Calculating Volumes - Cylindrical Shell Method

Calculating Volumes - Cylindrical Shells Method

We have just looked at the method of using disks/washers to calculate a solid of revolution. We are now going to look at a new technique involving cylindrical shells.

Suppose that we had a function $y = f(x)$, and we wanted to find the volume of the solid of revolution formed from rotating the area trapped between $f(x)$, the $x$-axis, and the lines $x = a$ and $x = b$ where $0 ≤ a < b$. The idea behind cylindrical shells is to "stack" multiple cylindrical shells within each other to form the solid. Recall that the volume of a cylinder can be obtained by the formula $v = \pi r^2 h$. Now let's calculate an equation for the volume of a single cylindrical shell with inner radius $r_1$ and outer radius $r_2$ as the diagram below illustrates


We can calculate this volume by taking the volume of the larger cylinder, $V_{\mathrm{larger cylinder}} = \pi r_2^2 h$ and subtracting the volume of the smaller cylinder, $V_{\mathrm{smaller cylinder}} = \pi r_1^2 h$. We will also do some algebraic manipulation to be able to recognize some important components of our equation as follows:

\begin{align} V_{\mathrm{shell}} = \pi r_2^2 h - \pi r_1^2h \\ V_{\mathrm{shell}} = \pi(r_2^2 - r_1^2)h \\ V_{\mathrm{shell}} = \pi(r_2 + r_1)(r_2 - r_1)h \\ V_{\mathrm{shell}} = 2\pi\frac{r_2 + r_1}{2}(r_2 - r_1)h \end{align}

We will note that $\frac{r_2 + r_1}{2}$ represents the average radius of our cylindrical shell which we will now denote simply as $r$, while $r_2 - r_1$ represents the change in the radius, or rather, $r_2 - r_1 = \Delta r$. Thus we obtain the formula $V_{\mathrm{shell}} = 2\pi r h \Delta r$ for any cylindrical shell. From this formula, we can easily identify the circumference of the circle defining the shell ($2\pi r$), the height of the shell ($h$), and the width of the shell ($\Delta r$).

Now if we convert this formula in terms of our problem with calculating the solid of revolution with cylindrical shells, we let $\bar{x}$ will represent the average radius of a shell, $f(\bar{x})$ represents the height of our shell, and $\Delta x$ will represent the change in thickness between our inner and outer radii. Thus, we get the volume formula of the $k^{\mathrm{th}}$ single shell from our solid as $V = 2\pi \bar{x}_k f(\bar{x}_k) \Delta x$. If we divide our interval $[a, b]$ into $n$ subintervals $[x_{k-1}, x_{k}$, then we will let $\bar{x}_k$ be the midpoint of this $k^{\mathrm{th}}$ subinterval. Now, the volume of the entire solid can be approximated by taking these $n$ cylindrical shells and summing their volumes, that is, $V \approx \sum_{i=1}^{n} 2\pi \bar{x}_k f(\bar{x}_k) \Delta x$. As $n \to \infty$ (that is, the number of subintervals of $[a, b]$ so to infinity and thus, the number of cylindrical shells composing our solid of revolution goes to infinity), our approximation gets better and better, and thus:

\begin{align} V = \lim_{n \to \infty} \sum_{i=1}^{n} 2\pi \bar{x}_k f(\bar{x}_k) \Delta x \\ V = \int_{a}^{b} 2\pi x f(x) \: dx \end{align}

The diagram below illustrates the technique we are implementing:

Remark: We will note that a more concise formula for using this method by cylindrical shells would be that $V = \int_{a}^{b} (2 \pi r) h \: dx$ where $r$ is the radius of our shell, and $h$ is the height of our shell. In our general formula, we had $r = x$ and $h = f(x)$.

We will now look at some examples of applying this method.

Example 1

Determine the solid of revolution determined by the region bounded by the function $f(x) = 2x^2 - x^3$ around the $y$-axis.

First let's find the $x$-intercepts of the function $f$ which defines the rotated region of area:

\begin{equation} f(x) = 2x^2 - x^3 0 = x^2(2 - x) \end{equation}

So there exists x-intercepts at $x = 0$ and $x = 2$. We can now apply the formula for calculating the volume of revolution by cylindrical shells:

\begin{align} \mathbf{Volume} = \int_0^2 2\pi \: x \: (2x^2 - x^3) \: dx \\ \mathbf{Volume} = 2 \pi \int_0^2 2x^3 - x^4 \: dx \\ \mathbf{Volume} = 2 \pi [ \frac{x^4}{2} - \frac{x^5}{5} ] \bigg |_{0}^{2} \\ \mathbf{Volume} = 2 \pi [ \frac{2^4}{2} - \frac{2^5}{5} ] = 2 \pi (8 - \frac{32}{5}) = \frac{16 \pi}{5} \end{align}

Example 2

Determine the volume of the solid of revolution formed by rotating the region of the function $f(x) = \sqrt{x}$ bounded by the $x$-axis on the interval $[0, 2]$ about the $y$-axis.

Setting up the integral and evaluating we obtain:

\begin{align} \mathbf{Volume} = \int_0^2 2\pi x \sqrt{x} \: dx \\ \mathbf{Volume} = 2\pi \int_0^2 x^{3/2} \: dx \\ \mathbf{Volume} = 2\pi [ \frac{2}{5}x^{5/2} ] \bigg |_{0}^{2} \\ \mathbf{Volume} = \frac{4 \pi \sqrt{32}}{5} \end{align}
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