Hello! In Lesson 23 (link here) we discussed how to find volumes of solids of revolution by integrating areas of cross sections which were disks or washers. Today we’ll discuss how to find volumes of solids of revolution by integrating areas of cylindrical “cross sections.”

Lesson 24: Volumes of Revolution: Cylindrical Shells

#### Learning goals:

- Determine the volume of a solid of revolution using the method of cylindrical shells.

**Topic**:

- Volume 2, Section 2.3 Volumes of Revolution: Cylindrical Shells (link to textbook section)

**WeBWorK**:

- Applications – Volumes of Revolution

**Motivating question**

How can we use integration to find volumes of more irregular 3-dimensional shapes?

#### Warmup exercise 1

Evaluate the integral

\[ \int_0^1 2\pi x^3\]

#### Show answer 1

\[ \int_0^1 2\pi x^3 = \left. 2 \pi \frac{x^4}{4} \right|_0^1 = 2 \pi \left(\frac{1^4}{4} – \frac{0^4}{4} \right) = \2 \pi \frac{1}{4} = \frac{\pi}{2} \]

## Calculating volumes of revolution using cylindrical shells

#### Rotating vertical segments around a horizontal axis: the disk and washer methods

We saw in Lesson 23 (link here) that if $A(x)$ is the area of the cross section of a solid at position $x$. If $a$ is the lowest value of $x$ and $b$ is the highest value of $x$, then the volume of the solid is given by

\[V = \int_a^b A(x) dx.\]

In the case that the three-dimensional shape in question is a solid of revolution, then the disk method gives $A(x) = \pi (f(x))^2$ and the washer method gives $A(x) = \pi ((g(x))^2 – (f(x))^2)$. The disk and the washer were swept out by rotating a vertical segment (infinitely thin rectangle) at position $x$ around the horizontal axis of rotation. Analogous formulas work when the axis of rotation is vertical and we are integrating with respect to the variable $y$.

#### Rotating vertical segments around a vertical axis: the method of cylindrical shells

What if we watch what happens to the same vertical segment when we rotate around a vertical axis instead of a horizontal axis? First we’ll rotate around the $y$-axis.

Now, instead of a flat shape like a disk or a washer, we get a shape that lives in three-dimensional space: a cylindrical shell. This cylindrical shell is hollow and it has no top or bottom; you can make a model of it by taking a piece of paper and taping the two sides of it together to get a tube. The center of the tube is the axis of rotation.

If we call the area of this tube $A(x)$, then the formula for volume is still

\[V = \int_a^b A(x) dx\]

but now $a$ represents the lowest value of $x$ in the region being rotated and $b$ represents the highest value of $x$ in this region. So to get a formula we can use to determine the volume, we need to figure out what $A(x)$ is.

First, we can figure out what shape this “cross section” is. The “area” of a tube is really the area of the flat rectangle we get when we cut it open to flatten it (think about the piece of paper again). The area of the rectangle is the base times its height, but keep in mind that the base of the rectangle is equal to the circumference of the tube and $C = 2 \pi r$. So we need to figure out the radius and height of the tube given by the vertical segment at position $x$.

If the vertical segment runs from the $x$ axis to the graph $y=f(x)$, then the height of the tube is just $f(x)$. When we say that the vertical segment is “at position $x$” we just mean that its distance from the $y$-axis is just $x$. Which means that the radius of the tube given by this vertical segment is just $x$!

Since $A = 2 \pi r \cdot h$, we have that $A(x) = 2 \pi x f(x)$. This gives us the following formula for the volume of a solid of revolution calculated in terms of cylindrical shells.

**Method of cylindrical shells**

Assume that $f(x) \geq 0$ and that $0 \leq a < b$. Consider the region bounded above by the graph $y=f(x)$, below by the $x$-axis, on the left by $x=a$, and on the right by $x=b$. Rotate this region around the $y$-axis to obtain a three-dimensional solid of revolution. The volume of this solid is given by

\[V = \int_a^b 2 \pi x f(x) dx\]

## Examples

*Video 1* explains this formula with more pictures and shows an application of this formula with rotation around the $y$-axis.

*Video 2* and *Video 3 *each show another such example.

*Video 4 *shows an example where rotation is now around the $x$-axis. Notice what is different now that we are integrating with respect to $y$ instead of $x$.

*Video 5 *starts with a quick review of the application of the method of cylindrical shells when the axis of rotation is the $x$-axis or the $y$-axis. It then shows how to modify the formula for an example where the axis of rotation is a horizontal line that is not the $x$-axis and for an example where the axis of rotation is a vertical line that is not the $y$-axis.

## Comparison of methods

When should we use the disk/washer method and when should we use the method of cylindrical shells? It’s helpful to keep track using the segments that represent infinitely thin rectangles in a Riemann sum. Usually, if the function(s) give(s) $y$ as a function of $x$, the segment will be vertical and if the function(s) give $x$ as a function of $y$, the segment will be horizontal, though this is not the case 100% of the time.

- If the
**segment is vertical**, we’ll integrate with respect to $x$.- If the axis of rotation is horizontal, the segment sweeps out a disk or washer.
- If the axis of rotation is vertical, the segment sweeps out a cylindrical shell.

- If the
**segment is horizontal**, we’ll integrate with respect to $y$.- If the axis of rotation is horizontal, the segment sweeps out a cylindrical shell.
- If the axis of rotation is vertical, the segment sweeps out a disk or washer.

Often, one method is much easier than the other and, sometimes, only one method is possible. *Video 6* shows two examples, one where the axis of rotation is horizontal and one where the axis of rotation is vertical. Each is computed in two different ways: once using the disk/washer method and once using the method of cylindrical shells.

Also see the table in Figure 2.34 in your textbook here.

#### Exit ticket

Consider the region bounded by the graph $y=1- x^2$, $y=0$, $x=0$, and $x=1$. Determine the volume of the solid obtained by rotating this region around the $y$-axis.