Similar to using the disk or washer method, we will use the cylinder method to find the volume of a solid. Specifically, it’s used when we rotate a function or region around an axis of rotation. **In fact, most problems that require finding the volume of a solid of rotation can use the disk/washer method or the cylinder method.** However, one will usually be significantly easier.

I’ll explain what I mean by this with an example.

## Example 1

Find the volume of the solid resulting from rotating the area bound between \(y=x^2-2x+2\), \(y=0\), \(x=1\), and \(x=2\) about the *y-axis*.

Although we are using a different method here we will follow the same 4 step process as I did with the disk method and washer method.

### 1. Graph the 2-D functions

This is generally a good idea with any type of problem: draw out whatever is being described in the problem. This helps to visualize whats going on in the problem and what exactly we are trying to measure. It can also help us decide if the answer we end up with is actually a reasonable one.

I would recommend trying to graph all of the functions listed by hand, but I’ll do this using Desmos. You can see the graph of the functions below with the bounded region shaded gray.

### 2. Rotate the 2-D area around the given axis

Again, we want to visualize the 3-D figure whose volume we are trying to find. To do this we want to imagine rotating the described area around the axis of rotation.

I like to imagine the area actually coming off the page and rotating around our axis of rotation, which is the y-axis in this case. Doing this would create a cylinder-like, round figure. Try sketching this rotation and the resulting figure. It may be helpful in the next step to have this sketch.

If needed, Wolfram Alpha can always be used to create a visualization of this figure.

### 3. Setting up the integral

This is the part where things start to get a bit different using the cylinder method than they were with the disk/washer method.

In order to make sense of the integral we need to set up here, let’s think about what we’re doing differently. With the disk/washer methods we were stacking many very thin disks on top of each other with the same thickness and varying radii to create our figure. This created a stack of cylinders whose volume we could find and add together.

The cylinder shell method is a bit different. Here we need to imagine just the outer shell of a cylinder that is very very very thin. We will stack many of these very thin shells inside of each other to create our figure. Each shell will have the same thickness, but all with different heights depending on where it is in the figure.

What we need to figure out is a formula for the volume of one of these shells, then the integral will be able to go through each shell and add up all of their volumes.

#### What would this look like?

Before we think about creating a function that we will need to integrate I want to take a moment to describe what one of these shells would look like. Imagine forming the outer shell of a cylinder with a piece of paper. All you would need to do is roll up the paper and it would create a cylinder shell. But what’s interesting about this is the cylinder shell came from a rectangle, or more accurately a very very thin rectangular prism. It might look something like this.

Since we are imagining finding the volume of an infinitely thin cylinder, these three figures would have the same volume. The fact that they’re infinitely thin means that the curvature won’t impact the volume of the shell. So to find the volume of the cylindrical shell, we can instead find the volume of a rectangular prism with the same dimensions. This is a much easier exercise to imagine since the volume of a rectangular prism is simply $$V= \ length \cdot width \cdot height.$$

#### But how does this relate to the cylinder?

What we need to think about now is how the dimensions of the rectangular prism translate to dimensions on the cylindrical shell. Looking back up to the drawing above, if you imagine the rectangle curling into a cylinder you can see **the long side of the rectangle** bends into the top and bottom of the cylinder. This would be the **circumference of the cylinder.**

In the above drawing we can also see that the **shorter side of the rectangle** lines up nicely with the **height of the cylinder.**

And lastly, the very very very thin **thickness of the shell and the rectangular prism would clearly correspond with each other.**

So as a result, the $$V= \ length \cdot width \cdot height$$ of the rectangular prism would be the same as $$V = circumference \cdot thickness \cdot height$$ for the cylindrical shell.

#### Putting it into an integral

Now that we know how to find the volume of each shell we need to come up with a way to put that into an integral. The reason for this is that the integral adds up the volume of all of the shells that make up the figure to find the total volume.

In order to do this, we will need to think about the formula for the volume in terms of measurments of the cylinders. We know the three pieces we need to find the volume of one of the shells are the **circumference, thickness, and height** of the cylinders. Typically when we describe a cylinder, we need two measurements to do this: **height and radius**. **So we want to represent the circumference, thickness, and height in terms of height and radius.**

First let’s think about the circumference. We know that the circumference of a circle is always going to be $$circumference=2 \pi r$$ where *r* is the radius.

The thickness of each shell is a bit strange. As we go from one shell to the next throughout the integral we want to think about what is changing. When we are doing this, we will always want to think about **integrating throughout the radius**. It’s as if we are starting at the center of the figure and integrating in the direction toward the edge of the figure. So when we go from one shell to the next **we travel throughout the radius**. Therefore, the thickness of each cylindrical shell with be the distance we travel between each shell. This will just be the **change in radius** between each shell. Therefore we’ll say the the thickness is $$thickness = dr.$$

Lastly, the height will still be described as the height. We don’t really need to do anything here.

Therefore, using these three conversions we know that the volume of the whole figure can be found with the following integral. $$\int 2 \pi r h \ dr$$

#### Relating it back to our figure

So we have a general outline to set up our integral, but now we need to figure out how our figure fits into these pieces. The easiest way to do this is draw it out with one of the cylindrical shells that makes up our 3-D figure.

So remember earlier I said that when we use this method to find the volume, we are integrating **in the direction of the radius of the cylinders**. We can see in our drawing that if we start in the center of our cylinder and move toward the edge, we would be **going in the x direction**. Therefore, we need to think about how to represent our integral

**in terms of**

*x**so we can integrate with respect to*

*x*.

So we have 3 pieces of our integral that we need to put in terms of *x*: *r*, *h*, and *dr*.

#### Finding *r*

You can see in the drawing above I drew an example of one of the cylindrical shells within our figure. There is also a smaller version of this shell in the upper left hand corner which has a few points labeled which will lie on the various functions that created our bounded region.

We can see the radius of our cylinder would be the distance between its center and edge, which is the distance between the two points labeled \((0, \ y)\) and \((x, \ y)\). The x-coordinate of that first point will always be *0 *because that point lies on the y-axis. And \((x, y)\) is some point that lies on the function \(y=x^2-2x+2\). Since these points have the same *y *value, the distance between them will just be the distance between their *x *values. So $$r=x-0$$ $$r=x.$$

#### Finding *h*

Looking at the labeled cylindrical shell in the drawing above, we can see that the height of the cylinder will be the distance between the points labeled \((x, \ y)\) and \((x, \ 0)\). Again, \((x, \ y)\) is some point on the function \(y=x^2-2x+2\). And the y-coordinate of that second point will always be *0 *because it sits on the x-axis.

It is also important to notice that **these two points will sit on the top and bottom edges of every single shell that makes up this figure.** This will be true for the inner most shell, the outer most shell, and every shell between them.

Since these these two points have the same *x *value, the distance between them will simply be the distance between their *y *values. So $$h=y-0$$ $$h=y$$

But remember we need everything in terms of *x*, not *y*. So we need to think about how to represent this height using *x *instead. Since our *y *is just the y-coordinate of some point that lies on the function \(y=x^2-2x+2\), we can replace the *y *with \(x^2-2x+2\). So $$h=x^2-2x+2.$$

#### Finding *dr*

This is actually the simplest part to find. The *dr *represents the change in the cylinder’s radius as we go from each shell to the next. Since we move in the same direction of the radius as we integrate to find our volume, the change in *r *should be the same as the change in *x *between each step. Therefore, we can say that $$dr=dx.$$

#### Putting it all back into an integral

We know that the volume of our figure can be found by using the integral $$\int 2 \pi r h \ dr.$$ And we just found how to represent all of these pieces in terms of x by $$r=x$$ $$h=x^2-2x+2$$ $$dr=dx.$$ So we can substitute all of these pieces into the integral and get something in terms of *x *that will tell us exactly how to find the volume of our figure. $$V=\int 2 \pi x \big( x^2-2x+2 \big) \ dx$$

But there is actually one more thing we need to consider. Our integral needs bounds. Since we are integrating with respect to *x* we need to figure out all the *x* values we want to consider when finding our volume.

To do this we just need to look at the original 2-D region we had bounded by all of our functions. Looking back at our drawings we can see that the entire region goes from \(x=1\) to \(x=2\). Therefore, these will be our bounds, telling us that $$V=\int_1^2 2 \pi x \big( x^2-2x+2 \big) \ dx.$$

### 4. Solve the integral

Now that we got our integral set up, all we need to go is evaluate the integral and find the volume it represents.

Before doing that I will simplify things a bit by pulling out the constant \(2 \pi\) and then simplifying the function by distributing.

$$V=\int_1^2 2 \pi x \big( x^2-2x+2 \big) \ dx$$ $$V=2 \pi \int_1^2 x^3-2x^2+2x \ dx$$ $$V=2 \pi \Bigg[ \frac{1}{4}x^4 – \frac{2}{3}x^3+ x^2 \Bigg]_1^2$$ $$V=2 \pi \Bigg[ \bigg( \frac{1}{4}(2)^4 – \frac{2}{3}(2)^3+ (2)^2 \bigg) \ – \ \bigg( \frac{1}{4}(1)^4 – \frac{2}{3}(1)^3+ (1)^2 \bigg) \Bigg]$$ $$V=2 \pi \Bigg[ \bigg( 4 – \frac{16}{3}+ 4 \bigg) \ – \ \bigg( \frac{1}{4} – \frac{2}{3} + 1 \bigg) \Bigg]$$ $$V=2 \pi \bigg[ \frac{8}{3} – \frac{7}{12} \bigg]$$ $$V=2 \pi \bigg[ \frac{25}{12} \bigg]$$ $$V= \frac{25 \pi}{6}$$

And that’s it! The volume of our 3-D figure is \(\frac{25 \pi}{6}\) cubic units.

I do quickly want to circle back to a comment I made a while ago. We could have found this volume using the washer method. However, we would have had to split it into two separate integrals to do so. The reason for this is that the inner radius of the washers on the lower half of the figure is formed by \(x=1\). But the inner radius of the washers on the upper half of our figure is formed by \(y=x^2-2x+2\). So we’d have to set up one integral for the lower half and another for the upper half, then add the resulting volumes to find the total volume of our figure.

Clearly using the cylindrical shell method is much easier in this case.

**If you want more practice on finding volumes of rotation using the shell method, ****you can find another example here****.**

Hopefully all of this helps you gain a bit of a better understanding of this method, but as always I’d love to hear your questions if you have any. Just email me at jakesmathlessons@gmail.com and I’ll see if I can help provide a bit more clarification. ** You can also use the form below to subscribe to my email list and I’ll send you my FREE calc 1 study guide!** Just put in your name and email address and I’ll be sure to let you know when I post new content! Feel free to go check out my other lessons and solutions about integrals as well.