# Shell Method

The **shell method** is a technique for finding the volumes of solids of revolutions. It considers vertical slices of the region being integrated rather than horizontal ones, so it can greatly simplify certain problems where the vertical slices are more easily described.

## Summary

The **shell method** is a method of finding volumes by decomposing a solid of revolution into cylindrical shells. Consider a region in the plane that is divided into thin vertical strips. If each vertical strip is revolved about the \(x\)-axis, then the vertical strip generates a disk, as we showed in the disk method. However, if this thin vertical strip is revolved about the \(y\)-axis, we obtain a different object of revolution,
one that looks like a cylindrical shell, or an empty tin can with the top and bottom removed. The resulting volume of the cylindrical shell is the surface area of the cylinder times the thickness of the cylinder wall, or

\[ \Delta V = 2 \pi x y \Delta x.\]

The shell method calculates the volume of the full solid of revolution by summing the volumes of these thin cylindrical shells as the thickness \(\Delta x \) goes to \( 0\) in the limit:

\[ V = \int dV = \int_a^b 2 \pi x y \, dx = \int_a^b 2 \pi x f(x) \, dx.\]

Notice that the axis of rotation and the variable of integration are on a different axis, i.e., even though we are integrating about the \(x\)-axis, we are actually rotating about the \(y\)-axis.

## Basic Examples

## Use the shell method to find the volume generated by revolving the region bounded by \(y = \frac{1}{x}\), \(y=0\), \(x=1\), and \(x=4\) about the \(y\)-axis.

(The shell method corresponds to rotating the red rectangles about the \(y\)-axis.)

In this case, the radius of the cylindrical shell is \( x \) and the height is \(y = \frac{1}{x}\). Since the endpoints of the \(x\)-interval of the solid are \(x=1\) and \(x=4\), the volume of the solid is

\[ \begin{align} V &= \int_1^4 2\pi x \cdot \frac{1}{x}dx \\ &= 2\pi \int_1^4 1 dx\\ &= 2\pi \left[ x \right]^4_1 \\ &= 2\pi (4 - 1) \\ &= 6\pi. \ _\square \end{align}\]

## Use the shell method to find the volume generated by revolving the region bounded by \( y = x \) and \( y = x^2 \) about the \(y\)-axis.

(The shell method corresponds to rotating the red rectangles about the \(y\)-axis.)

These 2 graphs intersect at the points \( (0,0) \) and \( (1,1) \).

The radius of the cylindrical shell is \( x \) and the height is \( x - x ^2 \). Hence, the volume of the solid is

\[\begin{align} \int_0^1 2 \pi x ( x - x^2 ) \, dx &= \int_0^1 2 \pi ( x^2 - x^3) \, dx \\ &= \left[ 2 \pi ( \frac{ x^3}{3} - \frac{ x^4} { 4} ) \right]_0^1 \\ &= \frac{\pi}{6}. \ _\square \end{align}\]

## Using the disk method, we saw how to find the volume of a sphere of radius \(r\). Use the shell method to find the volume of a sphere of radius \(r\) with a vertical hole of radius \(a < r\) bored through the center of the sphere.

The cylindrical shell radius is \(2\pi x \) and the cylindrical shell height is \(2y = 2 \sqrt{r^2 - x^2} .\) Then the volume of the cylindrical shell is

\[ \Delta V = 2 \pi x (2y) \Delta x = 4 \pi x \sqrt{r^2 - x^2} \Delta x.\]

Now, since a vertical hole of radius \(a\) is bored through the center, the endpoints of the \(x\)-interval of the solid are \(x=a\) and \(x=r\), which gives

\[\begin{align} V &=\int_a^r 4 \pi x \sqrt{r^2 - x^2} dx\\ &= - \left[ \frac{4\pi}{3} (r^2 - x^2)^{3/2} \right]^{r}_a\\ &= \frac{4\pi}{3}\left( r^2 - a^2 \right)^{3/2}. \ _\square \end{align}\]

## When to use the shell method?

It is sometimes difficult to distinguish between using the disk method or the shell method. Many people do not like the shell method, because they do not understand what is happening. As such, they try and use the disk method almost exclusively.

However, there are instances when the shell method is much simpler:

- When the function \(f(x) \) is rotated around the \(y\)-axis.
- (rotated around the \(x\)-axis) When the graph is not a function on \(x\), but is a function on \(y\).
- When \( f(x) ^2 \) is hard to integrate but \(x f(x) \) is easy to integrate (especially by parts).

Let's look at some examples.

**1.** When the function \(f(x) \) is rotated about the \(y\)-axis:

The shell method will yield a direct answer, but the disk method requires us to figure out how to evaluate the corresponding volume.

## Determine the volume of the solid obtained when the region bounded by \( y = \sqrt{x} \), the line \( x = 1 \), and the \(x\)-axis is rotated about the \(y\)-axis.

(The shell method corresponds to using the red rectangles, while the disk method corresponds to using the blue rectangles.)

By the shell method, as \(x\) ranges from 0 to 1, the corresponding radius is \(x\) and the height is \( \sqrt{x} \). Hence, the volume is

\[ \int_0^1 2 \pi x \sqrt{x} \, dx = \left[ 2 \pi \times \frac{2}{5} x ^ { \frac{5}{2} } \right]_0^1 = \frac{4 \pi}{5}.\]

If we wanted to use the disk method, we will take the cylinder obtained by rotating the unit square, and then subtract off the region bounded by \( y = 1, y = \sqrt{x} \) and the \(y\)-axis, rotated about the \(y\)-axis. This gives us

\[ \pi \times 1^2 \times 1 - \int_0^1 \pi x^2 \, dy = \pi - \int_0^1 \pi y^4 \, dy = \pi - \left[ \pi \frac{y^3}{5} \right]_0^1 = \frac{4 \pi}{5} . \ _\square\]

**2.** When the graph is not a function on \(x\), but is a function on \(y\):

The shell method will yield a direct answer, but the disk method requires us to figure out all the corresponding functions.

## Determine the volume of the solid obtained when the region bounded by \( x = y^2 - 4y + 4, x = 0, x = 1 \) is rotated about the \(x\)-axis.

(The shell method corresponds to using the red rectangles, while the disk method corresponds to using the blue rectangles minus the yellow rectangles.)

When \(x = 0 , y = 2 \). When \( x = 1, y = 1, 3 \). As \(x\) ranges from 0 to 1, \(y\) ranges from 1 to 3.

Notice that we have a function in terms of \(y\). Using the shell method, as \(y\) ranges from 1 to 3, the corresponding radius is \(y\), and the height of the shell is \( 1 - x \). Hence, the volume is

\[\begin{align} \int_1^3 2 \pi y (1-x) \, dy &= \int_1^3 2 \pi y ( 1 - y^2 + 4y - 4) \ dy \\ &= \int_1^3 2 \pi ( - y^3 + 4y^2 - 3y) \, dy \\ &= \left[ 2 \pi \left( - \frac{ y^4 } { 4} + \frac{ 4 y^3 } { 3} - \frac{3y^2} { 2}\right) \right] _1^3 \\ &= \frac{ 16 \pi }{3}. \end{align}\]

If we wanted to use the disk method, we will have to take the volume generated by the "top curve" and then subtract away the volume generated by the "bottom curve." We will have to calculate the functions as follows: \( x = (y-2)^2 \), or \( \pm \sqrt{x} = y-2 \), so \( y = 2 \pm x \). So, the "top curve" is \( y = 2 + \sqrt{x} \) and the "bottom curve" is \( y = 2 - \sqrt{x} \).

Then, the volume would be equal to

\[\begin{align} \int_0^1 \pi ( 2 + \sqrt{x} ) ^2 \, dx - \int_0^1 \pi ( 2 - \sqrt{x} ) ^2 \, dx

&= \int_0^1 \pi 8 \sqrt{x} \, dx \\ &= \left[ \frac{ 16 \pi } { 3} x ^ { \frac{3}{2} } \right]_0^1 \\ &= \frac{16}{3}. \ _\square \end{align}\]

Using the shell method to find the volume of revolution will allow you to approach problems like this:

## Intermediate Examples

The shell method can also be applied to compute volumes of revolution around other axes of rotation.

## Use the shell method to find the volume generated by revolving the region bounded by \(y = \sqrt{x-1}\), \(y=0\), and \(x=10\) about the line \(y=5\).

Since the region is revolved about the line \(y=5\), we consider cylindrical shells with center axis line \(y=5\). Then the cylindrical shell radius is \(2\pi (5-y) \), the cylindrical shell height is \( 10-x \), and the cylindrical shell volume is

\[ \Delta V = 2\pi (5 -y)(10-x) \Delta y = 2 \pi (5-y)(10-y^2 - 1) \Delta y.\]

Since the endpoints are given by \(y=0\) and \(y = \sqrt{10-1} = 3\), we have

\[ \begin{align} V &= \int_0^3 2 \pi (5-y)(10-y^2 - 1)\, dy \\ &= \int_0^3 2 \pi (5-y)(9-y^2 )\, dy \\ &= \int_0^3 2\pi (45 - 9y - 5y^2+ y^3)\, dy \\ &= 2\pi \left[ 45y - \frac{9}{2} y^2 - \frac{5}{3}y^3 + \frac{y^4}{4} \right]^3_0\\ &= 2\pi \left( 45 \cdot 3 - \frac{9}{2} \cdot 9 - \frac{5}{3} \cdot 27 + \frac{81}{4} \right)\\ & = \frac{279\pi}{2}. \ _\square \end{align}\]

**in kg**)?

This problem is part of Calvin's set Fun In Multiple Dimensions.

Hint: We are integrating along the radius. What is the corresponding area element?