Cylindrical Shells Method

Learning Outcomes

Calculate the volume of a solid of revolution by using 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 Figure 1(a). We then revolve this region around the $y$ -axis, as shown in Figure 1(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.

This figure has two graphs. The first graph is labeled “a” and is an increasing curve in the first quadrant. The curve is labeled “y=f(x)”. The curve starts on the y-axis at y=a. Under the curve, above the x-axis is a shaded region labeled “R”. The shaded region is bounded on the right by the line x=b. The second graph is a three dimensional solid. It has been created by rotating the shaded region from “a” around the y-axis.

Figure 1. (a) A region bounded by the graph of a function of $x.$ (b) The solid of revolution formed when the region is revolved around the $y\text{-axis}\text{.}$

As we have done many times before, partition the interval $\left[a,b\right]$ using a regular partition, $P=\left\{{x}_{0},{x}_{1}\text{,…},{x}_{n}\right\}$ and, for $i=1,2\text{,…},n,$ choose a point ${x}_{i}^{*}\in \left[{x}_{i-1},{x}_{i}\right].$ Then, construct a rectangle over the interval $\left[{x}_{i-1},{x}_{i}\right]$ of height $f({x}_{i}^{*})$ and width $\text{Δ}x.$ A representative rectangle is shown in Figure 2(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.

This figure has two images. The first is a cylindrical shell, hollow in the middle. It has a vertical axis in the center. There is also a curve that meets the top of the cylinder. The second image is a set of concentric cylinders, one inside of the other forming a nesting of cylinders.

Figure 2. (a) A representative rectangle. (b) When this rectangle is revolved around the $y\text{-axis},$ the result is a cylindrical shell. (c) When we put all the shells together, we get an approximation of the original solid.

To calculate the volume of this shell, consider Figure 3.

This figure is a graph in the first quadrant. The curve is increasing and labeled “y=f(x)”. The curve starts on the y-axis at f(x*). Below the curve is a shaded rectangle. The rectangle starts on the x-axis. The width of the rectangle is delta x. The two sides of the rectangle are labeled “xsub(i-1)” and “xsubi”.

Figure 3. Calculating the volume of the shell.

Notice that the rectangle we are using is parallel to the axis of revolution (y axis), not perpendicular like the disk and washer method. This could be very useful, particularly for y axis revolutions.

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{matrix} V_{shell} & = f (x_{i}^{*}) (π x_{i}^{2} - π x_{i - 1}^{2}) = π f (x_{i}^{*}) (x_{i}^{2} - x_{i - 1}^{2}) = π f (x_{i}^{*}) (x_{i} + x_{i - 1}) (x_{i} - x_{i - 1}) = 2 π f (x_{i}^{*}) (\frac{x_{i} + x_{i - 1}}{2}) (x_{i} - x_{i - 1}) . \end{matrix}

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

Note that ${x}_{i}-{x}_{i-1}=\text{Δ}x,$ so we have

V_{shell} = 2 π f (x_{i}^{*}) (\frac{x_{i} + x_{i - 1}}{2}) Δ x

${V}_{\text{shell}}=2\pi f({x}_{i}^{*})(\frac{{x}_{i}+{x}_{i-1}}{2})\text{Δ}x$

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

V_{shell} \approx 2 π f (x_{i}^{*}) x_{i}^{*} Δ x

${V}_{\text{shell}}\approx 2\pi f({x}_{i}^{*}){x}_{i}^{*}\text{Δ}x$

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 (Figure 4).

This figure has two images. The first is labeled “a” and is of a hollow cylinder around the y-axis. On the front of this cylinder is a vertical line labeled “cut line”. The height of the cylinder is “y=f(x)”. The second figure is labeled “b” and is a shaded rectangular block. The height of the rectangle is “f(x*), the width of the rectangle is “2pix*”, and the thickness of the rectangle is “delta x”.

Figure 4. (a) Make a vertical cut in a representative shell. (b) Open the shell up to form a flat plate.

In reality, the outer radius of the shell is greater than the inner radius, and hence the back edge of the plate would be slightly longer than the front edge of the plate. However, we can approximate the flattened shell by a flat plate of height $f({x}_{i}^{*}),$ width $2\pi {x}_{i}^{*},$ and thickness $\text{Δ}x$ (Figure 4). The volume of the shell, then, is approximately the volume of the flat plate. Multiplying the height, width, and depth of the plate, we get

V_{shell} \approx f (x_{i}^{*}) (2 π x_{i}^{*}) Δ x,

${V}_{\text{shell}}\approx f({x}_{i}^{*})(2\pi {x}_{i}^{*})\text{Δ}x,$

which is the same formula we had before.

To calculate the volume of the entire solid, we then add the volumes of all the shells and obtain

V \approx n \sum i = 1 (2 π x_{i}^{*} f (x_{i}^{*}) Δ x)

$V\approx \underset{i=1}{\overset{n}{\text{∑}}}(2\pi {x}_{i}^{*}f({x}_{i}^{*})\text{Δ}x)$

Here we have another Riemann sum, this time for the function $2\pi xf(x).$ Taking the limit as $n\to \infty$ gives us

V = lim n \to \infty n \sum i = 1 (2 π x_{i}^{*} f (x_{i}^{*}) Δ x) = \int_{a}^{b} (2 π x f (x)) d x

$V=\underset{n\to \infty }{\text{lim}}\underset{i=1}{\overset{n}{\text{∑}}}(2\pi {x}_{i}^{*}f({x}_{i}^{*})\text{Δ}x)={\displaystyle\int }_{a}^{b}(2\pi xf(x))dx$

This leads to the following rule for the method of cylindrical shells.

The Method of Cylindrical Shells

Let $f(x)$ be continuous and nonnegative. Define $R$ as the region bounded above by the graph of $f(x),$ below by the $x\text{-axis},$ on the left by the line $x=a,$ and on the right by the line $x=b.$ Then the volume of the solid of revolution formed by revolving $R$ around the $y$ -axis is given by

V = \int_{a}^{b} (2 π x f (x)) d x

$V={\displaystyle\int }_{a}^{b}(2\pi xf(x))dx$

Now let’s consider an example.

Example: The Method of Cylindrical Shells 1

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 $\left[1,3\right].$ Find the volume of the solid of revolution formed by revolving $R$ around the $y\text{-axis}.$

Show Solution

First we must graph the region $R$ and the associated solid of revolution, as shown in the following figure.

This figure has three images. The first is a solid that has been formed by rotating the curve y=1/x about the y-axis. The solid begins on the x-axis and stops where y=1. The second image is labeled “a” and is the graph of y=1/x in the first quadrant. Under the curve is a shaded region labeled “R”. The region is bounded by the curve, the x-axis, to the left at x=1 and to the right at x=3. The third image is labeled “b” and is half of the solid formed by rotating the shaded region about the y-axis.

Figure 5. (a) The region $R$ under the graph of $f(x)=1\text{/}x$ over the interval $\left[1,3\right].$ (b) The solid of revolution generated by revolving $R$ about the $y\text{-axis}.$

Then the volume of the solid is given by

\begin{matrix} V & = \int_{a}^{b} (2 π x f (x)) d x = \int_{1}^{3} (2 π x (\frac{1}{x})) d x = \int_{1}^{3} 2 π d x = {2 π x |}_{1}^{3} = 4 π {units}^{3} . \end{matrix}

$\begin{array}{cc}\hfill V& ={\displaystyle\int }_{a}^{b}(2\pi xf(x))dx\hfill \\ & ={\displaystyle\int }_{1}^{3}(2\pi x(\frac{1}{x}))dx\hfill \\ & ={\displaystyle\int }_{1}^{3}2\pi dx={2\pi x|}_{1}^{3}=4\pi {\text{units}}^{3}\text{.}\hfill \end{array}$

Try It

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

Show Solution

$\frac{15\pi }{2}$ units³

Hint

Use the procedure from the previous example.

example: The Method of Cylindrical Shells 2

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 $\left[0,2\right].$ Find the volume of the solid of revolution formed by revolving $R$ around the $y\text{-axis}.$

Show Solution

First graph the region $R$ and the associated solid of revolution, as shown in the following figure.

This figure has two graphs. The first graph is labeled “a” and is the curve f(x)=2x-x^2. It is an upside down parabola intersecting the x-axis at the origin ant at x=2. Under the curve the region in the first quadrant is shaded and is labeled “R”. The second figure is a graph of the same curve. On the graph is a solid that is formed by rotation the region from “a” about the y-axis.

Figure 6. (a) The region $R$ under the graph of $f(x)=2x-{x}^{2}$ over the interval $\left[0,2\right].$ (b) The volume of revolution obtained by revolving $R$ about the $y\text{-axis}.$

Then the volume of the solid is given by

\begin{matrix} V & = \int_{a}^{b} (2 π x f (x)) d x = \int_{0}^{2} (2 π x (2 x - x^{2})) d x = 2 π \int_{0}^{2} (2 x^{2} - x^{3}) d x = {2 π [\frac{2 x^{3}}{3} - \frac{x^{4}}{4}] |}_{0}^{2} = \frac{8 π}{3} {units}^{3} . \end{matrix}

$\begin{array}{cc}\hfill V& ={\displaystyle\int }_{a}^{b}(2\pi xf(x))dx\hfill \\ & ={\displaystyle\int }_{0}^{2}(2\pi x(2x-{x}^{2}))dx=2\pi {\displaystyle\int }_{0}^{2}(2{x}^{2}-{x}^{3})dx\hfill \\ & ={2\pi \left[\frac{2{x}^{3}}{3}-\frac{{x}^{4}}{4}\right]|}_{0}^{2}=\frac{8\pi }{3}{\text{units}}^{3}\text{.}\hfill \end{array}$

Try It

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 $\left[0,2\right].$ Find the volume of the solid of revolution formed by revolving $R$ around the $y\text{-axis}.$

Show Solution

$8\pi$ units³

Hint

Use the process from the last example.

As with the disk method and the washer method, we can use the method of cylindrical shells with solids of revolution, revolved around the $x\text{-axis},$ when we want to integrate with respect to $y.$ The analogous rule for this type of solid is given here.

The Method of Cylindrical Shells for Solids of Revolution around the $x$ -axis

Let $g(y)$ be continuous and nonnegative. Define $Q$ as the region bounded on the right by the graph of $g(y),$ on the left by the $y\text{-axis},$ below by the line $y=c,$ and above by the line $y=d.$ Then, the volume of the solid of revolution formed by revolving $Q$ around the $x\text{-axis}$ is given by

V = \int_{c}^{d} (2 π y g (y)) d y

$V={\displaystyle\int }_{c}^{d}(2\pi yg(y))dy$

example: The Method of Cylindrical Shells for a Solid Revolved around the $x$ -axis

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 $y\in \left[0,4\right].$ Find the volume of the solid of revolution formed by revolving $Q$ around the $x$ -axis.

Show Solution

First, we need to graph the region $Q$ and the associated solid of revolution, as shown in the following figure.

This figure has two graphs. The first graph is labeled “a” and is the curve g(y)=2squareroot(y). It is an increasing curve in the first quadrant beginning at the origin. Between the y-axis and the curve, there is a shaded region labeled “Q”. The shaded region is bounded above by the line y=4. The second graph is the same curve in “a” and labeled “b”. It also has a solid region that has been formed by rotating the curve in “a” about the x-axis. The solid starts at the y-axis and stops at x=4.

Figure 7. (a) The region $Q$ to the left of the function $g(y)$ over the interval $\left[0,4\right].$ (b) The solid of revolution generated by revolving $Q$ around the $x\text{-axis}.$

Label the shaded region $Q.$ Then the volume of the solid is given by

\begin{matrix} V & = \int_{c}^{d} (2 π y g (y)) d y = \int_{0}^{4} (2 π y (2 \sqrt{y})) d y = 4 π \int_{0}^{4} y^{3 / 2} d y = {4 π [\frac{2 y^{5 / 2}}{5}] |}_{0}^{4} = \frac{256 π}{5} {units}^{3} . \end{matrix}

$\begin{array}{cc}\hfill V& ={\displaystyle\int }_{c}^{d}(2\pi yg(y))dy\hfill \\ & ={\displaystyle\int }_{0}^{4}(2\pi y(2\sqrt{y}))dy=4\pi {\displaystyle\int }_{0}^{4}{y}^{3\text{/}2}dy\hfill \\ & ={4\pi \left[\frac{2{y}^{5\text{/}2}}{5}\right]|}_{0}^{4}=\frac{256\pi }{5}{\text{units}}^{3}\text{.}\hfill \end{array}$

Try It

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 $y\in \left[1,3\right].$ Find the volume of the solid of revolution formed by revolving $Q$ around the $x\text{-axis}.$

Show Solution

$12\pi$ units³

Hint

Use the process from the previous example.

Watch the following video to see the worked solution to the above Try It.

Closed Captioning and Transcript Information for Video

For the next example, we look at a solid of revolution for which the graph of a function is revolved around a line other than one of the two coordinate axes. To set this up, we need to revisit the development of the method of cylindrical shells. Recall that we found the volume of one of the shells to be given by

\begin{matrix} V_{shell} & = f (x_{i}^{*}) (π x_{i}^{2} - π x_{i - 1}^{2}) = π f (x_{i}^{*}) (x_{i}^{2} - x_{i - 1}^{2}) = π f (x_{i}^{*}) (x_{i} + x_{i - 1}) (x_{i} - x_{i - 1}) = 2 π f (x_{i}^{*}) (\frac{x_{i} + x_{i - 1}}{2}) (x_{i} - x_{i - 1}) . \end{matrix}

This was based on a shell with an outer radius of ${x}_{i}$ and an inner radius of ${x}_{i-1}.$ If, however, we rotate the region around a line other than the $y\text{-axis},$ we have a different outer and inner radius. Suppose, for example, that we rotate the region around the line $x=\text{−}k,$ where $k$ is some positive constant. Then, the outer radius of the shell is ${x}_{i}+k$ and the inner radius of the shell is ${x}_{i-1}+k.$ Substituting these terms into the expression for volume, we see that when a plane region is rotated around the line $x=\text{−}k,$ the volume of a shell is given by

\begin{matrix} V_{shell} & = 2 π f (x_{i}^{*}) (\frac{(x_{i} + k) + (x_{i - 1} + k)}{2}) ((x_{i} + k) - (x_{i - 1} + k)) = 2 π f (x_{i}^{*}) ((\frac{x_{i} + x_{i - 2}}{2}) + k) Δ x . \end{matrix}

$\begin{array}{cc}\hfill {V}_{\text{shell}}& =2\pi f({x}_{i}^{*})(\frac{({x}_{i}+k)+({x}_{i-1}+k)}{2})(({x}_{i}+k)-({x}_{i-1}+k))\hfill \\ & =2\pi f({x}_{i}^{*})((\frac{{x}_{i}+{x}_{i-2}}{2})+k)\text{Δ}x.\hfill \end{array}$

As before, we notice that $\frac{{x}_{i}+{x}_{i-1}}{2}$ is the midpoint of the interval $\left[{x}_{i-1},{x}_{i}\right]$ and can be approximated by ${x}_{i}^{*}.$ Then, the approximate volume of the shell is

V_{shell} \approx 2 π (x_{i}^{*} + k) f (x_{i}^{*}) Δ x

${V}_{\text{shell}}\approx 2\pi ({x}_{i}^{*}+k)f({x}_{i}^{*})\text{Δ}x$

The remainder of the development proceeds as before, and we see that

V = \int_{a}^{b} (2 π (x + k) f (x)) d x

$V={\displaystyle\int }_{a}^{b}(2\pi (x+k)f(x))dx$

We could also rotate the region around other horizontal or vertical lines, such as a vertical line in the right half plane. In each case, the volume formula must be adjusted accordingly. Specifically, the $x\text{-term}$ in the integral must be replaced with an expression representing the radius of a shell. To see how this works, consider the following example.

Example: A Region of Revolution Revolved around a Line

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

Show Solution

First, graph the region $R$ and the associated solid of revolution, as shown in the following figure.

This figure has two graphs. The first graph is labeled “a” and is the line f(x)=x, a diagonal line through the origin. There is a shaded region above the x-axis under the line labeled “R”. This region is bounded to the left by the line x=1 and to the right by the line x=2. There is also the vertical line x=-1 on the graph. The second figure has the same graphs as “a” and is labeled “b”. Also on the graph is a solid formed by rotating the region “R” from the first graph about the line x=-1.

Figure 8. (a) The region $R$ between the graph of $f(x)$ and the $x\text{-axis}$ over the interval $\left[1,2\right].$ (b) The solid of revolution generated by revolving $R$ around the line $x=-1.$

Note that the radius of a shell is given by $x+1.$ Then the volume of the solid is given by

\begin{matrix} V & = \int_{1}^{2} (2 π (x + 1) f (x)) d x = \int_{1}^{2} (2 π (x + 1) x) d x = 2 π \int_{1}^{2} (x^{2} + x) d x = {2 π [\frac{x^{3}}{3} + \frac{x^{2}}{2}] |}_{1}^{2} = \frac{23 π}{3} {units}^{3} . \end{matrix}

$\begin{array}{cc}\hfill V& ={\displaystyle\int }_{1}^{2}(2\pi (x+1)f(x))dx\hfill \\ & ={\displaystyle\int }_{1}^{2}(2\pi (x+1)x)dx=2\pi {\displaystyle\int }_{1}^{2}({x}^{2}+x)dx\hfill \\ & ={2\pi \left[\frac{{x}^{3}}{3}+\frac{{x}^{2}}{2}\right]|}_{1}^{2}=\frac{23\pi }{3}{\text{units}}^{3}\text{.}\hfill \end{array}$

Try It

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 $\left[0,1\right].$ Find the volume of the solid of revolution formed by revolving $R$ around the line $x=-2.$

Show Solution

$\frac{11\pi }{6}$ units³

Hint

Use the process from the last example.

Watch the following video to see the worked solution to the above Try It.

Closed Captioning and Transcript Information for Video

For our final example in this section, let’s look at the volume of a solid of revolution for which the region of revolution is bounded by the graphs of two functions.

Example: A Region of Revolution Bounded by the Graphs of Two Functions

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 $g(x)=\frac{1}{x}$ over the interval $\left[1,4\right].$ Find the volume of the solid of revolution generated by revolving $R$ around the $y\text{-axis}.$

Show Solution

First, graph the region $R$ and the associated solid of revolution, as shown in the following figure.

This figure has two graphs. The first graph is labeled “a” and has two curves. The curves are the graphs of f(x)=squareroot(x) and g(x)=1/x. In the first quadrant the curves intersect at (1,1). In between the curves in the first quadrant there is a shaded region labeled “R”, bounded to the right by the line x=4. The second graph is labeled “b” and is the same as the graphs in “a”. Also on this graph is a solid that has been formed by rotating the region “R” from the figure “a” about the y-axis.

Figure 9. (a) The region $R$ between the graph of $f(x)$ and the graph of $g(x)$ over the interval $\left[1,4\right].$ (b) The solid of revolution generated by revolving $R$ around the $y\text{-axis}.$

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{matrix} V & = \int_{1}^{4} (2 π x (f (x) - g (x))) d x = \int_{1}^{4} (2 π x (\sqrt{x} - \frac{1}{x})) d x = 2 π \int_{1}^{4} (x^{3 / 2} - 1) d x = {2 π [\frac{2 x^{5 / 2}}{5} - x] |}_{1}^{4} = \frac{94 π}{5} {units}^{3} . \end{matrix}

$\begin{array}{cc}\hfill V& ={\displaystyle\int }_{1}^{4}(2\pi x(f(x)-g(x)))dx\hfill \\ & ={\displaystyle\int }_{1}^{4}(2\pi x(\sqrt{x}-\frac{1}{x}))dx=2\pi {\displaystyle\int }_{1}^{4}({x}^{3\text{/}2}-1)dx\hfill \\ & ={2\pi \left[\frac{2{x}^{5\text{/}2}}{5}-x\right]|}_{1}^{4}=\frac{94\pi }{5}{\text{units}}^{3}.\hfill \end{array}$

Try It

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 $\left[0,1\right].$ Find the volume of the solid of revolution formed by revolving $R$ around the $y\text{-axis}.$

Show Solution

$\frac{\pi }{6}$ units³

Hint

Use the process from the previous example.

Module 6: Applications of Integration