 Hi everyone, we're going to take a look at an example of a very typical washer method problem for finding the volume of a solid of revolution. Here we are asked to find the volume of the solid formed by revolving the region bounded by y equals x squared and y equals square root of x about the x-axis. So if you want to go ahead and graph those on your calculator, I've already done so and I did adjust my window a little bit. I think I did, each of those hash marks is one, I do believe. I know this point of intersection because we will be needing that. That is one comma one. You can always find that on your calculator if needed. So if you read the problem again, we want to find that wedge-shaped region. We're going to take that region and revolve it about the x-axis. So when you do that, imagine that it kind of comes down and this is not going to be a really perfect picture, okay? Apologize now. Alright, but imagine taking the yellow region that you see and rotating that yellow region about the x-axis. So everyone turn your head to the right. Sort of looks like tulip leaves. At least that's what it looks like to me. So in order to find the volume of the solid that's generated, this is a great example of a classic washer problem. If we think about our representative rectangle, which could be drawn anywhere in this yellow region, I'm going to put it up here. Notice that the rectangle, when we draw the rectangle, we want it to be hitting both curves. So because of that, we know the rectangle has to go vertically. Therefore, making it a DX problem, what makes it a washer method problem are two major characteristics. That representative rectangle is perpendicular to your axis of revolution, which in this case we were told was the x-axis itself, and it also is not touching it. That is in contrast to the disc method in which the representative rectangle was touching the axis of revolution. The fact that your representative rectangle is not touching the axis of revolution, that's where your hole comes from. So if you think about your basic setup for the formula, that's going to become the definite integral, our integrand for our definite integral. It's going to be pi times big R squared minus little r squared. Big R is always the distance from your axis of revolution to the far side of your rectangle. It's basically that outer circle of the washer. So in this case, big R is going to be this distance right here. Alright, notice from the axis of revolution to that far side of the rectangle. So let's go ahead and get an expression for that. So we'll use the top minus bottom idea, so on the top, notice we're hitting the blue curve. Now that's the curve for square root of x, and on the bottom we're simply hitting the x-axis. So big R is going to be square root of x. Now for little r. Little r is essentially the radius of the hole itself. So little r is this part right in here. Once again though, we can think top minus bottom. So on the top it's hitting the red curve, which is our quadratic x squared, and on the bottom it's hitting the axis of revolution, the x-axis. So we simply have x squared for that. So these are the pieces that are going to become the integrand for our definite integral. Our limits of integration, since it's a dx problem, do need to be x values. So in this case we're going from the origin at zero up to x equals one. So let's go back to the previous page and we'll get that set up. So our volume then is going to be the integral from zero to one of pi. Now remember we need big R, which was the square root of x. We need to square that, so that's big R squared. Of course that becomes simply x once you simplify it. Minus little r squared, which of course becomes x to the fourth. You are most welcome to simplify it if you'd like. This one might not be a bad idea simply because it'll be easier to type it into your calculator, but you don't necessarily have to. And you are welcome to just evaluate that in your calculator if you just do the integrand part without the pi. You would get as your answer 0.3 pi. If you multiply the pi in to get a decimal answer, a pure decimal answer, you would get approximately 0.942. If you did need to attach units of measure, of course that would be cubic units since this is a volume problem.