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Volumes of Revolution: Cylindrical Shells

Calculate the volume of a solid of revolution by using the method of cylindrical shells.

The Method of Cylindrical Shells

Again, we are working with a solid of revolution. As before, we define a region \(R,\) bounded above by the graph of a function \(y=f(x),\) below by the \(x\text{-axis,}\) and on the left and right by the lines \(x=a\) and \(x=b,\) respectively, as shown in (a). We then revolve this region around the y-axis, as shown in (b). Note that this is different from what we have done before. Previously, regions defined in terms of functions of \(x\) were revolved around the \(x\text{-axis}\) or a line parallel to it.

As we have done many times before, partition the interval \([a,b]\) using a regular partition, \(P=\{{x}_{0},{x}_{1}\text{,\ldots },{x}_{n}\}\) and, for \(i=1,2\text{,\ldots },n,\) choose a point \({x}_{i}^{*}\in [{x}_{i-1},{x}_{i}].\) Then, construct a rectangle over the interval \([{x}_{i-1},{x}_{i}]\) of height \(f({x}_{i}^{*})\) and width \(\text{\Delta }x.\) A representative rectangle is shown in (a). When that rectangle is revolved around the y-axis, instead of a disk or a washer, we get a cylindrical shell, as shown in the following figure.

To calculate the volume of this shell, consider .

The shell is a cylinder, so its volume is the cross-sectional area multiplied by the height of the cylinder. The cross-sections are annuli (ring-shaped regions—essentially, circles with a hole in the center), with outer radius \({x}_{i}\) and inner radius \({x}_{i-1}.\) Thus, the cross-sectional area is \(\pi {x}_{i}^{2}-\pi {x}_{i-1}^{2}.\) The height of the cylinder is \(f({x}_{i}^{*}).\) Then the volume of the shell is

\[\begin{array}{ll}{V}_{\text{shell}} & =f({x}_{i}^{*})(\pi {x}_{i}^{2}-\pi {x}_{i-1}^{2}) \\ & =\pi f({x}_{i}^{*})({x}_{i}^{2}-{x}_{i-1}^{2}) \\ & =\pi f({x}_{i}^{*})({x}_{i}+{x}_{i-1})({x}_{i}-{x}_{i-1}) \\ & =2\pi f({x}_{i}^{*})(\frac{{x}_{i}+{x}_{i-1}}{2})({x}_{i}-{x}_{i-1}).\end{array}\]

Note that \({x}_{i}-{x}_{i-1}=\text{\Delta }x,\) so we have

\[{V}_{\text{shell}}=2\pi f({x}_{i}^{*})(\frac{{x}_{i}+{x}_{i-1}}{2})\text{\Delta }x.\]

Furthermore, \(\frac{{x}_{i}+{x}_{i-1}}{2}\) is both the midpoint of the interval \([{x}_{i-1},{x}_{i}]\) and the average radius of the shell, and we can approximate this by \({x}_{i}^{*}.\) We then have

\[{V}_{\text{shell}}\approx 2\pi f({x}_{i}^{*}){x}_{i}^{*}\text{\Delta }x.\]

Another way to think of this is to think of making a vertical cut in the shell and then opening it up to form a flat plate ().

\[{V}_{\text{shell}}\approx f({x}_{i}^{*})(2\pi {x}_{i}^{*})\text{\Delta }x,\]\[V\approx \sum _{i=1}^{n}(2\pi {x}_{i}^{*}f({x}_{i}^{*})\text{\Delta }x).\]\[V=\underset{n\to \infty }{\text{lim}}\sum _{i=1}^{n}(2\pi {x}_{i}^{*}f({x}_{i}^{*})\text{\Delta }x)={\int }_{a}^{b}(2\pi xf(x))dx.\]\[V={\int }_{a}^{b}(2\pi (x+k)f(x))dx.\]

Condensed — the full section is in OpenStax Calculus Volume 2.

Which Method Should We Use?

We have studied several methods for finding the volume of a solid of revolution, but how do we know which method to use? It often comes down to a choice of which integral is easiest to evaluate. describes the different approaches for solids of revolution around the \(x\text{-axis}.\) It’s up to you to develop the analogous table for solids of revolution around the \(y\text{-axis}.\)

Let’s take a look at a couple of additional problems and decide on the best approach to take for solving them.

Example

Try it.

For each of the following problems, select the best method to find the volume of a solid of revolution generated by revolving the given region around the \(x\text{-axis},\) and set up the integral to find the volume (do not evaluate the integral).

The region bounded by the graphs of \(y=x,\) \(y=2-x,\) and the \(x\text{-axis}.\)
The region bounded by the graphs of \(y=4x-{x}^{2}\) and the \(x\text{-axis}.\)

Solution

First, sketch the region and the solid of revolution as shown.

Looking at the region, if we want to integrate with respect to \(x,\) we would have to break the integral into two pieces, because we have different functions bounding the region over \([0,1]\) and \([1,2].\) In this case, using the disk method, we would have
\[V={\int }_{0}^{1}(\pi {x}^{2})dx+{\int }_{1}^{2}(\pi {(2-x)}^{2})dx.\]
If we used the shell method instead, we would use functions of \(y\) to represent the curves, producing
\[\begin{array}{ll}V & ={\int }_{0}^{1}(2\pi y[(2-y)-y])dy \\ & ={\int }_{0}^{1}(2\pi y[2-2y])dy.\end{array}\]
Neither of these integrals is particularly onerous, but since the shell method requires only one integral, and the integrand requires less simplification, we should probably go with the shell method in this case.
First, sketch the region and the solid of revolution as shown.

Looking at the region, it would be problematic to define a horizontal rectangle; the region is bounded on the left and right by the same function. Therefore, we can dismiss the method of shells. The solid has no cavity in the middle, so we can use the method of disks. Then
\[V={\int }_{0}^{4}\pi {(4x-{x}^{2})}^{2}dx.\]

Key Concepts

The method of cylindrical shells is another method for using a definite integral to calculate the volume of a solid of revolution. This method is sometimes preferable to either the method of disks or the method of washers because we integrate with respect to the other variable. In some cases, one integral is substantially more complicated than the other.
The geometry of the functions and the difficulty of the integration are the main factors in deciding which integration method to use.

Volumes of Revolution: Cylindrical Shells

For the following exercises, find the volume generated when the region between the two curves is rotated around the given axis. Use both the shell method and the washer method. Use technology to graph the functions and draw a typical slice by hand.

For the following exercises, use shells to find the volumes of the given solids. Note that the rotated regions lie between the curve and the \(x\text{-axis}\) and are rotated around the \(y\text{-axis}.\)

For the following exercises, use shells to find the volume generated by rotating the regions between the given curve and \(y=0\) around the \(x\text{-axis}.\)

For the following exercises, find the volume generated when the region between the curves is rotated around the given axis.

For the following exercises, use technology to graph the region. Determine which method you think would be easiest to use to calculate the volume generated when the function is rotated around the specified axis. Then, use your chosen method to find the volume.

For the following exercises, use the method of shells to approximate the volumes of some common objects, which are pictured in accompanying figures.

The Method of Cylindrical Shells

To calculate the volume of this shell, consider .

Note that \({x}_{i}-{x}_{i-1}=\text{\Delta }x,\) so we have

\[{V}_{\text{shell}}=2\pi f({x}_{i}^{*})(\frac{{x}_{i}+{x}_{i-1}}{2})\text{\Delta }x.\]

\[{V}_{\text{shell}}\approx 2\pi f({x}_{i}^{*}){x}_{i}^{*}\text{\Delta }x.\]

Another way to think of this is to think of making a vertical cut in the shell and then opening it up to form a flat plate ().

Condensed — the full section is in OpenStax Calculus Volume 1.

Which Method Should We Use?

Let’s take a look at a couple of additional problems and decide on the best approach to take for solving them.

Example

Try it.

The region bounded by the graphs of \(y=x,\) \(y=2-x,\) and the \(x\text{-axis}.\)
The region bounded by the graphs of \(y=4x-{x}^{2}\) and the \(x\text{-axis}.\)

Solution

First, sketch the region and the solid of revolution as shown.

Looking at the region, if we want to integrate with respect to \(x,\) we would have to break the integral into two pieces, because we have different functions bounding the region over \([0,1]\) and \([1,2].\) In this case, using the disk method, we would have
\[V={\int }_{0}^{1}(\pi {x}^{2})dx+{\int }_{1}^{2}(\pi {(2-x)}^{2})dx.\]
If we used the shell method instead, we would use functions of \(y\) to represent the curves, producing
\[\begin{array}{ll}V & ={\int }_{0}^{1}(2\pi y[(2-y)-y])dy \\ & ={\int }_{0}^{1}(2\pi y[2-2y])dy.\end{array}\]
Neither of these integrals is particularly onerous, but since the shell method requires only one integral, and the integrand requires less simplification, we should probably go with the shell method in this case.
First, sketch the region and the solid of revolution as shown.

Looking at the region, it would be problematic to define a horizontal rectangle; the region is bounded on the left and right by the same function. Therefore, we can dismiss the method of shells. The solid has no cavity in the middle, so we can use the method of disks. Then
\[V={\int }_{0}^{4}\pi {(4x-{x}^{2})}^{2}dx.\]

Key Concepts

The method of cylindrical shells is another method for using a definite integral to calculate the volume of a solid of revolution. This method is sometimes preferable to either the method of disks or the method of washers because we integrate with respect to the other variable. In some cases, one integral is substantially more complicated than the other.
The geometry of the functions and the difficulty of the integration are the main factors in deciding which integration method to use.

Volumes of Revolution: Cylindrical Shells

For the following exercises, use shells to find the volume generated by rotating the regions between the given curve and \(y=0\) around the \(x\text{-axis}.\)

For the following exercises, find the volume generated when the region between the curves is rotated around the given axis.

For the following exercises, use the method of shells to approximate the volumes of some common objects, which are pictured in accompanying figures.

Practice (40)

Try each one on paper first. Reveal the answer to check; verified ones can be opened in the solver for every step.

Define \(R\) as the region bounded above by the graph of \(f(x)=1\text{/}x\) and below by the \(x\text{-axis}\) over the interval \([1,3].\) Find the volume of the solid of revolution formed by revolving \(R\) around the \(y\text{-axis}.\)

i

First we must graph the region \(R\) and the associated solid of revolution, as shown in the following figure.
Then the volume of the solid is given by
\[\begin{array}{ll}V & ={\int }_{a}^{b}(2\pi xf(x))dx \\ & ={\int }_{1}^{3}(2\pi x(\frac{1}{x}))dx \\ & ={\int }_{1}^{3}2\pi \ dx={2\pi x|}_{1}^{3}=4\pi \ {\text{units}}^{3}\text{.}\end{array}\]
Define R as the region bounded above by the graph of \(f(x)={x}^{2}\) and below by the x-axis over the interval \([1,2].\) Find the volume of the solid of revolution formed by revolving \(R\) around the \(y\text{-axis}.\)

i

\(\frac{15\pi }{2}\) units³
Define R as the region bounded above by the graph of \(f(x)=2x-{x}^{2}\) and below by the \(x\text{-axis}\) over the interval \([0,2].\) Find the volume of the solid of revolution formed by revolving \(R\) around the \(y\text{-axis}.\)

i

First graph the region \(R\) and the associated solid of revolution, as shown in the following figure.
Then the volume of the solid is given by
\[\begin{array}{ll}V & ={\int }_{a}^{b}(2\pi xf(x))dx \\ & ={\int }_{0}^{2}(2\pi x(2x-{x}^{2}))dx=2\pi {\int }_{0}^{2}(2{x}^{2}-{x}^{3})dx \\ & ={2\pi [\frac{2{x}^{3}}{3}-\frac{{x}^{4}}{4}]\ |}_{0}^{2}=\frac{8\pi }{3}\ {\text{units}}^{3}\text{.}\end{array}\]
Define \(R\) as the region bounded above by the graph of \(f(x)=3x-{x}^{2}\) and below by the \(x\text{-axis}\) over the interval \([0,2].\) Find the volume of the solid of revolution formed by revolving \(R\) around the \(y\text{-axis}.\)

i

\(8\pi\) units³
Define \(Q\) as the region bounded on the right by the graph of \(g(y)=2\sqrt{y}\) and on the left by the \(y\text{-axis}\) for \(y\in [0,4].\) Find the volume of the solid of revolution formed by revolving \(Q\) around the x-axis.

i

First, we need to graph the region \(Q\) and the associated solid of revolution, as shown in the following figure.
Label the shaded region \(Q.\) Then the volume of the solid is given by
\[\begin{array}{ll}V & ={\int }_{c}^{d}(2\pi yg(y))dy \\ & ={\int }_{0}^{4}(2\pi y(2\sqrt{y}))dy=4\pi {\int }_{0}^{4}{y}^{3\text{/}2}dy \\ & ={4\pi [\frac{2{y}^{5\text{/}2}}{5}]\ |}_{0}^{4}=\frac{256\pi }{5}\ {\text{units}}^{3}\text{.}\end{array}\]
Define \(Q\) as the region bounded on the right by the graph of \(g(y)=3\text{/}y\) and on the left by the \(y\text{-axis}\) for \(y\in [1,3].\) Find the volume of the solid of revolution formed by revolving \(Q\) around the \(x\text{-axis}.\)

i

\(12\pi\) units³
Define \(R\) as the region bounded above by the graph of \(f(x)=x\) and below by the \(x\text{-axis}\) over the interval \([1,2].\) Find the volume of the solid of revolution formed by revolving \(R\) around the line \(x=-1.\)

i

First, graph the region \(R\) and the associated solid of revolution, as shown in the following figure.
Note that the radius of a shell is given by \(x+1.\) Then the volume of the solid is given by
\[\begin{array}{ll}V & ={\int }_{1}^{2}(2\pi (x+1)f(x))dx \\ & ={\int }_{1}^{2}(2\pi (x+1)x)dx=2\pi {\int }_{1}^{2}({x}^{2}+x)dx \\ & ={2\pi [\frac{{x}^{3}}{3}+\frac{{x}^{2}}{2}]\ |}_{1}^{2}=\frac{23\pi }{3}\ {\text{units}}^{3}\text{.}\end{array}\]
Define \(R\) as the region bounded above by the graph of \(f(x)={x}^{2}\) and below by the \(x\text{-axis}\) over the interval \([0,1].\) Find the volume of the solid of revolution formed by revolving \(R\) around the line \(x=-2.\)

i

\(\frac{11\pi }{6}\) units³
Define \(R\) as the region bounded above by the graph of the function \(f(x)=\sqrt{x}\) and below by the graph of the function \(g(x)=1\text{/}x\) over the interval \([1,4].\) Find the volume of the solid of revolution generated by revolving \(R\) around the \(y\text{-axis}.\)

i

First, graph the region \(R\) and the associated solid of revolution, as shown in the following figure.
Note that the axis of revolution is the \(y\text{-axis},\) so the radius of a shell is given simply by \(x.\) We don’t need to make any adjustments to the x-term of our integrand. The height of a shell, though, is given by \(f(x)-g(x),\) so in this case we need to adjust the \(f(x)\) term of the integrand. Then the volume of the solid is given by
\[\begin{array}{ll}V & ={\int }_{1}^{4}(2\pi x(f(x)-g(x)))dx \\ & ={\int }_{1}^{4}(2\pi x(\sqrt{x}-\frac{1}{x}))dx=2\pi {\int }_{1}^{4}({x}^{3\text{/}2}-1)dx \\ & ={2\pi [\frac{2{x}^{5\text{/}2}}{5}-x]\ |}_{1}^{4}=\frac{94\pi }{5}\ {\text{units}}^{3}.\end{array}\]
Define \(R\) as the region bounded above by the graph of \(f(x)=x\) and below by the graph of \(g(x)={x}^{2}\) over the interval \([0,1].\) Find the volume of the solid of revolution formed by revolving \(R\) around the \(y\text{-axis}.\)

i

\(\frac{\pi }{6}\) units³
For each of the following problems, select the best method to find the volume of a solid of revolution generated by revolving the given region around the \(x\text{-axis},\) and set up the integral to find the volume (do not evaluate the integral).
1. The region bounded by the graphs of \(y=x,\) \(y=2-x,\) and the \(x\text{-axis}.\)
2. The region bounded by the graphs of \(y=4x-{x}^{2}\) and the \(x\text{-axis}.\)
i
1. First, sketch the region and the solid of revolution as shown.
  
  Looking at the region, if we want to integrate with respect to \(x,\) we would have to break the integral into two pieces, because we have different functions bounding the region over \([0,1]\) and \([1,2].\) In this case, using the disk method, we would have
  \[V={\int }_{0}^{1}(\pi {x}^{2})dx+{\int }_{1}^{2}(\pi {(2-x)}^{2})dx.\]
  If we used the shell method instead, we would use functions of \(y\) to represent the curves, producing
  \[\begin{array}{ll}V & ={\int }_{0}^{1}(2\pi y[(2-y)-y])dy \\ & ={\int }_{0}^{1}(2\pi y[2-2y])dy.\end{array}\]
  Neither of these integrals is particularly onerous, but since the shell method requires only one integral, and the integrand requires less simplification, we should probably go with the shell method in this case.
2. First, sketch the region and the solid of revolution as shown.
  
  Looking at the region, it would be problematic to define a horizontal rectangle; the region is bounded on the left and right by the same function. Therefore, we can dismiss the method of shells. The solid has no cavity in the middle, so we can use the method of disks. Then
  \[V={\int }_{0}^{4}\pi {(4x-{x}^{2})}^{2}dx.\]
Select the best method to find the volume of a solid of revolution generated by revolving the given region around the \(x\text{-axis},\) and set up the integral to find the volume (do not evaluate the integral): the region bounded by the graphs of \(y=2-{x}^{2}\) and \(y={x}^{2}.\)

i

Use the method of washers; \(V={\int }_{-1}^{1}\pi [{(2-{x}^{2})}^{2}-{({x}^{2})}^{2}]dx\)
[T] Bounded by the curves \(y=3x,x=0,\) and \(y=3\) rotated around the \(y\text{-axis}.\)
[T] Bounded by the curves \(y=3x,y=0,\ \text{and}\ x=3\) rotated around the \(y\text{-axis}.\)

i

\(54\pi\) units³
[T] Bounded by the curves \(y=3x,y=0,\ \text{and}\ y=3\) rotated around the \(x\text{-axis}.\)
[T] Bounded by the curves \(y=3x,y=0,\ \text{and}\ x=3\) rotated around the \(x\text{-axis}.\)

i

\(81\pi\) units³
[T] Bounded by the curves \(y=2{x}^{3},y=0,\ \text{and}\ x=2\) rotated around the \(y\text{-axis}.\)
[T] Bounded by the curves \(y=2{x}^{3},y=0,\ \text{and}\ x=2\) rotated around the \(x\text{-axis}.\)

i

\(\frac{512\pi }{7}\) units³
\(y=1-{x}^{2},x=0,\ \text{and}\ x=1\)
\(y=5{x}^{3},x=0,\ \text{and}\ x=1\)

i

\(2\pi\) units³
\(y=\frac{1}{x},x=1,\ \text{and}\ x=100\)
\(y=\sqrt{1-{x}^{2}},x=0,\ \text{and}\ x=1\)

i

\(\frac{2\pi }{3}\) units³
\(y=\frac{1}{1+{x}^{2}},x=0,\ \text{and}\ x=3\)
\(y=\text{sin}{x}^{2},x=0,\ \text{and}\ x=\sqrt{\pi }\)

i

\(2\pi\) units³
\(y=\frac{1}{\sqrt{1-{x}^{2}}},x=0,\ \text{and}\ x=\frac{1}{2}\)
\(y=\sqrt{x},x=0,\ \text{and}\ x=1\)

i

\(\frac{4\pi }{5}\) units³
\(y={(1+{x}^{2})}^{3},x=0,\ \text{and}\ x=1\)
\(y=5{x}^{3}-2{x}^{4},x=0,\ \text{and}\ x=2\)

i

\(\frac{64\pi }{3}\) units³
\(y=\sqrt{1-{x}^{2}},x=0,\ x=1\) and the x-axis
\(y={x}^{2},x=0,\ x=2\) and the x-axis

i

\(\frac{32\pi }{5}\) units³
\(y=\frac{{x}^{3}}{2},\ x=0,\ x=2,\) and the x-axis
\(y=\frac{2}{{x}^{2}},\ x=1,\ x=2,\) and the x-axis

i

\(\frac{7\pi }{6}\)
\(x=\frac{1}{1+{y}^{2}},y=4\)
\(x=\frac{1+{y}^{2}}{y},y=1,\ y=4,\) and the y-axis

i

\(48\pi\)
\(x=\sqrt{4-{y}^{2}}\text{,}x=0\text{,}y=0\)
\(x={y}^{3}-2{y}^{2},\ x=0,\ x=9\)

i

\(\frac{308\pi }{5}\)
\(x=\sqrt{y}+1,\ x=1,\ x=3,\) and the x-axis
\(x=\sqrt[3]{27y}\text{and}\ x=\frac{3y}{4}\)

i

\(\frac{512\pi }{7}\)
\(y=3-x,y=0,x=0,\ \text{and}\ x=2\) rotated around the \(y\text{-axis}.\)
\(y={x}^{3},x=0,\ \text{and}\ y=8\) rotated around the \(y\text{-axis}.\)

i

\(\frac{8\pi }{5}\)

Symbols used here

\sum_{k=1}^{n} a_k

summation
Add a_k for k = 1 up to n.

\int f(x)\,dx,\ \int_a^b

integral
Antiderivative (indefinite) or signed area from a to b (definite).

\sqrt{x},\ \sqrt[n]{x}

square root, n-th root
The non-negative number whose square (n-th power) is x.

\infty

infinity
Not a number: "grows without bound" in limits and intervals.

\pi

pi
Ratio of a circle's circumference to its diameter, 3.14159…

x \in A,\ A \subseteq B

element of, subset
x belongs to A; every element of A is in B.

imaginary unit
i² = −1.

\approx

approximately equal
Equal to the precision shown, not exactly.

\sup,\ \inf

supremum, infimum
Least upper bound, greatest lower bound.

Euler's number
2.71828…, the base whose exponential is its own derivative.

\sin,\ \cos,\ \tan

sine, cosine, tangent
Ratios of sides in a right triangle; coordinates on the unit circle.

\log_b x,\ \ln x

logarithm, natural log
The exponent b must be raised to for x; ln uses base e.

\lim_{x \to a} f(x)

limit
The value f(x) approaches as x approaches a.

f'(x),\ \frac{dy}{dx}

derivative
Instantaneous rate of change; slope of the graph.

y',\ y''

first and second derivative of y
Prime notation for derivatives with respect to x (or t).

C,\ C_1,\ C_2

arbitrary constants
Constants of integration fixed by initial conditions.

How to: Volumes of Revolution: Cylindrical Shells

Calculate the volume of a solid of revolution by using the method of cylindrical shells.
Compare the different methods for calculating a volume of revolution.
The region bounded by the graphs of
The region bounded by the graphs of
First, sketch the region and the solid of revolution as shown.
First, sketch the region and the solid of revolution as shown.
The geometry of the functions and the difficulty of the integration are the main factors in deciding which integration method to use.

Questions people ask

What is a derivative in one sentence?

The slope of the graph at a point — the rate at which the output is changing there. Speed is the derivative of position.

What is an integral in one sentence?

The accumulated total of a rate — the area under the curve. Distance travelled is the integral of speed.

Why are derivatives and integrals opposites?

That is the fundamental theorem of calculus: accumulating a rate and then measuring how fast the accumulation grows gets you back the rate. Integration undoes differentiation up to a constant.

When do I use substitution and when integration by parts?

Substitution when part of the integrand is the derivative of another part (u and du both present). Parts when the integrand is a product of two unrelated kinds of function — a polynomial times an exponential, log or trig function.

Kuri Gukoresha

Parts of this page are adapted from OpenStax Calculus Volume 1 (CC BY-NC-SA 4.0), OpenStax Calculus Volume 2 (CC BY-NC-SA 4.0). Condensed and re-explained here; errors are ours.

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