Lesson Explainer: Volumes of Solids of Revolution | Nagwa Lesson Explainer: Volumes of Solids of Revolution | Nagwa

Lesson Explainer: Volumes of Solids of Revolution Mathematics

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 𝑦=𝑓(𝑥) in some interval [𝑎,𝑏], as depicted in the diagram.

If we take the part of the curve between 𝑥=𝑎 and 𝑥=𝑏 and rotate it about the 𝑥-axis by a complete revolution (i.e., by 360 or 2𝜋), 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 𝑓 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 𝑥=𝑎 and 𝑥=𝑏, whose cross-sectional area in the plane through 𝑥 and perpendicular to the 𝑥-axis is 𝐴(𝑥). If 𝐴(𝑥) is continuous on the interval [𝑎,𝑏], we can divide the interval into 𝑛 subintervals of equal width, Δ𝑥, and choose a point, 𝑥, in each interval.

The volume of the solid formed by revolving the region bounded by the curve and the 𝑥-axis between 𝑥=𝑎 and 𝑥=𝑏 about the 𝑥-axis is given by 𝑉=𝐴(𝑥)Δ𝑥=𝐴(𝑥)𝑥,limd where 𝑥=𝑎+𝑖Δ𝑥,Δ𝑥=𝑏𝑎𝑛=𝑥𝑥.

Similarly, the volume of the solid formed by revolving the region bounded by the curve and the 𝑦-axis between 𝑦=𝑐 and 𝑦=𝑑 about the 𝑦-axis is given by 𝑉=𝐴(𝑦)𝑦.d

This can also be visualized in a diagram.

For a curve 𝑦=𝑓(𝑥), the cross-sectional shape is a solid disk or circle whose area is 𝐴=𝜋𝑅, where 𝑅 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 𝑅=𝑓(𝑥), and we have 𝐴(𝑥)=𝜋[𝑓(𝑥)].

Therefore, the volume of the solid formed by revolving the region bounded by the curve 𝑦=𝑓(𝑥) and the 𝑥-axis between 𝑥=𝑎 and 𝑥=𝑏 about the 𝑥-axis can alternatively be written as 𝑉=𝜋[𝑓(𝑥)]𝑥.d

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 𝑦-axis. The volume of the solid formed by revolving the region bounded by the curve 𝑥=𝑓(𝑦) and the 𝑦-axis between 𝑦=𝑐 and 𝑦=𝑑 about the 𝑦-axis is given by 𝑉=𝜋[𝑓(𝑦)]𝑦.d

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 𝑟 is given by 𝑥+𝑦=𝑟 or, equivalently, written as a functions of 𝑥 as 𝑦=±𝑟𝑥.

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

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

The cross-sectional area of this solid is given by 𝐴(𝑥)=𝜋𝑦=𝜋𝑟𝑥.

Therefore, the volume of the solid of revolution is given by 𝑉=𝐴(𝑥)𝑥=𝜋𝑟𝑥𝑥=𝜋𝑟𝑥𝑥3=𝜋𝑟𝑟3𝑟+𝑟3=43𝜋𝑟.dd

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 𝑥-axis, described by 𝑦=𝑟𝑥.

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

We can rotate this line between 𝑥=0 and 𝑥= by a complete revolution about the 𝑥-axis to form a cone of radius 𝑟 and vertical height , as shown in the diagram.

The cross-sectional area of this solid is given by 𝐴(𝑥)=𝜋𝑦=𝜋𝑟𝑥=𝜋𝑟𝑥.

Therefore, the volume of the solid of revolution is given by 𝑉=𝐴(𝑥)𝑥=𝜋𝑟𝑥𝑥=𝜋𝑟𝑥𝑥=𝜋𝑟𝑥3=𝜋𝑟30=13𝜋𝑟.ddd

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 𝑥-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 𝑦=𝑥+4, 𝑦=0, 𝑥=0, and 𝑥=3. Determine the volume of the solid of revolution created by rotating this region about the 𝑥-axis.

Answer

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

We can visualize the region bounded by the curves 𝑦=𝑥+4, 𝑦=0, 𝑥=0, and 𝑥=3 as follows:

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

Since we rotated the region about the 𝑥-axis, the volume is given by 𝑉=𝐴(𝑥)𝑥.d

The vertical cross-sectional shape of our solid will be a disk, and the radius, 𝑅, of each disk will be the value of the function at that point, 𝑅=𝑥+4. Thus, the cross-sectional area is given by 𝐴(𝑥)=𝜋𝑅=𝜋(𝑥+4).

The limits of the integral, 𝑎 and 𝑏, are determined by the vertical bounds of 𝑥 in the region bounded by the curves, 0𝑥3; this can also be visualized from the plot of the region bounded by the curves. Thus, we have 𝑎=0 and 𝑏=3, and the volume of the solid of revolution is given by 𝑉=𝐴(𝑥)𝑥=𝜋(𝑥+4)𝑥=𝜋𝑥+8𝑥+16𝑥=𝜋𝑥3+8𝑥2+16𝑥=93𝜋.ddd

Thus, the volume of the solid of revolution is 93𝜋.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 𝑥-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 𝑦=𝑥+1 and the lines 𝑦=0 and 𝑥=4 about the 𝑥-axis.

Answer

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

We can visualize the region bounded by the curve 𝑦=𝑥+1 and lines 𝑦=0 and 𝑥=4 as follows:

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

Since we are rotating this region about the 𝑥-axis, the volume is given by 𝑉=𝐴(𝑥)𝑥.d

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, 𝑅=𝑥+1. Thus, the cross-sectional area is given by 𝐴(𝑥)=𝜋𝑅=𝜋𝑥+1=𝜋(𝑥+1).

The limits of the integral, 𝑎 and 𝑏, are determined by the vertical bounds of 𝑥 in the region. The upper bound is 𝑥=4 and the lower bound is determined by the intersection of the curves 𝑦=𝑥+1 and 𝑦=0, which is 𝑥=1; this can also be visualized from the plot of the region bounded by the curves. Thus, we have 𝑎=1 and 𝑏=4, and the volume of the solid of revolution is given by 𝑉=𝐴(𝑥)𝑥=𝜋(𝑥+1)𝑥=𝜋𝑥2+𝑥=25𝜋2.dd

Thus, the volume of the generated solid is 25𝜋2.cubicunits

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 𝑥-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 𝑦=𝑥+2𝑥 and the 𝑥-axis complete revolution about the 𝑥-axis.

Answer

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

We can visualize the region bounded by the curve 𝑦=𝑥+2𝑥 and the 𝑥-axis as follows:

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

Since we are rotating this region about the 𝑥-axis, the volume is given by 𝑉=𝐴(𝑥)𝑥.d

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, 𝑅=𝑥+2𝑥. Thus, the cross-sectional area is given by 𝐴(𝑥)=𝜋𝑅=𝜋𝑥+2𝑥=𝜋𝑥4𝑥+4𝑥.

The limits of the integral, 𝑎 and 𝑏, are determined by the vertical bounds of 𝑥 in the region, which in this case occurs when the curve 𝑦=𝑥+2𝑥 intersects the 𝑥-axis or when 𝑦=0: 𝑥+2𝑥=𝑥(𝑥2)=0.

The solutions are given by 𝑥=0 and 𝑥=2, which can also be visualized from the plot of the region bounded by the curves. Thus, we have 𝑎=0 and 𝑏=2, and the volume of the solid of revolution is given by 𝑉=𝐴(𝑥)𝑥=𝜋𝑥4𝑥+4𝑥𝑥=𝜋𝑥5𝑥+4𝑥3=16𝜋15.dd

Thus, the volume of the generated solid is 16𝜋15.cubicunits

Now, let’s look at an example where we determine the volume of a solid of revolution, but this time rotated about the 𝑦-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 𝑦-axis, the region bounded by the curve 9𝑥𝑦=0 and the lines 𝑥=0, 𝑦=9, and 𝑦=0.

Answer

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

We can visualize the region bounded by the curves 9𝑥𝑦=0 and the lines 𝑥=0, 𝑦=9, and 𝑦=0 as follows:

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

Since we rotated this region about the 𝑦-axis, the volume is given by 𝑉=𝐴(𝑦)𝑦.d

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, 𝑅=𝑦9. Thus, the cross-sectional area is given by 𝐴(𝑦)=𝜋𝑅=𝜋𝑦9=𝜋𝑦81.

The limits of the integral, 𝑐 and 𝑑, are determined by the horizontal bounds of 𝑦 in the region, 9𝑦0; this can also be visualized from the plot of the region bounded by the curves. Thus, we have 𝑐=9 and 𝑑=0, and the volume of the solid of revolution is given by 𝑉=𝐴(𝑦)𝑦=𝜋81𝑦𝑦=𝜋81𝑦3=3𝜋.dd

Thus, the volume of the generated solid is 3𝜋.cubicunits

The volume 𝑉 of a solid generated by revolving the region bounded by 𝑦=𝑓(𝑥) and 𝑦=𝑔(𝑥) on the interval [𝑎,𝑏], where 𝑓(𝑥)𝑔(𝑥), about the 𝑥-axis is 𝑉=𝜋[𝑓(𝑥)][𝑔(𝑥)]𝑥,d which we can also visualize in a diagram:

Similarly, the volume 𝑉 of a solid generated by revolving the region bounded by 𝑥=𝑓(𝑦) and 𝑥=𝑔(𝑦) on the interval [𝑐,𝑑], where 𝑓(𝑦)𝑔(𝑦), about the 𝑦-axis is 𝑉=𝜋[𝑓(𝑦)][𝑔(𝑦)]𝑥,d 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 𝑦, a pair of polynomials of degrees 2 and 4, about the 𝑦-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 𝑥=65𝑦 and 𝑥=𝑦 about the 𝑦-axis.

Answer

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

We can visualize the region bounded by the curves 𝑥=65𝑦 and 𝑥=𝑦 as follows:

The solid of revolution formed by rotating 𝑥=65𝑦 about the 𝑦-axis looks like this:

And the solid formed by 𝑥=𝑦 looks like this:

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

Since we rotated this region about the 𝑦-axis, the volume is given by 𝑉=𝐴(𝑦)𝑦.d

The cross-sectional shape of each solid will be a annulus, and the radius of each disk will be the value of the function at that point, 𝑅=65𝑦 and 𝑟=𝑦. Thus, the difference between the areas of the larger disk and the smaller disk is given by 𝐴(𝑦)=𝜋𝑅𝜋𝑟=𝜋65𝑦𝜋𝑦=𝜋3660𝑦+25𝑦𝑦.

The limits of the integral, 𝑐 and 𝑑, are determined by the horizontal bounds of 𝑦 in the region, which in this case is given by the values of 𝑦 at the points where the two curves intersect: 65𝑦=𝑦.

The solution is given by 𝑦=1 and 𝑦=1, and the two curves intersect at the points (1,1) and (1,1); this can also be visualized from the plot of the region bounded by the curves. Thus, we have 𝑐=1 and 𝑑=1, and the volume of the solid of revolution is given by 𝑉=𝐴(𝑦)𝑦=𝜋3660𝑦+25𝑦𝑦𝑦=𝜋36𝑦20𝑦+5𝑦𝑦9=376𝜋9.dd

Thus, the volume of the generated solid is 376𝜋9.cubicunits

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 𝑦-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 𝑦=𝑥 and the line 𝑥=3𝑦 about the 𝑦-axis.

Answer

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

We can visualize the region bounded by the curve 𝑥=𝑦 and the line 𝑥=3𝑦 as follows:

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

And the solid formed by 𝑥=3𝑦 looks like this:

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

Since we rotated the region about the 𝑦-axis, the volume is given by 𝑉=𝐴(𝑦)𝑦.d

The horizontal cross-sectional shape of our solid will be a annulus, and the radius of each disk will be the value of the function at that point, 𝑅=3𝑦 and 𝑟=𝑦. Thus, the cross-sectional area is given by 𝐴(𝑦)=𝜋𝑅𝜋𝑟=𝜋(3𝑦)𝜋𝑦=𝜋9𝑦𝑦.

The limits of the integral, 𝑐 and 𝑑, are determined by the horizontal bounds of 𝑦 in the region, which in this case is where the two curves intersect: 𝑦=3𝑦.

The solutions are given by 𝑦=0 and 𝑦=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 𝑐=0 and 𝑑=3, and the volume of the solid of revolution is given by 𝑉=𝐴(𝑦)𝑦=𝜋9𝑦𝑦𝑦=𝜋3𝑦𝑦5=162𝜋5.dd

Thus, the volume of the generated solid is 162𝜋5.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, 𝑅, of each disk will be the value of the function at that point. The cross-sectional area is determined by 𝐴=𝜋𝑅.
    The volume of the solid formed by revolving the region bounded by the curve 𝑦=𝑓(𝑥) and the 𝑥-axis between 𝑥=𝑎 and 𝑥=𝑏 about the 𝑥-axis is given by 𝑉=𝜋[𝑓(𝑥)]𝑥.d Similarly, the volume of the solid formed by revolving the region bounded by the curve 𝑥=𝑓(𝑦) and the 𝑦-axis between 𝑦=𝑐 and 𝑦=𝑑 about the 𝑦-axis is given by 𝑉=𝜋[𝑓(𝑦)]𝑦.d
  • 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 annulus, and the radii, 𝑅 and 𝑟, 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 𝐴=𝜋𝑅𝜋𝑟.
    The volume 𝑉 of a solid generated by revolving the region bounded by 𝑦=𝑓(𝑥) and 𝑦=𝑔(𝑥) between 𝑥=𝑎 and 𝑥=𝑏, where 𝑓(𝑥)𝑔(𝑥), about the 𝑥-axis is 𝑉=𝜋[𝑓(𝑥)][𝑔(𝑥)]𝑥.d Similarly, the volume 𝑉 of a solid generated by revolving the region bounded by 𝑥=𝑓(𝑦) and 𝑥=𝑔(𝑦) between 𝑦=𝑐 and 𝑦=𝑑, where 𝑓(𝑦)𝑔(𝑦), about the 𝑦-axis is 𝑉=𝜋[𝑓(𝑦)][𝑔(𝑦)]𝑥.d
  • The limits of the integrals are determined by the vertical or horizontal bounds of 𝑥 or 𝑦, respectively, depending on the axis of rotation and the region bounded by the curves.

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