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Current time:0:00Total duration:10:31

Video transcript

let's see if we can use our knowledge of greens theorem to solve some actual line integrals and actually before I show an example I want to make one clarification on greens theorem all of the examples that I did is I had a region like this 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 I went counterclockwise and so our region was to the left of if you imagine walking along the path in that direction it was always to our left and that's the situation which greens theorem would apply so if you were to take a line integral along this path the 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 d R 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 a reminder and this Q and P are coming from the components of F F in this situation F would be written as f of X Y is equal to P of X Y times the I component plus Q of X Y 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 if this error 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 reverse the direction it becomes the negative of that we saw it showed that I think four or five videos ago with that said it was convenient to write greens theorem up here let's actually solve a problem let's say I have the line in a row let's add 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 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 is the boundary of so the curve curve is boundary of the region given by all of the X all of the points X Y such that X is greater than or equal to 0 less than or equal to 1 and then Y is greater than or equal to 2 x 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 my y-axis sorry my y-axis and then my x-axis right there and let's the 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 it's that's all of the X values and Y varies it's above 2x squared and below 2x normally if you get too large enough numbers 2 x 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 and then this bottom curve 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 a counterclockwise direction I have to specify that counter counter clock wise direction so our curve we could start at any point really but we're gonna go in we're gonna go like that and then get to that point and then come back down along that curved 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 greens theorem we don't have to do the negative of it and let's define our region so our region let's just do our region if I the this integral right here it's going to go I'll just do it the way Y goes from what let me do it y varies from y is equal to 2x squared - y is equal to 2y is equal to 2x and so maybe we'll do the will integrate with respect to Y first and then X I'll do the outside X the boundary of X goes from 0 to 1 so they set it up set us up well to do an indefinite integral and now we just have to figure out what goes over here greens theorem this is the this right here is our F it would look like this in this situation F is f of X Y is going to be equal to x squared x squared minus y squared I plus 2 XY 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 XY that is our P of XY and this expression right here is our Q of XY Q of XY so inside of here we're just going to apply greens theorem straight up so the partial of Q with respect to X the partial of Q with respect to X so take the derivative of this with respect to X we're just gonna end up with the 2y we're gonna with a 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 a the derivative with respect to Y here is minus 2y alright minus 2y so you have any minus 2y just like that and so this simplifies to 2y minus minus 2i that's 2 I plus plus 2i I'm just subtracting a negative or this inside and just to save space this inside let me that's just for y so let me just I don't want to have to rewrite the boundaries oh that right there is the same thing as for Y the partial of Q with respect to y 2y minus the partial of P with respect to Y which was minus 2y you subtract a negative and I'm positive you have 4y so 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 right if you take the derivative of this with respect the partial with respect to Y you're going to get 4y and we're going to evaluate that we're going to evaluate that from Y is equal to from Y is equal to 2x squared 2 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 it 2x in here 2x squared is 4x squared right 2 squared x squared so 4x squared times 2 it's going to be 8x squared 8x squared - put this guy in there 2x squared squared is 4x to the fourth 4 X to the fourth times 2 is 8 X to the fourth 8 X to the 4 that do that right 2 x squared I'm going to put it there 4y substitute Y with it that squared is 4x to the fourth times 2 is 8 X to the fourth very good all right now DX now this is just a straightforward 1 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 3rd 8/3 X to the 3rd and then the anti-derivative the 8x to the fourth is minus 8/5 X to the fifth 8/5 X to the fifth and we're gonna have to evaluate that from 0 to 1 or give a little line there sometimes 0 to 1 when you put one in there you get under in a different color we get 8 fifths times one of the third was just eight fifths minus eight fifths minus 8/5 and then we're going to have minus when you put zero and you're just gonna get a bunch of zeros oh sorry I made a mistake it's not a been a blunder it's 8/3 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 zero so then minus you evaluate zero here just gonna get a bunch of zero 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 denominator by 5 that is 40 15 and then if we multiply this numerator denominator by 3 that's going to be 24 over 15 so minus 24 over 15 and we get it being equal to 16 right 16 over 15 and so using greens theorem we were able to 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