## Washer Method Practice Problem

Once you have the disk method down, the next step would be to find the volume of a solid using the washer method. The washer method for finding the volume of a solid is very similar to the disk method with one small added complexity.

You can think of the main difference between these two methods being that the washer method deals with a solid with a piece of it taken out. Exactly as you would expect from the name, a washer is just a disk with a hole taken out of its center. So let’s jump into an example and I’ll explain the difference as we go.

## Example

Find the volume obtained by rotating the area bounded by $$y=x$$ and $$y= \sqrt{x}$$ about the line $$y=1$$.

## Solution

Like I said, finding the volume of this solid is going to be very similar to finding the volume of a solid using the disk method. Therefore, I’m going to follow the same general process as I did when using the disk method.

### 1. Graph the 2-D functions

Just like before, I’ll do this using Wolfram Alpha. Below you can see the graph of $$y=x$$ and $$y=\sqrt{x}$$ and you can see the area that is bounded by these two functions.

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

Now that we have graphed our functions in the 2-D space, we will need to rotate them about our axis of rotation to create a 3-D figure. This is very similar to rotating a single function when using the disk method. The only difference is that the resulting figure will be round figure with something missing out of its center.

Remember, we need to rotate this area between the two functions around the line $$y=1$$, which is above our functions this time. Again, with Wolfram Alpha, we can see what this figure would look like.

### 3. Setting up the integral

This is where things get a little different as a result of using the washer method. We still want to think about taking infinitely thin cylinders from this figure and measuring their volumes.

But let’s think about what these infinitely thin cylinders look like.

Since our figure is a cone-like shape with a piece missing from the middle, our cylinders are also going to have a piece missing from the middle. This makes them look like washers, which is why we call it the washer method! Since the slices of our figure look like washers we can’t find their volume by just finding the volume of a cylinder. There will be something a little extra.

#### What’s different about a washer?

Let’s take a look at the washer below and consider how we might find its volume.

Clearly this figure takes on the shape of a cylinder. If we knew its radius and its height, we could use the formula for the volume of a cylinder to find its volume. However, there is a piece missing.

If we think about the piece that is missing from the cylinder for a moment you can see that this hole also has a cylindrical shape. In other words, we are starting with a cylinder and taking out a smaller cylinder from the middle of it.

So let’s say the distance from the center of these cylinders to the outer edge of this figure is the large radius, R. And we will say that the distance from the center of this figure to the inner edge is the small radius, r. We can think of r as the radius of the small cylinder that was taken out of the large cylinder.

Clearly both of these cylinders would have the same height, which we will call h, because this washer has the same thickness everywhere. So we can label all of these, giving us something like this.

Then we could use the volume of a cylinder formula to say that the volume of the large cylinder before the middle is taken out is $$\pi R^2 h.$$ And we can also say that the volume of the middle piece that’s taken out is $$\pi r^2 h.$$ So if we started with the volume of the large cylinder and take away the volume of the small cylinder, that means the volume of the washer is $$V= \pi R^2 h \ – \ \pi r^2 h.$$

Then we can simplify this a bit by factoring out like terms to get $$V = \pi h \big( R^2 – r^2 \big)$$

#### But what does this have to do with an integral?

Just like we did when we used the disk method, we will need to add up the volume of all of the washers. This is what the integral accomplishes. In order to write our volume as an integral, we first need to come up with the function we will integrate.

Consider the following drawing of the area described in this problem rotated around $$y=1$$ with one of the washer slices depicted.

The green washer in the figure represents one of the infinitely thin washers that we are slicing the figure into. Since we are trying to make this integral represent the sum of all of these washers, we need to think about the volume of each washer in particular.

You can imagine if we were to look at a different washer to the left or right of this one, the inner and outer radius would be different. As we change the x value where the washer sits, both radii would need to change. The inner radius would always reach to the part of this figure resulting from rotating $$y=\sqrt{x}$$ around $$y=1$$. And the outer radius would always reach out to the part of this figure resulting from rotating $$y=x$$ around $$y=1$$.

We already figured out that the volume of a washer would be $$V = \pi h \big( R^2 – r^2 \big).$$ But now we need to apply this to this specific washer in the drawing above.

Let’s look up at the green washer in the above drawing. In order to find its volume, we will need its inner radius, outer radius, and height (or thickness).

The inner radius of this washer would be the distance from the center of the washer to the inner edge. In the drawing above, we can see that this would be the distance between the axis of rotation (the green line labeled $$y=1$$) and the the function which is closer to this axis (the red function labeled $$y=\sqrt{x}$$).

Off to the side of these functions you can see a closer look at this washer with three important points labeled. In this larger version the inner radius is the distance between the point $$(x, \ 1)$$ and $$(x, \ y_1)$$. We know that the center of our disk will always have a y-coordinate of 1 because we rotated our function around the line $$\mathbf{y=1}$$. And we will leave the x-coordinate as the variable x because we are trying to find the volume of any washer along this figure with all different x values, not just the one disk drawn above.

Let’s take a second to think about the other point I mentioned. This point is labeled $$(x, \ y_1)$$. This is just meant to be any (x, y) combination that sits on our function $$\mathbf{y=\sqrt{x}}$$. This point will always have the same x-value as our other labeled point, so the distance between these two points will simply be the distance between their y-coordinates, which are $$y_1$$ and 1. To find the distance between these two values, we just need to do the larger value minus the smaller one. Looking at the drawing, you can see that $$y=1$$ is above $$y=\sqrt{x}$$ over the entire domain of the area we care about. Therefore, the larger y value minus the smaller one, is $$1-y_1.$$

But we need this to be in terms of x. Remember we realized earlier that the inner radius of our washer will depend on x, so we want everything in terms of x. We know that our point $$(x, \ y_1)$$ is some point that lies on the function $$y=\sqrt{x}$$. So we know $$\mathbf{y_1=\sqrt{x}}$$. Therefore, if the inner radius of our washer is $$1-y_1$$, we can also say that the radius is $$1-\sqrt{x}$$. So the inner radius of this washer will be $$r=1-\sqrt{x}.$$

We could plug in any x value into this to find the inner radius of the washer that corresponds with that specific x value. So when we integrate across a range of x values, we will be taking into account the inner radius of the washers with all of the x values in that range. But let’s not get ahead of ourselves. We have a bit more work to do before we can do that.

We will find the outer radius very similarly to how we found the inner radius. The outer radius of this washer would be the distance from the center of the washer to the outer edge. In the drawing above, we can see that this would be the distance between the axis of rotation (the green line labeled $$y=1$$) and the the function which is farther from this axis (the blue function labeled $$y=x$$).

Again, let’s take a look at the washer drawn off to the side of our figure. This time we are looking for the outer radius. The outer radius would be shown here as the distance between the points $$(x, \ 1)$$ and $$(x, \ y_2)$$.

Since these two points have the same x value, the distance between them will be the same as the distance between their y values. To find this we just need to take the larger one and subtract the smaller of the two. The point $$(x, \ 1)$$ will always lie on $$y=1$$ and the point $$(x, \ y_2)$$ will always lie on $$y=x$$. Since $$y=1$$ is above $$y=x$$ between $$x=0$$ and $$x=1$$, we know that 1 will be larger than $$y_2$$. Therefore, the distance between these two points is $$1-y_2.$$

But again, we need this radius to be written in terms of x. We know that our point $$(x, \ y_2)$$ is some point that lies on the function $$y=x$$. So we know $$\mathbf{y_2=x}$$. Therefore, if the outer radius of our washer is $$1-y_2$$, we can also say that the radius is $$1-x$$. So the outer radius of this washer will be $$R=1-x.$$

#### Finding the height (or thickness)

The height of our infinitely thin cylinders or washers is actually quite simple. Just like when we integrate a 2-D function to find the area under the curve, our slices here are all the same width. We don’t have to worry about each washer, or cylinder, having a different height.

The height of each cylinder will just be how far we always move over before taking another slice. Since we are moving over in the x direction as we imagine the next slice, this can simply be our change in x between the slices. Change in x is always represented as dx. So we can simply say the height of each cylinder is $$h=dx.$$

#### Back to the integral

Like I said before, all the integral will do is go through all the x values in our domain and add up the volumes of all of the infinitely thin washers. In order for it to achieve this, we need to put a function for the volume of each washer that depends on x. We already know that the volume of a washer in general would be $$V = \pi h \big( R^2 – r^2 \big).$$

This means that our integral might look something like this $$\int \pi h \big( R^2 – r^2 \big).$$

But this doesn’t really have any meaning on its own. In order to give this meaning we need to represent this volume in terms of x and give the integral a domain of x values to integrate over.

Remember we also found the inner radius, outer radius, and height of the washers that make up our figure to be $$r=1-\sqrt{x},$$ $$R=1-x,$$ $$h=dx.$$

Putting all these into our integral, along with the fact that our figure takes on all x values between $$x=0$$ and $$x=1$$ tells us that the volume of this 3-D figure is $$V = \int_0^1 \pi (dx) \Big( \big( 1-x \big)^2 – \big( 1-\sqrt{x} \big)^2 \Big).$$

But we should rewrite this in a form that is more in line with how integrals are usually formatted. And we can also pull our constant $$\pi$$ outside of the integral. $$V = \pi \int_0^1 \big( 1-x \big)^2 – \big( 1-\sqrt{x} \big)^2 \ dx$$

### 4. Solve the integral

We made it through the hard part! Now that we created our integral to represent the volume, we just need to evaluate the integral. Before we integrate this, let’s start with simplifying it.

In order to simplify this function that we need to integrate, the first step would be to F.O.I.L. out each portion. Remember $$(1-x)^2$$ is NOT $$(1^2-x^2)$$. Instead we need to treat $$(1-x)^2$$ as if it were $$(1-x)(1-x)$$.

$$V = \pi \int_0^1 \big( 1-x \big)^2 – \big( 1-\sqrt{x} \big)^2 \ dx$$ $$V = \pi \int_0^1 (1-x)(1-x) – \big(1-\sqrt{x})(1-\sqrt{x} \big) \ dx$$ $$V = \pi \int_0^1 \big(1-2x+x^2 \big) – \big(1-2\sqrt{x}+x \big) \ dx$$ $$V = \pi \int_0^1 1-2x+x^2 – 1+2\sqrt{x}-x \ dx$$ $$V = \pi \int_0^1 x^2 -3x +2\sqrt{x} \ dx$$ $$V = \pi \int_0^1 x^2 -3x +2x^{\frac{1}{2}} \ dx$$

Now we have a function that is fairly simple to integrate. All we need to do to integrate this is use the power rule.

$$V = \pi \Bigg[ \frac{1}{3}x^3 – \frac{3}{2}x^2 + \frac{4}{3}x^{\frac{3}{2}} \Bigg]_0^1$$ $$V = \pi \Bigg[ \bigg( \frac{1}{3}1^3 – \frac{3}{2}1^2 + \frac{4}{3}1^{\frac{3}{2}}\bigg) – \bigg( \frac{1}{3}0^3 – \frac{3}{2}0^2 + \frac{4}{3}0^{\frac{3}{2}} \bigg) \Bigg]$$ $$V = \pi \bigg( \frac{1}{3} – \frac{3}{2} + \frac{4}{3} \bigg)$$ $$V = \pi \bigg( \frac{2}{6} – \frac{9}{6} + \frac{8}{6} \bigg)$$ $$V = \frac{1}{6} \pi = \frac{\pi}{6}$$

So we know the volume of this solid is $$\frac{\pi}{6}$$! Hopefully this has helped you with the washer method, but if there’s still a topic you’d like to learn about take a look at some of my other lessons and problem solutions about integrals. You can also get some more practice with the washer method here. If you can’t find the topic or question you’re looking for just let me know by emailing me at jakesmathlessons@gmail.com!

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