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## Multivariable calculus

### Course: Multivariable calculus>Unit 4

Lesson 1: Line integrals for scalar functions

# Line integral example 2 (part 2)

Part 2 of an example of taking a line integral over a closed path. Created by Sal Khan.

## Video transcript

In the last video, we set out to figure out the surface area of the walls of this weird-looking building, where the ceiling of the walls was defined by the function f of xy is equal to x plus y squared, and then the base of this building, or the contour of its walls, was defined by the path where we have a circle of radius 2 along here, then we go down along the y-axis, and then we take another left, and we go along the x-axis, and that was our building. And in the last video, we figured out this first wall's surface area. In fact, you can think of it, our original problem is, we wanted to figure out the line integral along the closed path, so it was a closed line integral, along the closed path c of f of xy, and we're always multiplying f of xy times a little bit, a little, small distance of our path, ds. We're writing this in the most abstract way possible. And what we saw in the last video is, the easiest way to do this is to break this up into multiple paths, or into multiple problems. So you can imagine, this whole contour, this whole path we call c, but we could call this part, we figured out in the last video, c1. This part we can call, let me make a pointer, c2, and this point right here is c3. So we could redefine, or we can break up, this line integral, this closed-line integral, into 3 non-closed line integrals. This will be equal to the line integral along the path c1 of f of xy ds, plus the line integral along c2 of f of x y ds plus the line integral, you might have guessed it, along c3 of f of xy ds, and in the last video, we got as far as figuring out this first part, this first curvy wall all right here. Its surface area, we figured out, was 4 plus 2 pi. Now we've got to figure out the other 2 parts. So let's do C2, let's do this line integral next. And in order to do it, we need to do another parameterization of x and y. It's going to be different than what we did for this part. We're no longer along this circle, we're just along the y-axis. So as long as we're there, x is definitely going to be equal to 0. So that's my parameterization, x is equal to 0. If we're along the y-axis, x is definitely equal to 0. And then y, we could say it starts off at y is equal to 2. Maybe we'll say y is equal to 2 minus t, for t is between 0, t is greater than or equal to 0, less than or equal to 2. And that should work. When t is equal to 0, we're at this point right there, and then as t increases towards 2, we move down the y-axis, until eventually when t is equal to 2, we're at that point right there. So that's our parameterization. And so let's evaluate this line, and we could do our derivatives, too, if we like. What's the derivative, I'll write it over here. What's dx dt? Pretty straightforward. Derivative of 0 is 0, and dy dt is equal to the derivative of this. It's just minus 1, right? 2 minus t, derivative of minus t, is just minus 1. And so let's just break it up. So we have this thing right here, so we have the integral along c2. But let's, instead of writing c2, I'll leave c2 there, but we'll say were going from t is equal to 0 to 2 of f of xy. f of xy is this thing right here, is x plus y squared, and then times ds. And we know from the last several videos, ds can be rewritten as the square root of dx dt squared, so 0 squared, plus dy dt squared, so minus 1 squared is 1, all of that times dt. And obviously, this is pretty nice and clean. This is 0 plus 1, square root, this just becomes 1. And then what is x? x, if we write it in terms of our parameterization, is always going to be equal to 0, and then y squared is going to be 2 minus is t squared. So this is going to be 2 minus t squared. So this whole crazy thing simplified to, we're going to go from t is equal to 0 to t is equal to 2, the x disappears in our parameterization, x just stays 0, regardless of what t is, and then you have y squared, but y is the same thing as 2 minus t, so 2 minus t squared, and then you have your dt sitting out there. This is pretty straightforward. I always find it easier when you're finding an antiderivative of this, although you can do this in your head, I like to just actually multiply out this binomial. So this is going to be equal to the antiderivative from t is equal to 0 to t is equal to 2 of 4 minus 2 minus 4t plus t squared, plus t squared, just like that dt. And this is pretty straightforward. This is going to be, the antiderivative of this is 4 t minus 2 t squared, right? When you take the derivative, there's 2 times minus 2 is minus 4 t, and then you have plus 1/3 t to the third, right? These are just simple antiderivatives, and we need to evaluate it from 0 to 2. And so let's evaluate it at 2. 4 times 2 is 8, let me pick a new color. 4 times 2 is 8, minus 2 times 2 squared, so 2 times 4, so minus 8, plus 1/3 times 2 to the third power. So 1/3 times 8. So these cancel out. We have 8 minus 8, and we just have 8/3. So this just becomes 8/3. And then we have to put a 0 in, minus 0 evaluate here, but it's just going to be 0. We have 4 times 0, two times 0, all of these are going to be 0. So minus 0. So just like that, we found our surface area of our second wall. This turned out being, this right here is 8/3. And now we have our last wall, and then we can just add them up. So we have our last wall. I'll do another parameterization. I want to have the graph there. Well, maybe I can paste it again. Edit. So there's the graph again. And now we're going to do our last wall. So our last wall is this one right here, which is, we could write it, you know, this was c3. Let me switch colors here. So this is c, we're going to go along contour c3 of f of xy ds, which is the same thing as, let's do a parameterization. Along this curve, if we just say, x is equal to t, very straight forward, for t is greater than or equal to 0, less than or equal to 2, and this whole time that we're along the x-axis, y is going to be equal to 0. That's pretty straightforward parameterization. So this is going to be equal to, we're going to go from t is equal to 0 to t is equal to 2 of f of xy, which is, I'll write in terms of x right now, x and y, x plus y squared times ds. Now, what is dx-- well, let me write ds right here. Times ds. That's what we're dealing with. Now we know what ds is. ds is equal to the square root of dx dt squared plus dy dt squared times dt. We proved that in the first video. Or we didn't rigorously prove it, but we got the sense of why this is true. And what's the derivative of x with respect to t? Well, that's just 1, so this is just going to be a 1, 1 squared, same thing. And the derivative of y with respect to z is 0. So this is is 0, 1 plus 0 is 1, square root of 1 is 1. So this thing just becomes dt. ds is going to be equal to dt, in this case. So this just becomes a dt. And then our x is going to be equal to a t, that's part of our definition of our parameterization, and y is zero, so we can ignore it. So this was a super-simple integral. So this simplified down to, we're going to go from 0 to 2 of t dt, which is equal to the antiderivative of t is just 1/2 t squared, and we're going to go 0 to 2, which is equal to 1/2 times 2 squared. 2 squared is 4, times 1/2 is 2, and then minus 1/2 times 0 squared, minus 0. So this third wall's area right there is just 2. Pretty straightforward. So that right there, the area there, is just 2. And so to answer our question, what was this line integral evaluated over this closed path of f of xy? Well, we just add up these numbers. We have 4 plus 2 pi plus 8/3 plus 2, well, what is this. 8/3 is same thing as 2 and 2/3, so we have 4 plus 2 and 2/3 is 6 and 2/3, plus another 2 is 8 and 2/3, so this whole thing becomes 8 and 2/3, if we write it as a mixed number, plus 2 pi. And we're done! And we're done. Now we can start trying to do line integrals with vector-valued functions.