Volumes of Revolution: Cylindrical Shells: Learn It 1

Lumen Learning

Volumes of Revolution: Cylindrical Shells: Learn It 1

Determine the volume of a solid formed by rotating a region around an axis using cylindrical shells
Evaluate the benefits and limitations of different methods (disk, washer, cylindrical shells) for calculating volumes

Cylindrical Shells Method

Let’s explore the final method for finding the volume of a solid of revolution—the method of cylindrical shells. This method can be used on the same types of solids as the disk or washer method; however, we integrate along the coordinate axis parallel to the axis of revolution. This ability to choose which variable of integration to use can be a significant advantage with more complicated functions. Additionally, the specific geometry of the solid sometimes makes the method of using cylindrical shells more appealing than the washer method.

Again, we are working with a solid of revolution. We define a region $R,$ bounded above by the graph of a function $y=f(x),$ below by the $x$ -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).

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 $y\text{-axis}\text{.}$

Note that this is different from what we have done before. Previously, regions defined in terms of functions of $x$ were revolved around the $x$ -axis or a line parallel to it.

Next, 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 $\text{Δ}x.$

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 the following.

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:

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

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:

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

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:

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

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

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

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:

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

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 $y\text{-axis}.$

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:

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

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 $y\text{-axis}.$

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:

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