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Double integrals 3

Let's integrate dy first! Created by Sal Khan.

Video transcript

In the last a video we figured out the volume between this surface, which was xy squared and the xy-plane when x went from 0 to 2 and y went from 0 to 1. And the way we did it is we integrated with respect to x first. We said, pick a y, and let's just figure out the area under the curve. And so we integrated with respect to x first, and then we integrated with respect to y. But we could have done it the other way around. So let's do that and just make sure we got the right answer. So let me erase a lot of this. So remember, our answer was 2/3 when we integrated with respect to x first, and then with respect to y. But I will show you that we can integrate the other way around. That's good when you can get the same answer in two different ways. So let me redraw that graph because I want to give you the intuition again. So that's my x-axis, y-axis, z-axis. x, y, z. Then this is my xy-plane. down here. y goes from 0 to 1; x goes from 0 to 2, This is x equals 1, this is equals 2, this is y equals 1, and then the graph-- I will do my best to draw it. Looks something-- let me get some contrast going here. So the graph looks something like this. Let me see if I can draw it. On this side it looks something like that and then it comes down like that, straight. And then, the volume we care about. It's actually this volume underneath the graph. This is the top of the surface on that side. We care about this volume underneath the surface. And then when we draw the bottom of the surface-- let me do it in a darker color-- looks something like this. This is the bottom underneath the surface. I can even shade it a little bit just to show you that it's the underneath. Hopefully that's a decent rendering of it. Let's look back at what we had before. It's like a page that I just flipped up at this point, and we care about this volume, kind of the colored area under there. So let's figure out how to do it. Last time we integrated with respect to x first. Let's integrate with respect to y first. So let's hold x constant. So if we hold x constant what we could do is for a given x-- let's pick an x. So if we pick a given x, let's pick the x here. Then what we can do, for a given x, you can view that function of x and y. If x is a constant, let's say if x is 1 then z is just equal to y squared. That's easy to figure out the area under because we can see that x isn't the constant, but we can treat it as a constant. So for example, for any given x, we would have a curve like this. What we could do is we could try to figure out the area of this curve first. So how do we do that? Well, we just said, we could kind of view this function up here as z is equal to xy squared because that's exactly what it is. But we're holding x constant. We're treating it like a constant. To figure out that area we could take a dy, a change in y, multiply it by the height, which is xy squared. So we take xy squared, multiply it by dy, and then if we want this entire area we integrate it from y is equal to 0 to y is equal to 1. Fair enough. Now once we have that area, if you want the volume underneath this entire surface what we could do is we can multiply this area times dx and get some depth going. Let me pick a nice color, that's green. So that's our dx. So if we multiply that times dx we would get some depth. Let me do a darker color, get some contrast. Sometimes I feel like that guy who paints on PBS. So now we have the volume of this, you kind of view it-- the area under the curve times a dx, so we have some depth here. So it's time dx. And if we want to figure out the entire volume under this surface-- between the surface and the xy-plane given this constraint to our domain-- we just integrate from x is equal to 0 to 2. All right, so let's think about it. This area in green here that we started with, that should be a function of x. We held x constant, but depending on which x you pick this area is going to change. So when we evaluate this magenta inner integral with respect to y we should get a function of x. And then when you evaluate the whole thing we'll get our volumes. So let's do it. Let's evaluate this inner integral. Hold x constant. What's the antiderivative of y squared? It's y of the third over 3. The x is a constant, right? We're going to evaluate that at 1 and at 0. The outer integral is still with respect to x dx. This is equal to-- let's see. When you evaluate y is equal to 1 you get 1 to the third. That's 1. So it's x/3 minus when y is 0 then that whole thing just becomes 0. This purple expression is just x/3. And then we still have the outside integral from 0 to 2 dx. So given what x we have, the area of this green surface-- that was where we started. Given any given x, that area-- I wanted something with some contrast. This area is x/3 depending on which x you pick. If x is 1, this area right here is 1/3. But now we're going to integrate underneath the entire surface and get our volume. And like I said, when you integrate it, it's a function of x. So let's do that. And this is just plain old vanilla, standard integral. So what's the antiderivative of x? It's x squared over 2. We have a 1/3 there so it equals x squared over 2 times 3. So x squared over 6. And we're going to evaluate it at 2 and at 0. 2 squared over 6 is 4/6. Minus 0/6, which is equal to 0. Equals 4/6. What is 4/6? Well, that's just the same thing as 2/3. So the volume under the surface is 2/3, and if you watched the previous video you will appreciate the fact that when we integrated the other way around, when we did it with respect to x first and then y, we got the exact same answer. So the universe is in proper working order. And I've surprisingly, actually finished this video with extra time. So for fun, we can just spin this graph and just appreciate the fact that we have figured out the volume between this surface, xy squared and the xy-plane. Pretty neat. Anyway, I'll see you in the next video.