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Shell method for rotating around vertical line

Introducing the shell method for rotation around a vertical line. Created by Sal Khan.

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Video transcript

I've got the function y is equal to x minus 3 squared times x minus 1. And what I want to do is think about rotating the part of this function that sits right over here between x is equal to 1 and x equals 3. And x equals 3 and x equals 1 are clearly the zeroes of this function right over here. And I want to take this region and rotate it around the y-axis. And if I did that, I'd get a shape that looks something like that. And I want to figure out the volume of that shape. And what we're going to do is a new method called the shell method. And the reason we're going to use the shell method-- you might say, hey, in the past, we've rotated things around a vertical line before. We used the disk method. We wrote everything as a function of y, et cetera, et cetera. We created all of these disks. We figured out the volume of each of those disks. But the problem here is this is hard to express as a function of y. How do you solve explicitly for y right over here? So instead, we're going to keep things in terms of x and have a different geometric visualization for how we can come up with the volume. What we're going to imagine instead-- instead of constructing disks, we're going to construct shells. And what do I mean by a shell? So for each x at the interval, on this kind of cut of it, we can construct a rectangle. And what happens if we were to rotate this rectangle? So this is the rectangle right over here. What happens if we rotate this rectangle around the y-axis along with everything else? I'll try my best attempt to draw it. It's going to look something like this. This is challenging my art skills, but I think I can handle it. So it's going to look something not too dissimilar to that right over there. So it looks like a hollowed-out cylinder. I guess that's why we call it a shell. And it's going to have some depth. The depth is going to be dx. And the height right over here is going to be the value of my function. The height is f of x. In this case, f of x is x minus 3 squared times x minus 1. How do we figure out the volume of a cylinder like this? Well, if we can figure out the circumference of the cylinder, and then multiply that circumference times the height of the cylinder, we'd essentially figure out the area of the outside surface of our cylinder. And then if we multiply the area of the outside surface of our cylinder by that infinitesimally small depth, then that'll give us the volume-- I shouldn't say cylinder-- of our shell. So let's try to do it. What is the circumference of a shell? What is the circumference of one shell going to be? Well, it's going to be 2 pi times the radius of that shell. We need to express this as a function of x. And so what is that going to be? It's going to be 2 pi. So for a given x, what is the radius? Well, the radius right over here is just the horizontal distance between the y-axis and that x. So that's just x. So the circumference, in this case, is just going to be 2 pi times x. Now, what is the height going to be for any one of those shells? The height is going to be f of x. This is right over here. And so what is going to be the surface area of the outside? So let me put this in quotes-- "outside" surface area. I'm not worried about the depth right now, the dx. I'm not worried about this top part and the bottom part. I'm just worried about the outside surface area. Well, the outside surface area is just going to be the circumference times the height. It's going to be 2 pi x times f of x. And in this situation, in the situation we're looking at right over here, that's going to be 2 pi x times x minus 3 squared times x minus 1. Now, what's going to be the volume? So the volume of the shell-- shell volume-- is just going to be all this business times dx. So it's going to be 2 pi x times f of x dx. And so now we're ready to integrate over the interval. So the volume of our entire shape is going to be the definite integral. We're going to integrate over all the x's in the interval, from x is equal to 1 to x is equal to 3 of this thing. And we could take the 2 pi out front. So we'll put 2 pi out front. And on the inside, we have x times f of x, which in our situation, is this business. So it's going to be x times x minus 3 squared times x minus 1. And then, of course, we have our dx. So there you have it. Using the shell method, we have set up our definite integral for the volume of this strange-looking shape right over there.