Lesson Explainer: Volumes of Solids of Revolution Mathematics • Higher Education

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 ๐‘ฆ=๐‘“(๐‘ฅ) in some interval [๐‘Ž,๐‘], as depicted in the diagram.

If we take the part of the curve between ๐‘ฅ=๐‘Ž and ๐‘ฅ=๐‘ and rotate it about the ๐‘ฅ-axis by a complete revolution (i.e., by 360โˆ˜ or 2๐œ‹), 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 ๐‘“ 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 ๐‘ฅ=๐‘Ž and ๐‘ฅ=๐‘, whose cross-sectional area in the plane through ๐‘ฅ and perpendicular to the ๐‘ฅ-axis is ๐ด(๐‘ฅ). If ๐ด(๐‘ฅ) is continuous on the interval [๐‘Ž,๐‘], we can divide the interval into ๐‘› subintervals of equal width, ฮ”๐‘ฅ, and choose a point, ๐‘ฅ๏ƒ, in each interval.

The volume of the solid formed by revolving the region bounded by the curve and the ๐‘ฅ-axis between ๐‘ฅ=๐‘Ž and ๐‘ฅ=๐‘ about the ๐‘ฅ-axis is given by ๐‘‰=๏„š๐ด(๐‘ฅ)ฮ”๐‘ฅ=๏„ธ๐ด(๐‘ฅ)๐‘ฅ,limd๏Šโ†’โˆž๏Š๏ƒ๏Šฒ๏Šง๏ƒ๏Œป๏Œบ where ๐‘ฅ=๐‘Ž+๐‘–ฮ”๐‘ฅ,ฮ”๐‘ฅ=๐‘โˆ’๐‘Ž๐‘›=๐‘ฅโˆ’๐‘ฅ.๏ƒ๏ƒ๏ƒ๏Šฑ๏Šง

Similarly, the volume of the solid formed by revolving the region bounded by the curve and the ๐‘ฆ-axis between ๐‘ฆ=๐‘ and ๐‘ฆ=๐‘‘ about the ๐‘ฆ-axis is given by ๐‘‰=๏„ธ๐ด(๐‘ฆ)๐‘ฆ.๏Œฝ๏Œผd

This can also be visualized in a diagram.

For a curve ๐‘ฆ=๐‘“(๐‘ฅ), the cross-sectional shape is a solid disk or circle whose area is ๐ด=๐œ‹๐‘…๏Šจ, where ๐‘… 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 ๐‘…=๐‘“(๐‘ฅ), and we have ๐ด(๐‘ฅ)=๐œ‹[๐‘“(๐‘ฅ)].๏Šจ

Therefore, the volume of the solid formed by revolving the region bounded by the curve ๐‘ฆ=๐‘“(๐‘ฅ) and the ๐‘ฅ-axis between ๐‘ฅ=๐‘Ž and ๐‘ฅ=๐‘ about the ๐‘ฅ-axis can alternatively be written as ๐‘‰=๐œ‹๏„ธ[๐‘“(๐‘ฅ)]๐‘ฅ.๏Œป๏Œบ๏Šจd

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 ๐‘ฆ-axis. The volume of the solid formed by revolving the region bounded by the curve ๐‘ฅ=๐‘“(๐‘ฆ) and the ๐‘ฆ-axis between ๐‘ฆ=๐‘ and ๐‘ฆ=๐‘‘ about the ๐‘ฆ-axis is given by ๐‘‰=๐œ‹๏„ธ[๐‘“(๐‘ฆ)]๐‘ฆ.๏Œฝ๏Œผ๏Šจd

This can be visualized as follows:

As an example, letโ€™s derive the formulae for the volume of a sphere. The equation of a circle centered at the origin with radius ๐‘Ÿ is given by ๐‘ฅ+๐‘ฆ=๐‘Ÿ๏Šจ๏Šจ๏Šจ or, equivalently, written as a functions of ๐‘ฅ as ๐‘ฆ=ยฑโˆš๐‘Ÿโˆ’๐‘ฅ๏Šจ๏Šจ.

In fact, we only need to consider the semicircle in the top half-quadrant described by ๐‘ฆ=โˆš๐‘Ÿโˆ’๐‘ฅ๏Šจ๏Šจ as shown in the plot:

We can rotate this curve between ๐‘ฅ=โˆ’๐‘Ÿ and ๐‘ฅ=๐‘Ÿ about this ๐‘ฅ-axis by a complete revolution to form a sphere of radius ๐‘Ÿ centered at the origin.

The cross-sectional area of this solid is given by ๐ด(๐‘ฅ)=๐œ‹๐‘ฆ=๐œ‹๏€น๐‘Ÿโˆ’๐‘ฅ๏….๏Šจ๏Šจ๏Šจ

Therefore, the volume of the solid of revolution is given by ๐‘‰=๏„ธ๐ด(๐‘ฅ)๐‘ฅ=๏„ธ๐œ‹๏€น๐‘Ÿโˆ’๐‘ฅ๏…๐‘ฅ=๐œ‹๏–๐‘Ÿ๐‘ฅโˆ’๐‘ฅ3๏ข=๐œ‹๏–๏€พ๐‘Ÿโˆ’๐‘Ÿ3๏Šโˆ’๏€พโˆ’๐‘Ÿ+๐‘Ÿ3๏Š๏ข=43๐œ‹๐‘Ÿ.๏Ž๏Šฑ๏Ž๏Ž๏Šฑ๏Ž๏Šจ๏Šจ๏Šจ๏Šฉ๏Ž๏Šฑ๏Ž๏Šฉ๏Šฉ๏Šฉ๏Šฉ๏Šฉdd

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 ๐‘ฅ-axis, described by ๐‘ฆ=โˆ’โˆš๐‘Ÿโˆ’๐‘ฅ๏Šจ๏Šจ.

Now, letโ€™s do the same thing for the volume of a cone. Consider the equation of a straight line through the origin, ๐‘ฆ=๐‘Ÿโ„Ž๐‘ฅ, as shown in the plot.

We can rotate this line between ๐‘ฅ=0 and ๐‘ฅ=โ„Ž by a complete revolution about the ๐‘ฅ-axis to form a cone of radius ๐‘Ÿ and vertical height โ„Ž, as shown in the diagram.

The cross-sectional area of this solid is given by ๐ด(๐‘ฅ)=๐œ‹๐‘ฆ=๐œ‹๏€พ๐‘Ÿโ„Ž๏Š๐‘ฅ=๏€พ๐œ‹๐‘Ÿโ„Ž๏Š๐‘ฅ.๏Šจ๏Šจ๏Šจ๏Šจ๏Šจ๏Šจ๏Šจ

Therefore, the volume of the solid of revolution is given by ๐‘‰=๏„ธ๐ด(๐‘ฅ)๐‘ฅ=๏„ธ๏€พ๐œ‹๐‘Ÿโ„Ž๏Š๐‘ฅ๐‘ฅ=๐œ‹๐‘Ÿโ„Ž๏„ธ๐‘ฅ๐‘ฅ=๐œ‹๐‘Ÿโ„Ž๏–๐‘ฅ3๏ข=๐œ‹๐‘Ÿโ„Ž๏€พโ„Ž3โˆ’0๏Š=13๐œ‹๐‘Ÿโ„Ž.๏‚๏Šฆ๏‚๏Šฆ๏Šจ๏Šจ๏Šจ๏Šจ๏Šจ๏‚๏Šฆ๏Šจ๏Šจ๏Šจ๏Šฉ๏‚๏Šฆ๏Šจ๏Šจ๏Šฉ๏Šจddd

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

Now, letโ€™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 ๐‘ฅ-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 ๐‘ฅ-Axis

Consider the region bounded by the curves ๐‘ฆ=๐‘ฅ+4, ๐‘ฆ=0, ๐‘ฅ=0, and ๐‘ฅ=3. Determine the volume of the solid of revolution created by rotating this region about the ๐‘ฅ-axis.

Answer

In this example, we want to calculate the volume of a solid generated by the revolution of a particular region around the ๐‘ฅ-axis.

We can visualize the region bounded by the curves ๐‘ฆ=๐‘ฅ+4, ๐‘ฆ=0, ๐‘ฅ=0, and ๐‘ฅ=3 as follows:

The solid of revolution formed by rotating this region about the ๐‘ฅ-axis is a frustum, which looks like a cone cut off at the end.

Since we rotated the region about the ๐‘ฅ-axis, the volume is given by ๐‘‰=๏„ธ๐ด(๐‘ฅ)๐‘ฅ.๏Œป๏Œบd

The vertical cross-sectional shape of our solid will be a disk, and the radius, ๐‘…, of each disk will be the value of the function at that point, ๐‘…=๐‘ฅ+4. Thus, the cross-sectional area is given by ๐ด(๐‘ฅ)=๐œ‹๐‘…=๐œ‹(๐‘ฅ+4).๏Šจ๏Šจ

The limits of the integral, ๐‘Ž and ๐‘, are determined by the vertical bounds of ๐‘ฅ in the region bounded by the curves, 0โ‰ค๐‘ฅโ‰ค3; this can also be visualized from the plot of the region bounded by the curves. Thus, we have ๐‘Ž=0 and ๐‘=3, and the volume of the solid of revolution is given by ๐‘‰=๏„ธ๐ด(๐‘ฅ)๐‘ฅ=๐œ‹๏„ธ(๐‘ฅ+4)๐‘ฅ=๐œ‹๏„ธ๏€น๐‘ฅ+8๐‘ฅ+16๏…๐‘ฅ=๐œ‹๏–๐‘ฅ3+8๐‘ฅ2+16๐‘ฅ๏ข=93๐œ‹.๏Šฉ๏Šฆ๏Šฉ๏Šฆ๏Šจ๏Šฉ๏Šฆ๏Šจ๏Šฉ๏Šจ๏Šฉ๏Šฆddd

Thus, the volume of the solid of revolution is 93๐œ‹.cubicunits

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 ๐‘ฅ-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 ๐‘ฅ-Axis

Find the volume of the solid obtained by rotating the region bounded by the curve ๐‘ฆ=โˆš๐‘ฅ+1 and the lines ๐‘ฆ=0 and ๐‘ฅ=4 about the ๐‘ฅ-axis.

Answer

In this example, we want to calculate the volume of a solid generated by the revolution of a particular region around the ๐‘ฅ-axis.

We can visualize the region bounded by the curve ๐‘ฆ=โˆš๐‘ฅ+1 and lines ๐‘ฆ=0 and ๐‘ฅ=4 as follows:

The solid of revolution formed by rotating this region about the ๐‘ฅ-axis looks like this:

Since we are rotating this region about the ๐‘ฅ-axis, the volume is given by ๐‘‰=๏„ธ๐ด(๐‘ฅ)๐‘ฅ.๏Œป๏Œบd

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, ๐‘…=โˆš๐‘ฅ+1. Thus, the cross-sectional area is given by ๐ด(๐‘ฅ)=๐œ‹๐‘…=๐œ‹๏€ปโˆš๐‘ฅ+1๏‡=๐œ‹(๐‘ฅ+1).๏Šจ๏Šจ

The limits of the integral, ๐‘Ž and ๐‘, are determined by the vertical bounds of ๐‘ฅ in the region. The upper bound is ๐‘ฅ=4 and the lower bound is determined by the intersection of the curves ๐‘ฆ=โˆš๐‘ฅ+1 and ๐‘ฆ=0, which is ๐‘ฅ=โˆ’1; this can also be visualized from the plot of the region bounded by the curves. Thus, we have ๐‘Ž=โˆ’1 and ๐‘=4, and the volume of the solid of revolution is given by ๐‘‰=๏„ธ๐ด(๐‘ฅ)๐‘ฅ=๐œ‹๏„ธ(๐‘ฅ+1)๐‘ฅ=๐œ‹๏–๐‘ฅ2+๐‘ฅ๏ข=25๐œ‹2.๏Šช๏Šฑ๏Šง๏Šช๏Šฑ๏Šง๏Šจ๏Šช๏Šฑ๏Šงdd

Thus, the volume of the generated solid is 25๐œ‹2.cubicunits

Now, letโ€™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 ๐‘ฅ-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 ๐‘ฅ-Axis

Find the volume of the solid generated by rotating the region bounded by the curve ๐‘ฆ=โˆ’๐‘ฅ+2๐‘ฅ๏Šจ and the ๐‘ฅ-axis complete revolution about the ๐‘ฅ-axis.

Answer

In this example, we want to calculate the volume of a solid generated by the revolution of a particular region around the ๐‘ฅ-axis.

We can visualize the region bounded by the curve ๐‘ฆ=โˆ’๐‘ฅ+2๐‘ฅ๏Šจ and the ๐‘ฅ-axis as follows:

The solid of revolution formed by rotating this region about the ๐‘ฅ-axis looks like this:

Since we are rotating this region about the ๐‘ฅ-axis, the volume is given by ๐‘‰=๏„ธ๐ด(๐‘ฅ)๐‘ฅ.๏Œป๏Œบd

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, ๐‘…=โˆ’๐‘ฅ+2๐‘ฅ๏Šจ. Thus, the cross-sectional area is given by ๐ด(๐‘ฅ)=๐œ‹๐‘…=๐œ‹๏€นโˆ’๐‘ฅ+2๐‘ฅ๏…=๐œ‹๏€น๐‘ฅโˆ’4๐‘ฅ+4๐‘ฅ๏….๏Šจ๏Šจ๏Šจ๏Šช๏Šฉ๏Šจ

The limits of the integral, ๐‘Ž and ๐‘, are determined by the vertical bounds of ๐‘ฅ in the region, which in this case occurs when the curve ๐‘ฆ=โˆ’๐‘ฅ+2๐‘ฅ๏Šจ intersects the ๐‘ฅ-axis or when ๐‘ฆ=0: โˆ’๐‘ฅ+2๐‘ฅ=โˆ’๐‘ฅ(๐‘ฅโˆ’2)=0.๏Šจ

The solutions are given by ๐‘ฅ=0 and ๐‘ฅ=2, which can also be visualized from the plot of the region bounded by the curves. Thus, we have ๐‘Ž=0 and ๐‘=2, and the volume of the solid of revolution is given by ๐‘‰=๏„ธ๐ด(๐‘ฅ)๐‘ฅ=๐œ‹๏„ธ๏€น๐‘ฅโˆ’4๐‘ฅ+4๐‘ฅ๏…๐‘ฅ=๐œ‹๏–๐‘ฅ5โˆ’๐‘ฅ+4๐‘ฅ3๏ข=16๐œ‹15.๏Šจ๏Šฆ๏Šจ๏Šฆ๏Šช๏Šฉ๏Šจ๏Šซ๏Šช๏Šฉ๏Šจ๏Šฆdd

Thus, the volume of the generated solid is 16๐œ‹15.cubicunits

Now, letโ€™s look at an example where we determine the volume of a solid of revolution, but this time rotated about the ๐‘ฆ-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 ๐‘ฆ-Axis

Find the volume of the solid generated by turning, through a complete revolution about the ๐‘ฆ-axis, the region bounded by the curve 9๐‘ฅโˆ’๐‘ฆ=0 and the lines ๐‘ฅ=0, ๐‘ฆ=โˆ’9, and ๐‘ฆ=0.

Answer

In this example, we want to calculate the volume of a solid generated by the revolution of a particular region about the ๐‘ฆ-axis.

We can visualize the region bounded by the curves 9๐‘ฅโˆ’๐‘ฆ=0 and the lines ๐‘ฅ=0, ๐‘ฆ=โˆ’9, and ๐‘ฆ=0 as follows:

The solid of revolution formed by rotating this region about the ๐‘ฆ-axis looks like a cone.

Since we rotated this region about the ๐‘ฆ-axis, the volume is given by ๐‘‰=๏„ธ๐ด(๐‘ฆ)๐‘ฆ.๏Œฝ๏Œผd

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, ๐‘…=๐‘ฆ9. Thus, the cross-sectional area is given by ๐ด(๐‘ฆ)=๐œ‹๐‘…=๐œ‹๏€ป๐‘ฆ9๏‡=๐œ‹๐‘ฆ81.๏Šจ๏Šจ๏Šจ

The limits of the integral, ๐‘ and ๐‘‘, are determined by the horizontal bounds of ๐‘ฆ in the region, โˆ’9โ‰ค๐‘ฆโ‰ค0; this can also be visualized from the plot of the region bounded by the curves. Thus, we have ๐‘=โˆ’9 and ๐‘‘=0, and the volume of the solid of revolution is given by ๐‘‰=๏„ธ๐ด(๐‘ฆ)๐‘ฆ=๐œ‹81๏„ธ๐‘ฆ๐‘ฆ=๐œ‹81๏–๐‘ฆ3๏ข=3๐œ‹.๏Šฆ๏Šฑ๏Šฏ๏Šฆ๏Šฑ๏Šฏ๏Šจ๏Šฉ๏Šฆ๏Šฑ๏Šฏdd

Thus, the volume of the generated solid is 3๐œ‹.cubicunits

The volume ๐‘‰ of a solid generated by revolving the region bounded by ๐‘ฆ=๐‘“(๐‘ฅ) and ๐‘ฆ=๐‘”(๐‘ฅ) on the interval [๐‘Ž,๐‘], where ๐‘“(๐‘ฅ)โ‰ฅ๐‘”(๐‘ฅ), about the ๐‘ฅ-axis is ๐‘‰=๐œ‹๏„ธ๏€บ[๐‘“(๐‘ฅ)]โˆ’[๐‘”(๐‘ฅ)]๏†๐‘ฅ,๏Œป๏Œบ๏Šจ๏Šจd which we can also visualize in a diagram:

Similarly, the volume ๐‘‰ of a solid generated by revolving the region bounded by ๐‘ฅ=๐‘“(๐‘ฆ) and ๐‘ฅ=๐‘”(๐‘ฆ) on the interval [๐‘,๐‘‘], where ๐‘“(๐‘ฆ)โ‰ฅ๐‘”(๐‘ฆ), about the ๐‘ฆ-axis is ๐‘‰=๐œ‹๏„ธ๏€บ[๐‘“(๐‘ฆ)]โˆ’[๐‘”(๐‘ฆ)]๏†๐‘ฅ,๏Œฝ๏Œผ๏Šจ๏Šจd which can be visualized as:

Now, letโ€™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 ๐‘ฆ, a pair of polynomials of degrees 2 and 4, about the ๐‘ฆ-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 ๐‘ฆ-Axis

Find the volume of the solid obtained by rotating the region bounded by the curves ๐‘ฅ=6โˆ’5๐‘ฆ๏Šจ and ๐‘ฅ=๐‘ฆ๏Šช about the ๐‘ฆ-axis.

Answer

In this example, we want to calculate the volume of a solid generated by the revolution of a particular region around the ๐‘ฆ-axis.

We can visualize the region bounded by the curves ๐‘ฅ=6โˆ’5๐‘ฆ๏Šจ and ๐‘ฅ=๐‘ฆ๏Šช as follows:

The solid of revolution formed by rotating ๐‘ฅ=6โˆ’5๐‘ฆ๏Šจ about the ๐‘ฆ-axis looks like this:

And the solid formed by ๐‘ฅ=๐‘ฆ๏Šช looks like this:

Thus, the solid of revolution formed by rotating the region bounded by these curves about the ๐‘ฆ-axis looks like this:

Since we rotated this region about the ๐‘ฆ-axis, the volume is given by ๐‘‰=๏„ธ๐ด(๐‘ฆ)๐‘ฆ.๏Œฝ๏Œผd

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, ๐‘…=6โˆ’5๐‘ฆ๏Šจ and ๐‘Ÿ=๐‘ฆ๏Šช. Thus, the difference between the areas of the larger disk and the smaller disk is given by ๐ด(๐‘ฆ)=๐œ‹๐‘…โˆ’๐œ‹๐‘Ÿ=๐œ‹๏€น6โˆ’5๐‘ฆ๏…โˆ’๐œ‹๏€น๐‘ฆ๏…=๐œ‹๏€น36โˆ’60๐‘ฆ+25๐‘ฆโˆ’๐‘ฆ๏….๏Šจ๏Šจ๏Šจ๏Šจ๏Šช๏Šจ๏Šจ๏Šช๏Šฎ

The limits of the integral, ๐‘ and ๐‘‘, are determined by the horizontal bounds of ๐‘ฆ in the region, which in this case is given by the values of ๐‘ฆ at the points where the two curves intersect: 6โˆ’5๐‘ฆ=๐‘ฆ.๏Šจ๏Šช

The solution is given by ๐‘ฆ=1 and ๐‘ฆ=โˆ’1, and the two curves intersect at the points (1,โˆ’1) and (1,1); this can also be visualized from the plot of the region bounded by the curves. Thus, we have ๐‘=โˆ’1 and ๐‘‘=1, and the volume of the solid of revolution is given by ๐‘‰=๏„ธ๐ด(๐‘ฆ)๐‘ฆ=๐œ‹๏„ธ๏‘36โˆ’60๐‘ฆ+25๐‘ฆโˆ’๐‘ฆ๏๐‘ฆ=๐œ‹๏–36๐‘ฆโˆ’20๐‘ฆ+5๐‘ฆโˆ’๐‘ฆ9๏ข=376๐œ‹9.๏Šง๏Šฑ๏Šง๏Šง๏Šฑ๏Šง๏Šจ๏Šช๏Šฎ๏Šฉ๏Šซ๏Šฏ๏Šง๏Šฑ๏Šงdd

Thus, the volume of the generated solid is 376๐œ‹9.cubicunits

Now, letโ€™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 ๐‘ฆ-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 ๐‘ฆ-Axis

Find the volume of the solid obtained by rotating the region bounded by the curve ๐‘ฆ=๐‘ฅ๏Šจ and the line ๐‘ฅ=3๐‘ฆ about the ๐‘ฆ-axis.

Answer

In this example, we want to calculate the volume of the solid generated by the revolution of a particular region around the ๐‘ฆ-axis.

We can visualize the region bounded by the curve ๐‘ฅ=๐‘ฆ๏Šจ and the line ๐‘ฅ=3๐‘ฆ as follows:

The solid of revolution formed by rotating ๐‘ฅ=๐‘ฆ๏Šจ about the ๐‘ฆ-axis looks like this:

And the solid formed by ๐‘ฅ=3๐‘ฆ looks like this:

Thus, the solid of revolution formed by rotating the region bounded by these curves about the ๐‘ฆ-axis looks like this:

Since we rotated the region about the ๐‘ฆ-axis, the volume is given by ๐‘‰=๏„ธ๐ด(๐‘ฆ)๐‘ฆ.๏Œฝ๏Œผd

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, ๐‘…=3๐‘ฆ and ๐‘Ÿ=๐‘ฆ๏Šจ. Thus, the cross-sectional area is given by ๐ด(๐‘ฆ)=๐œ‹๐‘…โˆ’๐œ‹๐‘Ÿ=๐œ‹(3๐‘ฆ)โˆ’๐œ‹๏€น๐‘ฆ๏…=๐œ‹๏€น9๐‘ฆโˆ’๐‘ฆ๏….๏Šจ๏Šจ๏Šจ๏Šจ๏Šจ๏Šจ๏Šช

The limits of the integral, ๐‘ and ๐‘‘, are determined by the horizontal bounds of ๐‘ฆ in the region, which in this case is where the two curves intersect: ๐‘ฆ=3๐‘ฆ.๏Šจ

The solutions are given by ๐‘ฆ=0 and ๐‘ฆ=3, and the two curves intersect at the points (0,0) and (9,3); this can also be visualized from the plot of the region bounded by the curves. Thus, we have ๐‘=0 and ๐‘‘=3, and the volume of the solid of revolution is given by ๐‘‰=๏„ธ๐ด(๐‘ฆ)๐‘ฆ=๐œ‹๏„ธ๏€น9๐‘ฆโˆ’๐‘ฆ๏…๐‘ฆ=๐œ‹๏–3๐‘ฆโˆ’๐‘ฆ5๏ข=162๐œ‹5.๏Šฉ๏Šฆ๏Šฉ๏Šฆ๏Šจ๏Šช๏Šฉ๏Šซ๏Šฉ๏Šฆdd

Thus, the volume of the generated solid is 162๐œ‹5.cubicunits

Finally, letโ€™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 ๐‘ฆ-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 ๐‘ฆ-Axis

Consider the region bounded by the curves ๐‘ฆ=๐‘ฅ๏Šฉ, ๐‘ฆ=0, and ๐‘ฅ=2. Find the volume of the solid obtained by rotating this region about ๐‘ฅ=3.

Answer

In this example, we want to calculate the volume of a solid generated by the revolution of a particular region around the line ๐‘ฅ=3, which is parallel to the ๐‘ฆ-axis.

We can visualize the region bounded by the curves ๐‘ฆ=๐‘ฅ๏Šฉ, ๐‘ฆ=0, and ๐‘ฅ=2 as follows:

The solid of revolution formed by rotating ๐‘ฅ=โˆš๐‘ฆ๏Žข about ๐‘ฅ=3 looks like this:

And the solid formed by ๐‘ฅ=2 looks like this:

Thus, the solid of revolution formed by rotating the region bounded by these curves about the ๐‘ฆ-axis looks like this:

Since we rotated this region about the line ๐‘ฅ=3, which is parallel to the ๐‘ฆ-axis, the volume is given by ๐‘‰=๏„ธ๐ด(๐‘ฆ)๐‘ฆ.๏Œฝ๏Œผd

The cross-sectional shape of each solid will be a disk. However, since we are rotating about the line ๐‘ฅ=3 rather than the ๐‘ฆ-axis, or the line ๐‘ฅ=0, 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 ๐‘…=3โˆ’โˆš๐‘ฆ๏Žข and ๐‘Ÿ=3โˆ’2=1. Thus, the difference between the areas of the larger disk and the smaller disk is given by ๐ด(๐‘ฆ)=๐œ‹๐‘…โˆ’๐œ‹๐‘Ÿ=๐œ‹๏€บ3โˆ’โˆš๐‘ฆ๏†โˆ’๐œ‹(3โˆ’2)=๐œ‹๏€ฝ๐‘ฆโˆ’6๐‘ฆ+8๏‰.๏Šจ๏Šจ๏Šจ๏Šจ๏Žข๏Žก๏Žข๏Ž ๏Žข

The limits of the integral, ๐‘ and ๐‘‘, are determined by the horizontal bounds of ๐‘ฆ in the region. The lower bound is ๐‘ฆ=0 and the upper bound is determined by the intersection of the curves ๐‘ฆ=๐‘ฅ๏Šฉ and ๐‘ฅ=2, which is ๐‘ฆ=8; this can also be visualized from the plot of the region bounded by the curves. Thus, we have ๐‘=0 and ๐‘‘=8, and the volume of the solid of revolution is given by ๐‘‰=๏„ธ๐ด(๐‘ฆ)๐‘ฆ=๐œ‹๏„ธ๏€ฝ๐‘ฆโˆ’6๐‘ฆ+8๏‰๐‘ฆ=๐œ‹๏™3๐‘ฆ5โˆ’18๐‘ฆ4+8๐‘ฆ๏ฅ=56๐œ‹5.๏Šฎ๏Šฆ๏Šฎ๏Šฆ๏Šฎ๏Šฆdd๏Žก๏Žข๏Ž ๏Žข๏Žค๏Žข๏Žฃ๏Žข

Thus, the volume of the generated solid is 56๐œ‹5.cubicunits

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, ๐‘…, of each disk will be the value of the function at that point. The cross-sectional area is determined by ๐ด=๐œ‹๐‘…๏Šจ.
    The volume of the solid formed by revolving the region bounded by the curve ๐‘ฆ=๐‘“(๐‘ฅ) and the ๐‘ฅ-axis between ๐‘ฅ=๐‘Ž and ๐‘ฅ=๐‘ about the ๐‘ฅ-axis is given by ๐‘‰=๐œ‹๏„ธ[๐‘“(๐‘ฅ)]๐‘ฅ.๏Œป๏Œบ๏Šจd Similarly, the volume of the solid formed by revolving the region bounded by the curve ๐‘ฅ=๐‘“(๐‘ฆ) and the ๐‘ฆ-axis between ๐‘ฆ=๐‘ and ๐‘ฆ=๐‘‘ about the ๐‘ฆ-axis is given by ๐‘‰=๐œ‹๏„ธ[๐‘“(๐‘ฆ)]๐‘ฆ.๏Œฝ๏Œผ๏Šจd
  • 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, ๐‘… and ๐‘Ÿ, 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 ๐ด=๐œ‹๐‘…โˆ’๐œ‹๐‘Ÿ๏Šจ๏Šจ.
    The volume ๐‘‰ of a solid generated by revolving the region bounded by ๐‘ฆ=๐‘“(๐‘ฅ) and ๐‘ฆ=๐‘”(๐‘ฅ) between ๐‘ฅ=๐‘Ž and ๐‘ฅ=๐‘, where ๐‘“(๐‘ฅ)โ‰ฅ๐‘”(๐‘ฅ), about the ๐‘ฅ-axis is ๐‘‰=๐œ‹๏„ธ๏€บ[๐‘“(๐‘ฅ)]โˆ’[๐‘”(๐‘ฅ)]๏†๐‘ฅ.๏Œป๏Œบ๏Šจ๏Šจd Similarly, the volume ๐‘‰ of a solid generated by revolving the region bounded by ๐‘ฅ=๐‘“(๐‘ฆ) and ๐‘ฅ=๐‘”(๐‘ฆ) between ๐‘ฆ=๐‘ and ๐‘ฆ=๐‘‘, where ๐‘“(๐‘ฆ)โ‰ฅ๐‘”(๐‘ฆ), about the ๐‘ฆ-axis is ๐‘‰=๐œ‹๏„ธ๏€บ[๐‘“(๐‘ฆ)]โˆ’[๐‘”(๐‘ฆ)]๏†๐‘ฅ.๏Œฝ๏Œผ๏Šจ๏Šจd
  • The limits of the integrals are determined by the vertical or horizontal bounds of ๐‘ฅ or ๐‘ฆ, respectively, depending on the axis of rotation and the region bounded by the curves.

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