In this explainer, we will learn how to find the volume of a solid generated by revolving a region around either a horizontal or a vertical line using integration.

Suppose we have a curve in some interval , as depicted in the diagram.

If we take the part of the curve between and and rotate it about the -axis by a complete revolution (i.e., by or ), the curve will map out the surface of a solid as it rotated; this is called a solid of revolution, as depicted in the diagram.

If the curve defined by described a straight line, we would obtain a cone. If it described a circle, we would obtain a sphere. So, how do we determine the volume of such solids of revolution?

### Definition: Volume of Solids of Revolution

Suppose that a solid lies between the vertical lines and , whose cross-sectional area in the plane through and perpendicular to the -axis is . If is continuous on the interval , we can divide the interval into subintervals of equal width, , and choose a point, , in each interval.

The volume of the solid formed by revolving the region bounded by the curve and the -axis between and about the -axis is given by where

Similarly, the volume of the solid formed by revolving the region bounded by the curve and the -axis between and about the -axis is given by

This can also be visualized in a diagram.

For a curve , the cross-sectional shape is a solid disk or circle whose area is , where is the radius. The radius of each circle will be the value of the function at that point. Thus, the cross section perpendicular to the axis of revolution is a disk of radius , and we have

Therefore, the volume of the solid formed by revolving the region bounded by the curve and the -axis between and about the -axis can alternatively be written as

We can visualize this in a graph as follows:

We can also use a similar process to calculate the volume of a solid rotated about the -axis. The volume of the solid formed by revolving the region bounded by the curve and the -axis between and about the -axis is given by

This can be visualized as follows:

As an example, let’s derive the formulae for the volume of a sphere. The equation of a circle centered at the origin with radius is given by or, equivalently, written as a functions of as .

In fact, we only need to consider the semicircle in the top half-quadrant described by as shown in the plot:

We can rotate this curve between and about this -axis by a complete revolution to form a sphere of radius centered at the origin.

The cross-sectional area of this solid is given by

Therefore, the volume of the solid of revolution is given by

This is the standard result for the volume of a sphere, as expected. We would have obtained the same result if we had considered the semicircle in the region below the -axis, described by .

Now, let’s do the same thing for the volume of a cone. Consider the equation of a straight line through the origin, , as shown in the plot.

We can rotate this line between and by a complete revolution about the -axis to form a cone of radius and vertical height , as shown in the diagram.

The cross-sectional area of this solid is given by

Therefore, the volume of the solid of revolution is given by

This is the standard result for the volume of a cone, as expected.

Now, let’s look at a few examples to practice and deepen our understanding of how to calculate the volumes of solids of revolution using integration. In the first example, we will determine the volume of a solid generated by a complete revolution of a particular region about the -axis. This will be determined using integration.

### Example 1: Finding the Volume of the Solid Generated by the Revolution of the Region Bounded by a Given Line around the 𝑥-Axis

Consider the region bounded by the curves , , , and . Determine the volume of the solid of revolution created by rotating this region about the -axis.

### Answer

In this example, we want to calculate the volume of a solid generated by the revolution of a particular region around the -axis.

We can visualize the region bounded by the curves , , , and as follows:

The solid of revolution formed by rotating this region about the -axis is a frustum, which looks like a cone cut off at the end.

Since we rotated the region about the -axis, the volume is given by

The vertical cross-sectional shape of our solid will be a disk, and the radius, , of each disk will be the value of the function at that point, . Thus, the cross-sectional area is given by

The limits of the integral, and , are determined by the vertical bounds of in the region bounded by the curves, ; this can also be visualized from the plot of the region bounded by the curves. Thus, we have and , and the volume of the solid of revolution is given by

Thus, the volume of the solid of revolution is

In the next example, we will determine the volume generated by the revolution of a region, bounded by a radical function and other lines, about the -axis. This will be determined using integration.

### Example 2: Finding the Volume of the Solid Generated by the Revolution of the Region under the Curve of a Root Function about the 𝑥-Axis

Find the volume of the solid obtained by rotating the region bounded by the curve and the lines and about the -axis.

### Answer

In this example, we want to calculate the volume of a solid generated by the revolution of a particular region around the -axis.

We can visualize the region bounded by the curve and lines and as follows:

The solid of revolution formed by rotating this region about the -axis looks like this:

Since we are rotating this region about the -axis, the volume is given by

The vertical cross-sectional shape of our solid will be a disk, and the radius of each disk will be the value of the function at that point, . Thus, the cross-sectional area is given by

The limits of the integral, and , are determined by the vertical bounds of in the region. The upper bound is and the lower bound is determined by the intersection of the curves and , which is ; this can also be visualized from the plot of the region bounded by the curves. Thus, we have and , and the volume of the solid of revolution is given by

Thus, the volume of the generated solid is

Now, let’s consider an example where we have to find the volume of solid generated by the revolution of a region bounded by a parabola around the -axis. This will be determined using integration.

### Example 3: Finding the Volume of the Solid Generated by the Revolution of the Region Bounded by a Parabola around the 𝑥-Axis

Find the volume of the solid generated by rotating the region bounded by the curve and the -axis complete revolution about the -axis.

### Answer

In this example, we want to calculate the volume of a solid generated by the revolution of a particular region around the -axis.

We can visualize the region bounded by the curve and the -axis as follows:

The solid of revolution formed by rotating this region about the -axis looks like this:

Since we are rotating this region about the -axis, the volume is given by

The cross-sectional shape of our solid will be a disk, and the radius of each disk will be the value of the function at that point, . Thus, the cross-sectional area is given by

The limits of the integral, and , are determined by the vertical bounds of in the region, which in this case occurs when the curve intersects the -axis or when :

The solutions are given by and , which can also be visualized from the plot of the region bounded by the curves. Thus, we have and , and the volume of the solid of revolution is given by

Thus, the volume of the generated solid is

Now, let’s look at an example where we determine the volume of a solid of revolution, but this time rotated about the -axis. This will be determined using integration.

### Example 4: Finding the Volume of the Solid Generated by the Revolution of the Region Bounded by Given Lines around the 𝑦-Axis

Find the volume of the solid generated by turning, through a complete revolution about the -axis, the region bounded by the curve and the lines , , and .

### Answer

In this example, we want to calculate the volume of a solid generated by the revolution of a particular region about the -axis.

We can visualize the region bounded by the curves and the lines , , and as follows:

The solid of revolution formed by rotating this region about the -axis looks like a cone.

Since we rotated this region about the -axis, the volume is given by

The horizontal cross-sectional shape of our solid will be a disk, and the radius of each disk will be the value of the function at that point, . Thus, the cross-sectional area is given by

The limits of the integral, and , are determined by the horizontal bounds of in the region, ; this can also be visualized from the plot of the region bounded by the curves. Thus, we have and , and the volume of the solid of revolution is given by

Thus, the volume of the generated solid is

The volume of a solid generated by revolving the region bounded by and on the interval , where , about the -axis is which we can also visualize in a diagram:

Similarly, the volume of a solid generated by revolving the region bounded by and on the interval , where , about the -axis is which can be visualized as:

Now, let’s consider some examples about the curves of two functions rotated about an axis, where the cross-sectional shape of each solid generated is a disk.

We will begin by determining the volume of a solid generated by rotating a region bounded by two functions of , a pair of polynomials of degrees 2 and 4, about the -axis.

### Example 5: Finding the Volume of the Solid Generated by the Revolution of the Region Bounded by a Parabola and the Curve of a Power Function about the 𝑦-Axis

Find the volume of the solid obtained by rotating the region bounded by the curves and about the -axis.

### Answer

We can visualize the region bounded by the curves and as follows:

The solid of revolution formed by rotating about the -axis looks like this:

And the solid formed by looks like this:

Thus, the solid of revolution formed by rotating the region bounded by these curves about the -axis looks like this:

Since we rotated this region about the -axis, the volume is given by

The cross-sectional shape of each solid will be a disk, and the radius of each disk will be the value of the function at that point, and . Thus, the difference between the areas of the larger disk and the smaller disk is given by

The limits of the integral, and , are determined by the horizontal bounds of in the region, which in this case is given by the values of at the points where the two curves intersect:

The solution is given by and , and the two curves intersect at the points and ; this can also be visualized from the plot of the region bounded by the curves. Thus, we have and , and the volume of the solid of revolution is given by

Thus, the volume of the generated solid is

Now, let’s look at an example where we have to find the volume of a solid generated by the revolution of a region bounded by two curves, a parabola and a line, around the -axis. This will be determined using integration.

### Example 6: Finding the Volume of the Solid Generated by the Revolution of the Region Bounded by a Parabola and a Line around the 𝑦-Axis

Find the volume of the solid obtained by rotating the region bounded by the curve and the line about the -axis.

### Answer

In this example, we want to calculate the volume of the solid generated by the revolution of a particular region around the -axis.

We can visualize the region bounded by the curve and the line as follows:

The solid of revolution formed by rotating about the -axis looks like this:

And the solid formed by looks like this:

Thus, the solid of revolution formed by rotating the region bounded by these curves about the -axis looks like this:

Since we rotated the region about the -axis, the volume is given by

The horizontal cross-sectional shape of our solid will be a disk, and the radius of each disk will be the value of the function at that point, and . Thus, the cross-sectional area is given by

The limits of the integral, and , are determined by the horizontal bounds of in the region, which in this case is where the two curves intersect:

The solutions are given by and , and the two curves intersect at the points and ; this can also be visualized from the plot of the region bounded by the curves. Thus, we have and , and the volume of the solid of revolution is given by

Thus, the volume of the generated solid is

Finally, let’s consider an example where we have to find the volume of the solid generated by rotating a region bounded by curves about a line that is parallel to the -axis. This will be determined using integration.

### Example 7: Finding the Volume of the Solid Generated by the Revolution of a Region around a Line Parallel to the 𝑦-Axis

Consider the region bounded by the curves , , and . Find the volume of the solid obtained by rotating this region about .

### Answer

In this example, we want to calculate the volume of a solid generated by the revolution of a particular region around the line , which is parallel to the -axis.

We can visualize the region bounded by the curves , , and as follows:

The solid of revolution formed by rotating about looks like this:

And the solid formed by looks like this:

Thus, the solid of revolution formed by rotating the region bounded by these curves about the -axis looks like this:

Since we rotated this region about the line , which is parallel to the -axis, the volume is given by

The cross-sectional shape of each solid will be a disk. However, since we are rotating about the line rather than the -axis, or the line , we have to take this into account by subtracting the equations describing the curves from 3. The radii of each of the disks will be and . Thus, the difference between the areas of the larger disk and the smaller disk is given by

The limits of the integral, and , are determined by the horizontal bounds of in the region. The lower bound is and the upper bound is determined by the intersection of the curves and , which is ; this can also be visualized from the plot of the region bounded by the curves. Thus, we have and , and the volume of the solid of revolution is given by

Thus, the volume of the generated solid is

### Key Points

**The First Method**: When the curve of a single function is rotated about an axis, the cross-sectional shape of the solid will be a disk, and the radius, , of each disk will be the value of the function at that point. The cross-sectional area is determined by .

The volume of the solid formed by revolving the region bounded by the curve and the -axis between and about the -axis is given by Similarly, the volume of the solid formed by revolving the region bounded by the curve and the -axis between and about the -axis is given by**The Second Method**: When the curves of two functions are rotated about an axis, the cross-sectional shape of each solid generated will be a disk, and the radii, and , will be the value of each of the functions at that point. The cross-sectional area is determined by subtracting the inner area from the outer area as .

The volume of a solid generated by revolving the region bounded by and between and , where , about the -axis is Similarly, the volume of a solid generated by revolving the region bounded by and between and , where , about the -axis is- The limits of the integrals are determined by the vertical or horizontal bounds of or , respectively, depending on the axis of rotation and the region bounded by the curves.