Accumulated shells

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Some volumes of revolution are more easily computed with cylindrical shells.

More than one method

In this section, we will show you a new method for computing volumes of solids of revolution. Consider the following region bounded by $f(x)=x+1$ and $g(x)=(x-1)^2$ :

If this region is rotated around the $y$ -axis, it is possible, but inconvenient, to compute the volume of the resulting solid by the methods we have used so far. The issue is that there are two “kinds” of cylindrical cross-sections:

As we see above, some of the cylindrical cross sections are defined by the line that goes from $g$ to $f$ , and others are defined by the line that touches $g$ at both ends. To compute the volume using accumulated cross-sections, we need to break the problem into two integrals:

an integral that computes the volume of the region bounded by $g$ and the line $y=1$ , rotated about the $y$ -axis, and
an integral that computes the volume of the region bounded by $f$ , $g$ and the line $y=1$ , rotated about the $y$ -axis.

Since we are rotating around the $y$ -axis, we should look at $f^{-1}$ and $g^{-1}$ .

Write with me: $f^{-1}(y) = \answer [given]{y-1}$ On the other hand $g$ is not one-to-one, so we cannot invert it on the entire domain. Nevertheless, if we restrict the domain of $g$ we may write two separate functions $g^{-1}_1(y) = \answer [given]{1-\sqrt {y}}\qquad \text {when $0<x<1$}$ and $g^{-1}_2(y) = \answer [given]{1+\sqrt {y}}\qquad \text {when $1<x<3$}.$

With this in mind, we can compute our volume with: $\begin{align*} \int_0^1 &\pi(g^{-1}_2(y))^2-\pi(g_1^{-1}(y))^2\d y\\ &+ \int_1^4\pi(g^{-1}_2(y))^2-\pi(f^{-1}(y))^2\d y \end{align*}$

Substituting in, we find $\begin{align*} =\int_0^1 &\pi(1+\sqrt{y})^2-\answer[given]{\pi(1-\sqrt{y})^2}\d y\\ &+ \int_1^4 \pi(1+\sqrt{y})^2-\answer[given]{\pi(y-1)^2}\d y \end{align*}$ and so $\begin{align*} &=\frac{8\pi}{3} + \frac{65\pi}{6}\\ &=\answer[given]{\frac{27\pi}{2}}. \end{align*}$

While we have successfully solved this problem, it wasn’t easy. Let’s see another, perhaps easier method to solve this problem. If instead we consider a vertical rectangle of height $f(x)-g(x)$ (just like we did when we computed areas of regions between curves!) and width $\d x$ , and we additionally rotate this rectangle around the $y$ -axis, we get a thin “shell” or hollow-tube:

Here the infinitesimal change in volume is:

Integrating $\d V$ will give us our desired volume.

Shells around the axes

Let’s start by actually doing our motivating example above.

Consider the region below bounded by $f(x)=x+1$ and $g(x)=(x-1)^2$ :

Find the volume of the solid of revolution formed by rotating this region around the $y$ -axis.

We’ll solve this problem using accumulated shells. If we draw our shell, we see

Since the volume of each shell is $\d V = 2\pi x (f(x) -g(x) ) \d x$ we may write $\begin{align*} \int_{\answer[given]{0}}^{\answer[given]{3}} 2\pi x(f(x)-g(x))\d x &= \int_{\answer[given]{0}}^{\answer[given]{3}} 2\pi x(\answer[given]{x+1}-(x-1)^2)\d x\\ &= \eval{\answer[given]{2\pi x^3 -\frac{\pi x^4}{2}}}_{\answer[given]{0}}^{\answer[given]{3}}\\ &=\answer[given]{\frac{27\pi}{2}}. \end{align*}$

Comparing our work above to our earlier work, we see that using shells not only solves the problem with just one integral, we see that the integral itself is somewhat easier than those in the previous calculation! Things are not always so neat, but it is often the case that one of the two methods will be simpler than the other, so it is worth considering both methods.

Suppose the area bounded by $y=\sqrt {x}$ , the line $y = 2x-1$ , and the $x$ -axis is rotated around the $x$ -axis,

find the volume of this solid.

While we could use accumulated cross-sections to compute this volume, it would require two integrals, (the industrious young mathematician is encouraged to attempt to solve this problem using accumulated cross-sections). However, here we will use the method of shells. Let’s see a picture:

In this case, the volume of each shell is

Since we are integrating with respect to $y$ , we have $y=\sqrt {x} \qquad \Rightarrow \qquad x = \answer [given]{y^2}$ and $y = 2x-1 \qquad \Rightarrow \qquad x= \answer [given]{\frac {y+1}{2}}.$ Hence the volume of our shell is $\d V = 2\pi y \left (\answer [given]{\frac {y+1}{2} -y^2}\right ) \d y$ Thus the the total volume of our solid of revolution is given by $\begin{align*} \text{Volume} &= \int_{\answer[given]{0}}^{\answer[given]{1}} \answer[given]{2\pi y (\frac{y+1}{2} - y^2)} \d y\\ &= 2\pi\int_{\answer[given]{0}}^{\answer[given]{1}} \frac{y^2}{2}+\frac{y}{2} - y^3 \d y\\ &= 2\pi\eval{\answer[given]{\frac{y^3}{6}+\frac{y^2}{4} - \frac{y^4}{4}}}_{\answer[given]{0}}^{\answer[given]{1}}\\ &= \answer[given]{\frac{\pi}{3}}. \end{align*}$

Shells around other lines

What if we had wanted to rotate the region from the last example about the line $y = -1$ instead of the $x$ -axis? In this case, we draw a picture and work much the same way as before. Let’s see an example.

Suppose the area bounded by $y=\sqrt {x}$ , the line $y = 2x-1$ , and the $x$ -axis is rotated around the line $y= -1$ ,

find the volume of this solid.

Let’s see a picture:

Since we are integrating with respect to $y$ , we have $y=\sqrt {x} \qquad \Rightarrow \qquad x = \answer [given]{y^2}$ and $y = 2x-1 \qquad \Rightarrow \qquad x= \answer [given]{\frac {y+1}{2}}.$

The radius is now $y+1$ , so the circumference is $2\pi (y+1)$

Our cylindrical shells have the same width and height as in the previous problem, but the circumference would changes, as the radius is $\answer [given]{y+1}$ . Hence the volume of our shell is $\d V = 2\pi (y+1) \left (\answer [given]{\frac {y+1}{2} -y^2}\right ) \d y$ Thus the the total volume of our solid of revolution is given by $\begin{align*} \text{Volume} &= \int_{\answer[given]{0}}^{\answer[given]{1}} \answer[given]{2\pi (y+1) (\frac{y+1}{2} - y^2)} \d y\\ &= 2\pi\int_{\answer[given]{0}}^{\answer[given]{1}} \frac{1}{2} + y - \frac{y^2}{2} -y^3 \d y\\ &= 2\pi\eval{\answer[given]{\frac{y}{2}+\frac{y^2}{2}-\frac{y^3}{6} - \frac{y^4}{4}}}_{\answer[given]{0}}^{\answer[given]{1}}\\ &= \answer[given]{\frac{7 \pi}{6}}. \end{align*}$