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Using Green's Theorem to solve a line integral of a vector field. Created by Sal Khan.
Video transcript
Let's see if we can use our knowledge of Green's theorem to solve some actual line integrals. And actually, before I show an example, I want to make one clarification on Green's theorem. All of the examples that I did is I had a region like this, and the inside of the region was to the left of what we traversed. So all my examples I went counterclockwise and so our region was to the left of-- if you imagined walking along the path in that direction, it was always to our left. And that's the situation which Green's theorem would apply. So if you were to take a line integral along this path, a closed line integral, maybe we could even specify it like that. You'll see that in some textbooks-- along the curve c of f dot dr. This is what equals the double integral over this region r of the partial of q with respect to x minus the partial of p with respect to y d area. And just as reminder, now this q and p are coming from the components of f in the situation. f would be written as f of xy is equal to P of xy time the i component plus Q of xy times the j component. So this is a situation where the inside of the region is to the left of the direction that we're taking the path. If it was in the reverse, then we would put a minus sign right here. If this arrow went the other way, we'd put a minus sign and we can do that because we know that when we're taking line integrals through vector fields, if we were reverse the direction it becomes the negative of that. We showed that, I think, 4 or 5 videos ago. With that said, it was convenient to write Green's theorem up here. Let's actually solve a problem. Let's say I have the line integral and let's say we're over a curve. I'll define the curve in a second. But let's say that the integral we're trying to solve is x squared minus y squared dx plus 2xy dy. And then our curve-- they're giving us the boundary. The boundary is the region. I'll do it in a different color. So the curve is boundary of the region given by all of the points x,y such that x is a greater than or equal to 0, less than or equal to 1. And then y is greater than or equal to 2x squared and less than or equal to 2x. So let's draw this region that we're dealing with right now. So let me draw my x-axis or my y-axis, sorry. My y-axis and then my x-axis right there. And let's see. x goes from 0 to 1, so if we make-- that's obviously 0. Let's say that that is x is equal to 1, so that's all the x values. And y varies, it's above 2x squared and below 2x. Normally, if you get to large enough numbers, 2x squared is larger, but if you're below 1 this is actually going to be smaller than that. So the upper boundary is 2x, so there's 1 comma 2. This is a line y is equal to 2x, so that is the line y is-- let me draw a straighter line than that. The line y is equal to 2x looks something like that. That right there is y is equal to 2x. Maybe I'll do that in yellow. And then the bottom curve right here is y is going to be greater than 2x squared. It might look something like this. And of course, the region that they're talking about is this, but we're saying that the curve is the boundary of this region and we're going to go in I counterclockwise direction. I have to specify that. So our curve, we could start at any point really, but we're going to go like that. Then get to that point and then come back down along that top curve just like that. And so this met the condition that the inside of the region is always going to be our left, so we can just do the straight up Green's theorem, we don't have to do the negative of it. And let's define our region. Let's just do our region. This integral right here is going to go-- I'll just do it the way-- y varies from y is equal to 2x squared to 2y is equal to 2x. And so maybe we'll integrate with respect to y first. And then x, I'll do the outside. The boundary of x goes from 0 to 1. So they set us up well to do an indefinite integral. Now we just have to figure out what goes over here-- Green's theorem. Our f would look like this in this situation. f is f of xy is going to be equal to x squared minus y squared i plus 2xy j. We've seen this in multiple videos. You take the dot product of this with dr, you're going to get this thing right here. So this expression right here is our P of x y. And this expression right here is our Q of xy. So inside of here we're just going to apply Green's theorem straight up. So the partial of Q with respect to x-- so take the derivative of this with respect to x. We're just going to end up with the 2y. And then from that we're going to subtract the partial of P with respect to y. So if you take the derivative of this with respect to y that becomes 0 and then, here you have-- the derivative with respect to y here is minus 2y. Just like that. And so this simplifies to 2y minus minus 2y. That's 2y plus plus 2y. I'm just subtracting a negative. Or this inside, and just to save space, this inside-- that's just 4y. I don't want to have to rewrite the boundaries. That right there is the same thing as 4y. The partial of Q with respect to y, 2y minus the partial of P with respect to y. Which is minus 2y. You subtract a negative you get a positive. You have 4y. Let's take the antiderivative of the inside with respect to y and we're going to get 2y squared. Let me do it a little bit lower. We're going to get 2y squared if you take the partial with respect to y you're going to get 4y. We're going to evaluate that from y is equal to 2x squared to y is equal to 2x. And then of course, we still have the outside integral still there. x goes from 0 to 1 dx. This thing is going to be equal to the integral from 0 to 1 and then we evaluate it first at 2x. So you put a 2x in here, 2x squared is 4x squared. 2 squared, x squared, so 4x squared times 2 is going to be 8x squared. Minus-- put this guy in there. 2x squared squared is 2x to the fourth. 4x to the fourth times 2 is 8x to the fourth. Did I do that right? 2x squared-- going to put it there for y, substitute y with it. That squared is 4 X to the fourth. Times 2 is 8x x to the fourth. Very good. All right. Now dx. Now this is just a straightforward one-dimensional integral. This is going to be equal to-- I'll just do it here. This is going to be equal to the antiderivative of 8x squared is 8/3 x to the third. And then the antiderivative of 8x to the fourth is minus 8/5 x to the fifth. Then we're going to have to evaluate that from 0 to 1, or we can put a little line there. When you put 1 in there you get-- I'll do it in a different color. We get 8/5 times 1 to the third, which is 8/5 minus 8/5 minus 8/5. And then we're going to have minus-- when you put 0 in here you're just going to get a bunch of 0's. Oh sorry, I made a mistake. It would have been a blunder. It's 8/3. 8/3 times 1 to the third minus 8/5 times 1 to the fifth, so that's minus 8/5. And then, when you subtract the 0, you evaluate 0 here. you're just going to get a bunch of 0's so you don't have to do anything else. So now we just have to subtract these two fractions. So let's get a common denominator of 15. 8/3 is the same thing if we multiply the numerator and denominator by 5. That is 40/15. And then if we multiply this numerator and denominator by 3, that's going to be 24/15. So minus 24/15 and we get it being equal to 16/15. And so using Green's theorem we were able to find the answer to this integral up here. It's equal to 16/15. Hopefully you found that useful. I'll do one more example in the next video.