Hovedinnhold

Skallmetoden med to funksjoner av x

Video transkripsjon

This right here is a solid of revolution whose volume we were able to figure out in previous videos, actually in a different tutorial, using the disk method and integrating in terms of y. What we're going to do right now is we're going to find the same volume for the same solid of revolution, but we're going to do it using the shell method and integrating with respect to x. So what we do is we have the region between these two curves, y is equal to square root of x and y is equal to x squared. And we're going to rotate it around the vertical line x equals 2. And we're going to do that at the interval that we're going to rotate this space between these two curves is the interval when square root of x is greater than x squared. And so it's between 0 and 1. And so let's try to do it with the shell method. And so to do that, what we do is we want to construct a shell. Let me do this in a different color. Let's imagine a rectangle right over here. It has width dx. and its height is the difference of these two functions. And so if I were to draw it right over here, it would look something like this. It'd be there, and then it is a shell, it's kind of a hollowed-out cylinder. So it would look something like this. Just like that. And it has some depth, that's what the dx gives us. So we have the depth that looks something like that. And then let me shade it in a little bit, just so we can see a little bit of its depth. So when you rotate this rectangle around the line x equals 2, you get a shell like this. So let's think about how we can figure out the volume of this shell. Well, we've already done this several times. The first thing we might want to think about is the circumference of the top of the shell. We know circumference is 2 pi times radius. We just need to know what the radius of the shell is. What is that distance going to be? Well, it's the horizontal distance between x equals 2 and whatever the x value is right over here. So it's going to be 2 minus our x value. So this radius, this distance right over here, is going to be 2 minus x. And so the circumference is going to be that times 2 pi. 2 pi r gives us the circumference of that circle. So 2 pi times 2 minus x. And then if we want the surface area of the outside of our shell, so the area is going to be the circumference 2 pi times 2 minus x times the height of each shell. Now, what is the height of each shell? It's going to be the vertical distance expressed as functions of y. So it's going to be the top boundary is y is equal to square root of x, the bottom boundary is y is equal to x squared. So it's going to be square root of x minus x squared. Let me do this in the yellow. So it's going to be square root of x minus x squared. And so if you want the volume of a given shell-- I'll write all this in white-- it's going to be 2 pi times 2 minus x times square root of x minus x squared. So this whole expression, I just rewrote it, is the area, the outside surface area, of one of these shells. If we want the volume, we have to get a little bit of depth, multiply by how deep the shell is, so times dx. And if we want the volume of this whole thing, we just have to solve all the shells for all of the x's in this interval and take the limit as the dx's get smaller and smaller and we have more and more shells. And so, what's our interval? Well our x's are going to go between 0 and 1. So that right over there is the volume of this figure.