## Section3.4Volume of Revolution: Shell Method

In the previous section, we calculated the volume of a solid of revolution over a closed interval $[a,b]$ by adding up the cross-sectional areas, which we obtained by slicing through the solid with planes perpendicular to the axis of rotation over $[a,b]\text{.}$ There is another method, which instead of creating disk-slices will create cylindrical shell-slices. These cylindrical shell-slices are created by cutting through the solid with cylinders that wrap symmetrically around the axis of rotation as shown in Figure 3.15. This is similar to stacking paper towel rolls of increasing radii inside of each other.

Just like we were able to add up disks, we can also add up cylindrical shells, and therefore this method of integration for computing the volume of a solid of revolution is referred to as the Shell Method. We begin by investigating such shells when we rotate the area of a bounded region around the $y$-axis.

### Subsection3.4.1Shell Method: Integration w.r.t. $x$

Suppose the region bounded by $f(x)=\sqrt{x-1}+2$ with $x\in[1,5]$ is rotated around the $y$-axis as shown below to the right. It is possible, but inconvenient, to compute the volume of the resulting solid by the Washer Method we have used so far. The problem is that there are two “kinds” of typical washers: Those that go from the curve $f$ to the line $x=5$ and those that sit between the lines $x=1$ and $x=5$ as shown below to the left.

To compute the volume using this approach, we need to break the problem into two parts and compute two integrals:

\begin{align*} V \amp = \int_0^2 \pi(5^2-1^2)\,dy + \int_2^4 \pi \left[5^2 - \left((y-2)^2+1\right)^2\right]\,dy\\ \amp = \int_0^2 24\pi \,dy + \int_2^4 \pi\left(-y^4+8y^3-26y^2+40y\right)\,dy\\ \amp = 24\pi y\bigg\vert_0^2 + \pi\left[-\frac{y^5}{5}+2y^4-\frac{26y^3}{3}+20y^2\right]_2^4 = 84 \frac{4}{15}\pi \end{align*}

If instead we slice through the solid of revolution parallel to the axis of rotation using cylindrical shells with increasing radii over the interval $[1,5]$ along the $x$-axis, then any of the cylindrical shells has height $f$ as shown below.

Note that “washers” are related to the area of a circle, $\pi r^2\text{,}$ whereas “shells” are related to the surface area of an open cylinder, $2\pi rh\text{.}$ If we add up the volume of such thin shells we will get an approximation to the true volume. What is the volume of such a shell? Consider the shell at $\ds x_i\text{.}$ Imagine that we cut the shell vertically in one place and “unroll” the surface of the cylindrical shell into a thin, flat sheet, as shown below.

This sheet will be almost a rectangular prism that is $\Delta x$ thick, $h$ high, and $\ds 2\pi r$ wide (the circumference of the cylindrical shell). In terms of $x_i\text{,}$ notice that $h$ is the height of $f\text{,}$ and that the radius $r$ is precisely $x_i\text{.}$ The volume of one cylindrical shell will then be approximately the volume of a rectangular prism with these dimensions: $\ds 2\pi x_i f(x_i)\Delta x\text{.}$ If we add these up and take the limit as usual, we get the integral

\begin{equation*} V = \int_1^5 2\pi x f(x) \,dx = \int_1^5 2\pi x\left(\sqrt{x-1}+2\right)\,dx\text{.} \end{equation*}

Now use substitution to evaluate the integral:

\begin{equation*} u = x-1,\ x = u+1,\ du = dx,\ u(1)=0,\ u(5)=4\text{.} \end{equation*}

Therefore,

\begin{equation*} \begin{split} V \amp = \int_1^5 2\pi x\left(\sqrt{x-1}+2\right)\,dx \\[0.5ex] \amp =2\pi \int_0^4 (u+1)\left(\sqrt{u}+2\right)\,du \\[0.5ex] \amp =2\pi\int_0^4\left(u^{3/2}+2u+u^{1/2}+2\right)\,du\\[0.5ex] \amp =2\pi\left[\frac{2u^{5/2}}{5}+u^2-\frac{2u^{3/2}}{3}+2u\right]_0^4\\[0.5ex] \amp = 84 \frac{4}{15}\pi. \end{split} \end{equation*}

This accomplishes the task with only one integral. One may argue that in the end, we had to do the same amount of work. However, there are often situations when one of the two methods, washer or shell, is easier than the other. Therefore, it is worthwhile investigating this technique when computing volumes of solids of revolution. We capture our results in the following theorem.

We now provide one example of the Shell Method.

###### Example3.33. Shell Method — Integration w.r.t. $x$.

Suppose the area below the curve $f(x)=x+1$ for all $x$ in $[0,3]$ is rotated about the $y$-axis. Find the volume using the Shell Method.

Solution

We begin by graphing $f$ on the interval $[0,3]$ and identify an arbitrary cylindrical shell as shown below.

Then the volume is given by

\begin{equation*} \begin{split} V \amp = \int_0^3 2\pi x f(x) \,dx = \int_0^3 2\pi x (x+1)\,dx \\[0.5ex] \amp = 2\pi \int_0^3 (x^2+x)\,dx = 2\pi \left[\frac{x^3}{3}+\frac{x^2}{2}\right]_0^3 = 27\pi. \end{split} \end{equation*}

Note:

1. Once again, we can rotate the area of any region around an axis of rotation, including the area of a region bounded above by a function $y=f(x)$ and below by a function $y=g(x)$ on an interval $x\in[a,b]$ as we will investigate next.

2. The axis of rotation can be any axis parallel to the $y$-axis for this method to work. Consider again the region bounded by the function $f(x)=\sqrt{x-1}+2$ with $x \in[1,5]\text{,}$ but let's also include a second bound, namely the function $g(x)=\frac{1}{4}(x-1)^2\text{.}$ Algebra shows us that $f > g$ on $x \in[1,5]\text{.}$ We want to compute the volume of the solid obtained by rotating the bounded region about the $y$-axis. We begin by graphing the region and remind ourselves that the given region is the difference of two regions as shown below:

The above figure shows us that the volume of the solid of revolution obtained from rotating the given region is simply the difference between the volumes of the two solids of revolution that are created from rotating the area under the curve of $f$ and the area under the curve of $g$ respectively. An arbitrary cylindrical shell must therefore have height $f(x_i)-g(x_i)$ as shown below.
Therefore, the volume is given by

\begin{equation*} V =\int_1^5 2\pi x \left[f(x)-g(x)\right]\,dx = \int_1^5 2\pi x \left[\sqrt{x-1}+2-\frac{(x-1)^2}{4}\right]\,dx = \frac{208\pi}{5} \end{equation*}

We capture our result in the following theorem.

We now provide one more example of such a region bounded below and above by two functions $f$ and $g$ respectively.

###### Example3.35. Shell Method — Integration w.r.t $x$.

Suppose the region between $f(x)=x+1$ and $g(x)=(x-1)^2$ is rotated around the $y$-axis. Find the volume using the Shell Method.

Solution

We begin by finding the intersection points of the two functions $f$ and $g\text{:}$

\begin{equation*} \begin{gathered} f(x) = g(x) \\ x+1 = (x-1)^2 \\ 0 = x^2 - 3x = x(x-3) \end{gathered} \end{equation*}

and so $x=0$ and $x=3$ are the intersection points. We now use this information to graph $f$ and $g$ on the interval $[0,3]$ and identify an arbitrary cylindrical shell as shown below.

Then the volume is computed as follows:

\begin{equation*} \begin{split} V \amp = \int_0^3 2 \pi x \left[f(x)-g(x)\right]\,dx \\[0.5ex] \amp =\int_0^3 2 \pi x\left[(x+1)-(x-1)^2\right]\,dx \\[0.5ex] \amp = 2 \pi \int_0^3 (3x^2-x^3)\,dx \\[0.5ex] \amp = 2\pi\left[x^3-\frac{x^4}{4}\right]_0^3\\[0.5ex] \amp = \frac{27}{2}\pi \end{split} \end{equation*}

### Subsection3.4.2Shell Method: Integration w.r.t. $y$

So far, we have discussed three main manners of generating a solid of revolution and how to compute its volume, which are listed below. Remember that the Washer Method is replaced by the Disk Method when the lower or left curve is described by the $x$-axis or the $y$-axis respectively.

• Rotating an area that is bounded above and below by functions of $x$ as well as lines $x=a$ and $x=b$ around the $x$-axis, and then using the Washer Method for volume-computation.

• Rotating an area that is bounded right and left by functions of $y$ as well as lines $y=c$ and $y=d$ around the $y$-axis, and then using the Washer Method for volume-computation.

• Rotating an area that is bounded above and below by functions of $x$ as well as lines $x=a$ and $x=b$ around the $y$-axis, and then using the Shell Method for volume-computation.

There is only one case left:

• Rotating an area that is bounded right and left by functions of $y$ as well as lines $y=c$ and $y=d$ around the $y$-axis, and then using the Shell Method for volume-computation.

We are readily convinced that the volume of such a solid of revolution can be calculated using a Shell Method similar in manner as the one discussed earlier, which is summarized in the following theorem.

We provide one example to exemplify this method as well.

###### Example3.37. Comparing Methods.

Suppose the area under $\ds y=-x^2+1$ is rotated around the $x$-axis. Find the volume of the solid of rotation using

1. the Disk Method; and

2. the Shell Method.

Solution
1. We begin by graphing the area and indicate one arbitrary disk:

Therefore, the volume is given by

\begin{equation*} \begin{split} V \amp = \int_{-1}^1 \pi y^2 \, dx = \int_{-1}^1 \pi \left(1-x^2\right)^2 \,dx \\[0.5ex] \amp = 2\pi \int_{0}^1\left(x^4-2x^2+1\right)\,dx = 2\pi\left[\frac{x^5}{5}-\frac{2x^3}{3}+x\right]_0^1 = \frac{16}{15}\pi \end{split} \end{equation*}
2. We again begin by graphing the area and indicate one arbitrary cylindrical shell:

First, we have to express $x$ in terms of $y\text{:}$

\begin{equation*} y=1-x^2 \implies x= \pm \sqrt{1-y} \end{equation*}

Therefore, the volume is computed with

\begin{equation*} V = \int_0^1 2\pi y(2x)\,dy = 4\pi \int_0^1 y \sqrt{1-y}\,dy\text{.} \end{equation*}

Now we use substitution

\begin{equation*} u = 1-y,\ y = 1-u,\ du=-dy,\ u(0)=1,\ u(1)=0 \end{equation*}

and get

\begin{equation*} \begin{split} V \amp = 4\pi \int_0^1 y\sqrt{1-y}\,dy= 4\pi \int_1^0 (1-u)\sqrt{u}\,(-du)\\ \amp = 4\pi \int_0^1 \left(u^{1/2}-u^{3/2}\right)\,du\\ \amp = 4\pi \left[\frac{2u^{3/2}}{3} - \frac{2u^{5/2}}{5}\right]_0^1 = \frac{16}{15}\pi \end{split} \end{equation*}

### Subsection3.4.3Summary

There are many different scenarios in which the Shell Method can be employed, which are not discussed here; however, we provide a general guideline.

###### Guideline for Shell Method.

The following steps outline how to employ the Shell Method.

1. Graph the bounded region.

2. Construct an arbitrary cylindrical shell parallel to the axis of rotation.

3. Identify the radius and height of the cylindrical shell.

4. Determine the thickness of the cylindrical shell.

5. Set up the definite integral by making sure you are computing the volume of the constructed cylindrical shell.

##### Exercises for Section 3.4.

Use Shell Method to find the volumes of the solids generated by revolving the shaded region about the indicated axis.

1. The $y$-axis:

$12\pi\text{.}$
Solution
We use the Shell method and integrate with respect to $x\text{:}$
\begin{equation*} \begin{split} V \amp= \int_0^2 2\pi x (1+x^2)\,dx\\ \amp= 2\pi\int_0^2 (x+x^3)\,dx\\ \amp= 2\pi \left[\frac{x^2}{2}+\frac{x^4}{4}\right]_0^2\\ \amp= 12\pi. \end{split} \end{equation*}

2. The $y$-axis:

$\frac{3\pi}{2}\text{.}$
Solution
We use the Shell method and integrate with respect to $x\text{:}$
\begin{equation*} \begin{split} V \amp= \int_0^1 2\pi x (2-x^2)\,dx\\ \amp= 2\pi\int_0^1 (2x-x^3)\,dx\\ \amp= 2\pi \left[x^2-\frac{x^4}{4}\right]_0^1\\ \amp= \frac{3\pi}{2}. \end{split} \end{equation*}
3. The $x$-axis:

$\frac{\pi}{2}\text{.}$
Solution
We use the Shell method and integrate with respect to $y\text{:}$
\begin{equation*} \begin{split} V \amp= \int_0^1 2\pi y (y^2+1-1)\,dy\\ \amp= 2\pi\int_0^1 y^3\,dx\\ \amp= 2\pi \left[\frac{y^4}{4}\right]_0^1\\ \amp= \frac{\pi}{2}. \end{split} \end{equation*}
4. The $x$-axis:

$8\pi.$
Solution
We use the Shell method and integrate with respect to $y\text{:}$
\begin{equation*} \begin{split} V \amp= \int_0^2 2\pi y \left[4-(4-y^2)\right]\,dy\\ \amp= 2\pi\int_0^2 y^3\,dx\\ \amp= 2\pi \left[\frac{y^4}{4}\right]_0^2\\ \amp= 8\pi. \end{split} \end{equation*}

Use Shell Method to find the volumes of the solids generated by revolving the given bounded regions about the $y$-axis.

1. $y=2x, \ y=-x, \ x=1$

$2\pi\text{.}$
Solution
We begin by plotting the area between the curves:
We notice that, for $0 \leq x \leq 1\text{,}$ the area is bounded above by the curve $y=2x$ and below by the curve $y=-x\text{.}$ Therefore, we use the Shell method and integrate with respect to $x\text{:}$
\begin{equation*} V = \int_0^1 2 \pi x\left[2x-(-x)\right]\,dx= 6\pi \int_0^1 x^2\,dx= 6\pi \frac{x^3}{3}\bigg\vert_0^1= 2\pi. \end{equation*}
2. $y=x^2, \ y=x, \ y \geq 0$

$\frac{2\pi}{15}.$
Solution
We notice that the curves $x=y^2$ and $x=y$ intersect at $x=1\text{.}$ We now plot the curves for $0 \leq x \leq 1\text{:}$
We notice that the region is bounded above by the curve $=\sqrt{x}\text{,}$ and below by the curve $y=x\text{.}$ Therefore, we use the Shell method and integrate with respect to $x\text{:}$
\begin{equation*} V = \int_{0}^1 2 \pi x \left[\sqrt{x} - x\right]\,dx = 2\pi \int_0^1 \left(x^{3/2}-x^2\right)\, dx = 2\pi \left[\frac{2x^{5/2}}{5} - \frac{x^3}{3}\right]_0^1 = \frac{2\pi}{15}. \end{equation*}
3. $y=x^2, \ y=x+2, \ x\geq 0$

$\frac{16\pi}{3}$
Solution
We first find where the two curves intersect:
\begin{equation*} x^2=x+2 \implies x^2-x-2=0 \implies x=-1,2. \end{equation*}
We reject the negative solution and consider $0 \leq x \leq 2\text{:}$
The region is bounded above by the curve $y=x+2$ and below by the curve $y=x^2\text{.}$ We therefore use the Shell method and integrate with respect to $x\text{:}$
\begin{equation*} V = \int_{0}^{2} 2\pi x \left[x+2-x^2\right]\,dx = 2\pi \int_0^2 \left(x^2+2x-x^3\right)\,dx = 2\pi \left[\frac{x^3}{3} + x^2 - \frac{x^4}{4} \right]_0^2 = \frac{16\pi}{3}. \end{equation*}
4. $y=1-x^2, \ x^2, \ x\geq 0$

$\frac{\pi}{4}$
Solution

We begin by plotting the area between the curves $y=1-x^2$ and $y=x^2$ for $x \geq 0\text{.}$

The curves intersect at $x = \pm \frac{1}{\sqrt{2}}\text{,}$ and for $0 \leq x \leq \frac{1}{\sqrt{2}}\text{,}$ we see that $1-x^2 \geq x^2\text{.}$ Therefore, we calculate the volume of the solid generated by rotating the area about the $y$-axis using the shell method:

\begin{equation*} V =\int_0^{1/\sqrt{2}} 2\pi x \left[1-x^2-x^2\right]\,dx = 2\pi\left[\frac{x^2}{2}-\frac{x^4}{2}\right]_0^{1/\sqrt{2}} = \frac{\pi}{4}\text{.} \end{equation*}
5. $y=\sqrt{x},\ y=2-x, \ x=0$

$\frac{8\pi}{15}.$
Solution
We begin by plotting the region bounded by the given curves:
We notice that the region is bounded above by the curve $y=2-x\text{,}$ and below by the curve $y=\sqrt{x}\text{.}$ We now use the Shell method and integrate with respect to $x\text{:}$
\begin{equation*} V = \int_0^1 2\pi x \left[2-x-\sqrt{x}\right]\,dx = 2\pi \int_0^1 \left(2x-x^2-x^{3/2}\right)\,dx = \frac{8\pi}{15}. \end{equation*}
6. $y=\frac{1}{x}, \ x=2, \ x=3, \ y=0$

$2 \pi.$
Solution
We begin by plotting the region bounded by the curves:
We now use the Shell method and integrate with respect to $x\text{:}$
\begin{equation*} V = \int_2^3 2\pi x\frac{1}{x}\,dx = 2\pi \int_2^3 \,dx = 2 \pi. \end{equation*}

Use Shell Method to find the volumes of the solids generated by revolving the given bounded regions about the $x$-axis.

1. $x=2\sqrt{y}, \ x=-y, \ y=4$

$\frac{1408 \pi}{15}\text{.}$
Solution
We begin by plotting the area bounded by the given curves:
We notice that the region is bounded for $y\in[0,4]$ to the left by the curve $x=-y$ and to the right by the curve $x=\sqrt{y}\text{.}$ Therefore, we use the Washer method and integrate with respect to $y\text{:}$
\begin{equation*} \begin{split} V \amp = \int_0^4 2\pi y\left[2\sqrt{y}+y\right]\,dy \\ \amp=2\pi \int_0^4 2y^{3/2}+y^2 \,dy\\ \amp= 2\pi \left[\frac{4y^{5/2}}{5} + \frac{y^3}{3}\right]_0^4\\ \amp= \frac{1408 \pi}{15}. \end{split} \end{equation*}
2. $x=y^2, \ x=y, \ y \geq 0$

$\frac{\pi}{6}.$
Solution
We begin by plotting the area bounded by the curves:
The region is bounded for $y\in[0,1]$ to the left by the curve $x=y^2$ and to the right by the curve $x=y\text{.}$ Therefore, we use the Washer method and integrate with respect to $y\text{:}$
\begin{equation*} \begin{split} V \amp= \int_0^1 2\pi y\left[y-y^2\right]\,dy\\ \amp= 2\pi \int_0^1 y^2-y^3 \,dy \\ \amp= 2 \pi \left[\frac{y^3}{3}-\frac{y^4}{4}\right]_0^1\\ \amp= 2\pi \left[\frac{1}{3} - \frac{1}{4}\right]\\ \amp= \frac{\pi}{6}. \end{split} \end{equation*}
3. $x = y-y^2/4, \ x=0$

$\frac{32 \pi}{3}.$
Solution
Notice that the two curves intersect at $y=0$ and $y=4\text{.}$ We now plot the region contained between the curves:
The region is bounded for $y\in[0,4]$ to the left by the curve $x=0$ and to the right by the curve $x=y=y^2/4\text{.}$ Therefore, we use the Washer method and integrate with respect to $y\text{:}$
\begin{equation*} \begin{split} V \amp= \int_0^4 2\pi y \left[y-\frac{y^2}{4}\right]\,dy\\ \amp= 2\pi \int_0^4 \left[y^2-\frac{y^3}{4}\right]\,dy \\ \amp= 2\pi \left[\frac{y^3}{3} - \frac{y^4}{16}\right]_0^4\\ \amp= \frac{32 \pi}{3}. \end{split} \end{equation*}
4. $x = y-y^2/4,\ x=y/2$

$\frac{5\pi}{24}.$
Solution
The two curves intersect at $(2,1)\text{.}$ We now plot the region bounded by the curves:
The area is bounded for $y\in[0,1]$ to the left by the curve $x=y/2$ and to the right by the curve $x=y-y^2/4\text{.}$ Therefore, we use the Washer method and integrate with respect to $y\text{:}$
\begin{equation*} \begin{split} V \amp= \int_0^1 2\pi y \left[y-\frac{y^2}{4} - \frac{y}{2}\right]\,dy \\ \amp= 2\pi \int_0^1 \left[y^2 - \frac{y^3}{4} - \frac{y^2}{2}\right]\,dy\\ \amp= 2\pi \left[\frac{y^3}{3} - \frac{y^4}{16} - \frac{y^3}{6}\right]_0^1\\ \amp= \frac{5\pi}{24} \end{split} \end{equation*}
5. $y=|x|, \ y=2$

$\frac{32\pi}{3}$
Solution

The area contained between the curves $y=|x|$ and $y=2$ is shown in green below.

We see that $y$ is going from $0$ to $2\text{,}$ and that for $0 \leq y \leq 2\text{,}$ $y \geq -y\text{.}$ And so the volume of the solid generated by rotating the above area about the $x$-axis is given by

\begin{equation*} V =\int_0^2 2\pi y \left[y-(-y)\right]\,dy = \frac{2\pi (2y^3)}{3} \bigg\vert_0^2 = \frac{32\pi}{3}\text{.} \end{equation*}
6. $y=x+1, \ y=2x$

$\frac{4\pi}{3}$
Solution
Notice that the two curves intersect at the point $(1,2)\text{.}$ We now plot the region bounded by the curves:
The region is bounded for $y\in[0,2]$ to the left by the curve $y=x+1 \implies x = y-1\text{,}$ and to the right by the curve $y=2x \implies x=y/2\text{.}$ Therefore, we use the Washer method and integrate with respect to $y\text{:}$
\begin{equation*} \begin{split} V \amp= \int_0^2 2\pi y \left[\frac{y}{2} - y + 1 \right]\,dy \\ \amp= 2\pi \int_0^2 \left[\frac{y^2}{2} - y^2 + y\right]\,dy \\ \amp= 2\pi \left[\frac{y^3}{6} - \frac{y^3}{3} + \frac{y^2}{2}\right]_0^2\\ \amp= \frac{4\pi}{3}. \end{split} \end{equation*}

Let $R$ be the region of the $x$-$y$-plane bounded below by the curve $y=\sqrt{x}\text{,}$ and above by the line $y=3\text{.}$ Find the volume of the solid obtained by rotating $R$ around

1. the $x$-axis;

$\frac{81\pi}{2}$
Solution
The area bounded between the curves $y=\sqrt{x}$ and $y=3$ is shown below:
The region is bounded for $y\in[0,3]$ to the left by the curve $x=0$ and to the right by the curve $x=y^2\text{.}$ We use the Shell method and integrate with respect to $y\text{:}$
\begin{equation*} V= \int_0^3 2\pi y \left[y^2\right] = 2\pi \frac{y^4}{4} \big\vert_0^3 = \frac{81\pi}{2}. \end{equation*}
2. the line $y=3\text{;}$

$\frac{27\pi}{2}$
Solution
The area bounded between the curves $y=\sqrt{x}$ and $y=3$ is shown below:
We solve this problem in two ways. First, we shift the region down by three units, and then rotate about the $x$-axis:
\begin{equation*} V = \int_{-3}^0 2\pi y \left[3-y^2\right]\,dy =\frac{27\pi}{2}. \end{equation*}
Next, we consider using the Shell method directly, but now the radius of our cylinders will be $3-y\text{:}$
\begin{equation*} V = \int_{0}^3 2\pi (3-y) (y^2)\,dy =\frac{27\pi}{2}, \end{equation*}
which gives the same result.
3. the $y$-axis, and

$\frac{243\pi}{5}$
Solution
The area bounded between the curves $y=\sqrt{x}$ and $y=3$ is shown below:
We use the Shell method and integrate with respect to $x\text{:}$
\begin{equation*} V = \int_0^9 2\pi x\left[3-\sqrt{x}\right]\,dx = \frac{243\pi}{5}. \end{equation*}
4. the line $x=9\text{.}$

$\frac{576\pi}{5}$
Solution
The area bounded between the curves $y=\sqrt{x}$ and $y=3$ is shown below:
We notice that the radius of the shells will now be $9-x\text{.}$ Therefore, we have
\begin{equation*} V = \int_0^9 2\pi (9-x)(3-\sqrt{x})\,dx = \frac{576\pi}{5}. \end{equation*}

Let $S$ be the region of the $x$-$y$-plane bounded above by the curve $x^3y=64\text{,}$ below by the line $y=1\text{,}$ on the left by the line $x=2\text{,}$ and on the right by the line $x=4\text{.}$ Find the volume of the solid obtained by rotating $S$ around

1. the $x$-axis;

$114\pi/5$
Solution
The area bounded between the curves $x^3y=64\, y=1, \ x = 2, \ x=4$ is shown below:
We use the Shell method and integrate with respect to $y\text{:}$
\begin{equation*} V = \int_1^8 2\pi y\left[\left(\frac{64}{y}\right)^{1/3} -2 \right]\,dy = \frac{114\pi}{5}. \end{equation*}
2. the line $y=1\text{;}$

$74\pi/5$
Solution
The area bounded between the curves $x^3y=64\, y=1, \ x = 2, \ x=4$ is shown below:
If we rotate around the line $y=1\text{,}$ we notice that the radius of our shells will be $y-1\text{:}$
\begin{equation*} V = \int_1^8 2\pi (y-1)\left[\left(\frac{64}{y}\right)^{1/3} -2 \right]\,dy=\frac{74\pi}{5}. \end{equation*}
3. the $y$-axis, and

$20\pi$
The area bounded between the curves $x^3y=64\, y=1, \ x = 2, \ x=4$ is shown below:
We use the Shell method and integrate with respect to $x\text{:}$
4. the line $x=2\text{.}$
$4\pi$
The area bounded between the curves $x^3y=64\, y=1, \ x = 2, \ x=4$ is shown below:
If we rotate about the line $x=2\text{,}$ the radius of our shells will be $x-2\text{:}$