Washer Method – Rotate around a vertical line

Find the volume obtained by rotating the region bounded by $$y=\frac{1}{4} x^2,$$ $$x=2,$$ and $$y=0$$ about the y-axis.

To solve this problem, I’m going to use the same 4 step process as I did in my disk method lesson and my first washer method practice problem. There is one key difference this time around: here we are rotating the region around a vertical line. Previously, I have only shown examples of rotating around a horizontal line.

1. Graph the 2-D functions

As I did in the other examples mentioned above, the first thing we should always do is graph the given functions. This will help us visualize what we’re dealing with and will make it easier to come up with the function we’ll need to integrate later.

All I will do here is plug these functions into Desmos, but see if you can graph these without the help of a calculator. That a skill that may come in handy at some point.

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

This is another step that is mostly helpful for visualization. Visualizing each step required to create the 3-D figure we’re looking for will make things a lot easier when we come up with the function that we need to integrate.

Remember, the problem said that we will need to rotate the region trapped between these three functions around the y-axis. So imagine this 2-D region rotating off the page (or screen) and around the y-axis. Doing this would create a round 3-D figure. This is the figure whose volume we need to find.

I encourage you to imagine this happening on your page and try drawing a rough sketch of the resulting figure. I will do this using Wolfram Alpha.

3. Setting up the integral

This step is at the heart of these problems. All of the graphing and sketching is to help us visualize what is being described so we can correctly formulate our integral.

We could solve this problem using the cylinder method as well, but that’s for another lesson. For this example, we will proceed using the washer method. This is important to distinguish here because we need to imagine all of the washers that make up this 3-D figure. What we need to think about is a stack of very, very, very thin washers stacked one on top of the other, in the same shape as the figure shown a couple paragraphs ago.

You will notice that if we imagine this figure as a stack of washers, the washers would be stacked vertically, one on top of the other. This is different from the first washer method example I did, where the washers were all side by side.

This is an important difference because adding up the volume of all of these washers will require us to move vertically throughout this figure to get the next washer and add its volume to the total. As a result of this, we will be integrating with respect to y! Since we move in the y direction to get to the next washer, we need to integrate with respect to y. Therefore, when we create our integral, it will all need to be in terms of y rather than x.

How do we set the integral up with respect to y?

Take a look at the drawing below. You can see one of these infinitely thin washers drawn in the figure. Let’s take a minute to consider the dimensions of this particular washer. Remember, as we showed in the first washer method practice problem, the volume of a washer is given by $$V=\pi h(R^2-r^2)$$ where r is the inner radius and R is the outer radius.

The large washer in the middle of our graph is there to help you visualize where these washers would be if we were to stack them up to create this figure. Take a look at the smaller washer in the upper left section of our graph. This will be used to help us find the inner and outer radii of the washers.

The inner radius of a washer will be the distance between the center of the washer and the inner edge. In the drawing above, this is shown in the smaller washer off to the side as the distance between the points labeled $$(0, \ y)$$ and $$(x, \ y)$$.

$$(0, \ y)$$ is some point on the y-axis. The y will be different depending on which washer we’re looking at, but since it lies on the y-axis we know that the x-coordinate will always be 0.

$$(x, \ y)$$ is some point that lies on the function $$y=\frac{1}{4}x^2$$. But remember, I said earlier that we need to integrate with respect to y because our washers are stacked vertically so we move in the y direction to add up all of their volumes. Therefore, we need everything just in terms of y without having any x‘s around. So we need to rewrite $$(x, \ y)$$ just in terms of y. In order to do this we will need to think about how we can write x in terms of y.

Since we know that this point lies on the function $$y=\frac{1}{4}x^2$$ we can use this relationship to find x in terms of y. All we need to do is take that equation and solve for x. $$y=\frac{1}{4}x^2$$ $$4y=x^2$$ $$\pm \sqrt{4y}=x$$ Notice, in general when we take the square root of both sides of the equation we need the positive and negative square root. In this case, the positive square root is the right half of the parabola and the negative square root represents the left half. Since the right half of the parabola is the part that formed the region we’re looking at, we only need the positive square root. So, $$x=\sqrt{4y}.$$

Now that we know $$x=\sqrt{4y}$$ for any $$(x, \ y)$$ pair that lies on our function, we can use this to say that $$(x, \ y)$$ can instead be written as $$(\sqrt{4y}, \ y)$$. Therefore, to find the inner radius we need to find the distance between $$\mathbf{(0, \ y)}$$ and $$\mathbf{(\sqrt{4y}, \ y)}$$. To find this distance we simply need to find the difference between their x-values because they will always have the same y-coordinate. So, $$r= \sqrt{4y} \ – 0$$ $$r= \sqrt{4y}.$$

Finding the outer radius will be very similar to finding the inner radius. The only difference is that we now need to find the distance between the point in the center of the washer and the outer edge. This is shown in the labeled washer by the distance between the two points labeled $$(0, \ y)$$ and $$(2, \ y)$$.

We know that any point that lies on the line $$x=2$$ will have an x-coordinate of 2. No matter what the y-coordinate is, if it lies on $$\mathbf{x=2}$$ we know the x-coordinate must be 2. This is the reason why the point on the outer edge of the washer is labeled $$(2, \ y)$$. Although the y-coordinate changes as we move up the side of our figure, the x-coordinate stays equal to 2.

So we need to find the distance between the points $$(0, \ y)$$ and $$(2, \ y)$$. Clearly these two points will have the same y-coordinate. The y-coordinate changes depending on which washer we are looking at in our figure, but these two points will have the same y-coordinate when they are on the same washer. Since they have the same y-value, to find the distance between them, we just need to find the distance between their x-coordinates. Therefore, $$R=2-0$$ $$R=2.$$

Finding the height (or thickness)

In order to find the volume of a washer we will also need it’s height.

The height of our infinitely thin 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, having a different height.

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

Back to the integral

Like I said before, all the integral will do is go through all the y values that our figure covers 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 y. 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 y and give the integral a range of y 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 $$r=\sqrt{4y},$$ $$R=2,$$ $$h=dy.$$

Putting all of this into an integral along with the fact that this figure goes across all y-values between 0 and 1, give us $$V= \int_0^1 \pi (dy) \bigg( (2)^2 – \Big(\sqrt{4y}\Big)^2 \bigg).$$

Of course, this looks a little strange. Let’s simplify this integral and rearrange the pieces a bit. $$V= \pi \int_0^1 4 – 4y \ dy$$

4. Solve the integral

Now we’ve gotten through the hard part. All we need to do now is evaluate the volume integral by finding the anti-derivative and evaluating the bounds. All we need in this case is the power rule for integration. $$V= \pi \int_0^1 4 – 4y \ dy$$ $$V= \pi \bigg[ 4y – 2y^2 \bigg]_0^1$$ $$V= \pi \bigg[ \Big( 4(1) – 2(1)^2 \Big) – \Big( 4(0) – 2(0)^2 \Big) \bigg]$$ $$V=\pi(4-2)$$ $$V=2 \pi$$

So the volume of this solid is $$2 \pi$$ cubic units! I hope that helps, but if you are still looking for some practice with the washer method go check out my first washer method problem. That one explains the rational behind some of the steps in a bit more detail. You should also check out my other lessons and problems about integrals.

If you still have any questions, comments, or suggestions I’d love to hear them. Email me at jakesmathlessons@gmail.com or us the form below to submit your name and email and I’ll send you my calculus 1 study guide as a free gift!