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,

## Example 4

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

# Example 5

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

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