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Disk method for rotation around x-axis

Finding the volume of a solid of revolution that is defined between two functions. Created by Sal Khan.
Video transcript
Let's do some more volumes of solids of revolutions. So let's say that I have the graph y is equal to square root of x. So let's do it, so it looks something like this. So that right over there is y is equal to the square root of x. And let's say I also have the graph of y equals x. So let's say y equals x looks something like this. It looks just like that. y equals x. And what I care about now is the solid I get if I were to rotate the area between these two things around to the x-axis. So let's try to visualize it. So the outside is going to be kind of a truffle shape, it's going to look like a truffle shape. And then we hollow out a cone inside of it. So I'll make my best attempt to draw this shape. So it's going to look something like-- so the outside is going to look something like this. It's going to look something like that. And we care about the interval. We care about the interval between the points that they intersect. So between this point and this point here. So the outside is going to look something like this. So this is the base of the truffle. That is the base of the truffle, it's going to have this kind of truffle shape on the outside. But I guess maybe we're on some type of a diet, we don't want to eat to the entire truffle. So we carve out a cone on the inside. So the inside of it is essentially hollow except for this kind of shell part. So we carve out a cone in the center. So we rotate it around to the x-axis. Truffle on the outside, carved out a cone on the inside. So what's going to be the volume of that thing? So it's essentially, if we take a slice of our figure, it's going to be, this is going to be the wall. And we're essentially going to take the volume of this entire wall that we're rotating around the x-axis. So how do we do that? Well, it might dawn on you that if we found the volume of the truffle if it was not carved out, and then subtract from that the volume of the cone, we would essentially find out the volume of the space in between the outside of the truffle and the cone part of the truffle. So how would we do that? Well, so to find the volume of the outer shape. So let me draw it over here. So actually let me just draw it over here. If we think about the volume of the outer shape, once again, we can use the disk method. So at any given point in time our radius for one of our disks is going to be equal to the function. Let's rotate that disk around. Actually let me do it in a different color, it's hard to see that disk because it's in the same magenta. So this is a radius, and let's rotate it around. So I'm rotating the disk around. This is our face of the disk, that's our face of the disk. It's going to have a depth of dx. We've seen this multiple times. It's going to have a depth of dx. So the volume of this disk is going to be our depth, dx, times the area of the face. The area of the face is going to be pi times the radius squared. The radius is going to be equal to the value of the outer function, in this case the square root of x. So it's going to be pi times our radius squared, which is pi times the square root of x squared. And so if we want to find the volume of the entire outer thimble, or truffle, or whatever we want to call it. Before we even carve out the center, we just take a sum of a bunch of these, a bunch of these disks that we've created. So that's one disk, we would have another disk over here. Another disk over here. For each x we have another disk. And as we go, as x's get larger and larger, the disks have a larger and larger radius. So we're going to sum up all of those disks. And we're going to take the limit as each of those disks get infinitely thin, and we have an infinite number of them. But we have to figure out our boundaries of integration. So what are our boundaries of integration? What are the two points right over here where they intersect? Well we could just set these two things to be equal to each other. If you just said x is equal to square root of x, when does x equal the square root of x? I mean you can square both sides of this. You could say when does x squared equal x? You could, well we could keep it there. You could kind of solve this, there's multiple ways you could do it. But you could solve this kind of just thinking about it. If x is equal to 0, x squared is equal to x. And you see that on the graph right over here. x is equal to 0. And also 1 squared is equal to 1. 1 is equal to the square root of 1. You could have done other things. You could say OK, x squared minus x is equal to 0. You could factor out an x. You get x times x minus 1 is equal to 0. And so either one of these could be equal to 0, so x is equal to 0 or x minus 1 is equal to 0. And then you get x equals 0 or x is equal to 1. Which gives us our boundaries of integration. So this goes from x equals 0 to x equals 1. And so for the outside of our shape we can now figure out the volume. But we're not done. We also need to figure out the volume of the inside of our shape that we're going to take out. So we're going to subtract out that volume. So we're going to subtract out of volume. Our x values, once again, are going between 0 and 1. And so let's think about those disks. So let's think about, let's construct a disk on the inside, right over here. So if I construct a disk on the inside. So now I'm carving out the cone part of it. What is the area of the face of one of those disks? Well it's going to be pi times the radius squared. In this case, the radius is going to be equal to the value of this inner function, which is just x. And so this is just y is equal to x. And then we're going to multiply it times the depth, times the depth of each of these disks. And each of these disks are going to have a depth of dx. If you imagine a quarter that has an infinitely-thin depth right over here. So it's going to be dx. And so the volume our, kind of our truffle with a cone carved out, is going to be this integral minus this integral right over here. And we could evaluate it just like that. Or we could even say, OK we could factor out a pi out of both of them. There are actually, there's multiple ways that we could write it. But let's just evaluate it like this, and then I'll generalize it in the next video. So this is going to be equal to the definite integral from 0 to 1. You take the pi outside. Square root of x squared is going to be x dx minus the integral, we can factor the pi out. >From 0 to 1 of x squared dx. And we could say this is going to be equal to pi times the antiderivative of x, which is just x squared over 2 evaluated from 0 to 1, minus pi, times the antiderivative of x squared, which is x to the third over 3 evaluated from 0 to 1. This expression is equal to-- and I'm going to arbitrarily switch colors just because the green's getting monotonous-- pi times 1 squared over 2 minus 0 squared over 2. I could write squared. 1 squared over 2 minus 0 squared over 2, minus pi times 1 to the third over 3 minus 0 to the third over 3. And so we get, this is equal to-- let me do it in that same blue color-- so this is this simplified. This is just 0 right over here. This is 1 squared over 2, which is just 1/2. So it's just pi over 2, 1/2 times pi minus-- well this is just 0, this is 1/3, minus pi over 3. And then to simplify this, it's just really subtracting fractions. So we can find a common denominator. Common denominator is 6. This is going to be 3 pi over 6. This is 3 pi over 6 minus 2 pi over 6. pi over 3 is 2 pi over 6, pi over 2 is 3 pi over 6. And we end up with, we end up with 3 of something minus 2 of something, you end up with 1 of something. We end up with 1 pi over 6. And we are done. We were able to find the volume of that wacky kind of gutted-out truffle.