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# Green's theorem proof (part 2)

## Video transcript

let's say we have the same path that we had in the last video draw my y-axis that is my x-axis let's say the path looks like this looks something like this it's the same one we had in the last video might not look exactly like it let me see what I did in the last video it looked like that in the last video but close enough let's say we're dealing with the exact same curve as the last video we could call that curve C now the last video we dealt with a with a vector field that only had vectors in the I direction let's deal with another vector field it only has vectors in the J direction or the vertical direction so let's say that Q the vector field Q of X Y so it's equal to capital Q of X Y times J and we are going to concern ourselves we're going to concern ourselves with the closed line integral around the path C of Q of Q dot d R and we've seen it already dr dr can be rewritten as DX times I plus dy times J so if we were to take the dot product of these two this this line integral is going to be the exact same thing this is going to be the same thing as the closed line integral of over C of Q dot d R well Q only has a J component so if you take its zero I so 0 times DX that's just zero and then you're going to have Q X Y times dy they had no I component so this is just going to be Q switch to the same color again Q of X + y times dy that's the dot product there was no eye component that's why we don't we lose the DX now let's see if there's any way that we can solve this line integral without having to resort to a third parameter T just like we did in the last video it's actually we'll do it it'll be almost identical we're just dealing with Y's now instead of X's so we can do is we could say well what's our minimum Y and our maximum Y so our minimum Y that's it's right here it's a minimum why let's call that a let's say our maximum Y that we attained is right over there it's call that B oh and just like in the last I forgot to tell you the direction of the curve but this is the same path as last times we're going in a counterclockwise direction this is the exact same curve exact same path exact same path so we're going in that direction now in the last video we broke it up into two functions of X is 2 y is a function of X now we want to deal with Y let's break it up into two functions of Y so if we break this path into two paths let's go that's our those are kind of our extreme points let's call this path right here let's call that path right there let's call that Y let's call that X of so here along this path X is equal to I could just write I could write this path too or I'll call it c2 we could say it's X is equal to x2 of Y that's that path and then the first path all right doesn't have to be the first path depending where you start you can start anywhere let's say this one in magenta we'll call that path one and we could say that that's defined as X is equal to x1 of Y it's a little confusing when you have X as a function of Y but it's really completely analogous to what we did in the last video we're literally just swapping X's and why we're now expressing X as a function of Y instead of Y as a function of X so we have these two curves you can imagine just flipping it and we're doing the exact same thing that we did in the last video is just now in terms of Y but if you look at it this way this line integral can be rewritten this could be re-written as being equal to the integral the integral well let's just do c2 first let's just do c2 first this is the integral from B to a we start at B and we go to a right this is we're coming back from a high Y to a low Y the integral from B to a of Q of of leaving that gray color of queue of instead of having an X there we know along this curve right here X is equal to we want everything in terms of Y so here x is equal to x2 of Y so Q of X 2 of Y x2 of Y comma Y maybe I'm using too many colors here but I think you get the idea D Y D Y so this is the part of the line integral just over this left-hand curve and then we're going to add to that we're going to add to that the line integral the line integral or really just the regular integral now from Y is equal to a 2 y is equal to B y is equal to a 2 y is equal to V of Q of instead of X being equal to X 2 now X is equal to X 1 of Y it's equal to this curve this other function so X 1 of Y x1 of Y comma Y dy and then we can do exactly what we did in the previous video instead of we don't like the larger number on the bottom so let's swap these two around so if you swap these two if you make this into an A and this into a B that makes it the negative of the integral when you swap the two change the direction this is exactly what we did in the last video so hopefully it's nothing too fancy but now that we have the same boundaries of integration these two definite integrals we can just write them as one definite integral so this is going to be equal to the integral from A to B and I'll write this one first since it's positive of all right in this one Q of X 1 of Y comma Y minus this one right we have the minus sign here minus Q of X 2 of Y and Y dy let me do that in that neutral color dy that's multiplied by all of these things I distribute it out the dy I think you get the idea this is identical to what we did in the last video and then could be really rewritten as this is equal to the integral from A to B of and inside of the integral we're evaluating the function Q of X Y from the boundaries the upper boundary where the upper boundary is going to be from X is equal to x1 of Y and the lower boundary is X is equal to x2 of Y right where all the X's we substitute it with that and then we get some expression and then from that we subtract this with X substituted as x2 of Y that's exactly what we did and just like I said in the last video we're going to kind of the reverse direction that we normally go in definite integrals we normally get to this and then the next step is we get this but now we're going in the reverse direction but it's all the same difference and all of that times dy all of that times dy and just like we saw in the last video this let me do it in orange this expression right here actually let me draw that dy a little further out so it doesn't get all congested let me do that dy out here this expression this entire expression is the same exact thing that entire expression is the exact same thing as the integral the integral from x is equal to x is equal to I can just write it here let me write in the same colors x2 of y2 x1 of y2 x1 of Y of of the partial other than the I'll stay in that orange color of the partial of Q with respect to X DX this I want make it very clear this is at least in my mind the first part you might have first a little confusing but if you look just saw an integral like this and this is the inside of a double integral and it is the outside is what we saw there it's the integral from A to B D Y but if you just saw this in a double integral what you would do is you would take the antiderivative of this the antiderivative of this with respect to X the antiderivative of the partial of Q with respect to X with respect to X is going to be just Q of XY and then you since it's a definite integral you would evaluate it at x1 of Y and then subtract from that this function evaluated x2 of Y which is exactly what we did so hopefully you appreciate that and then we got our result which is very similar to the last result what is what is this double integral represent what does it represent it represents well anything if you have any double integral that goes from if you imagine this is some function let me draw it in three dimensions this is really almost a review of what we did in the last video if that's the y-axis that's our x axis that's our z axis this is some function of of x and y so it's some surface you could imagine on the XY plane it's some surface so we could call that a partial of Q with respect to X and what this double integral is this is a this is essentially defining a region and you can kind of view this DX times dy it's kind of a small differential of area so the region under question the boundary points are from Y going from Y goes from at the bottom it goes from X to of Y which we saw was a curve that looks something like this that's the lower Y and over here if we draw it in two dimensions this was the lower Y curve the upper Y curve is X 1 of Y so the upper Y curve looks something like that the upper Y curve goes something like that so X varies from the lower Y curve to the upper Y curve right that's what we're doing right here and then Y varies from A to B and so this is essentially saying let's take the double integral over this region over this region right here of this function of this function so it's essentially the volume if this is the ceiling and this boundary is essentially the wall it's the volume of that room and I don't know what it would look like when it comes up here but you can kind of imagine something like that it would be the volume of that so that's what we're taking this is the identical result we got in the last video and this a pretty neat thing so all of a sudden this vector that and Q of XY I didn't draw it out like I did the last time this vet Q of XY only has things in the J direction so it only has if I were to draw it's a vector field it only has the vectors only go up and down the vectors only go up and down like that they have no horizontal component to them but what we saw when you start with a vector field like this you take the line integral around this closed loop I'll rewrite it like right here you take this line integral around this closed loop of Q dot dr which is equal to the integral around the closed loop of Q of X Y dy we just figured out that that's equivalent to the double integral the double integral over the region over the region this is the region right that's exactly what we're doing over here if I just give you the region you have to define it you say well X is going from this is going from this function that function and Y is going from A to B and you might want to review the double integral videos if that confuses you so we're taking the double integral over the region of the partial of Q with respect to X D well you could write DX dy or you can even write a little da write the differential of area right that's we could imagine as a da which is the same thing as a DX dy and if we combine that with the last video and this is kind of the neat bringing it all together apart the result of the last video was this that if I had a function that's defined completely in terms of X we had this right here we had that result actually let me copy and paste both of these to a nice clean part of my whiteboard and then we can we can do the exciting conclusion so let me copy and paste that so that's what we got in the last video and then in this video in this video we got this result I like to just copy and paste that part right there you might already predict where this is going and then let me paste it over here this is the result from this video now let's think about arbitrary vector field that is defined as let's say I have an arbitrary vector field I'll do it in pink let's say F is a vector field defined over the XY plane and F is equal to P of XY I plus Q of X Y J you can almost imagine F being the addition of our vector fields P and Q that we did in the last two videos Q was this video and we did P in the video before that but this is really in any arbitrary vector field and let's say we wanted to take the vector field or sorry the line integral of this vector field along some path it could be the same one we've done which has been very arbitrary one so it's really any arbitrary path so let me draw some arbitrary path over here so let's say that is my arbitrary path my arbitrary curve let's say it goes in that clockwise counterclockwise direction just like that and I'm interested in what the line integral the closed line integral around that path of f dot d R is and we've seen it multiple times D R is equal to DX times I plus dy times J so this line integral can be re-written as this is equal to the line integral around the Patsey F dot d R that's going to be this term times DX so that's P of X Y times DX Plus this term Q of X Y times dy and this whole thing this is you could put this as the same thing as you could essentially this is the same thing as the line integral of P of X Y DX plus the line integral of Q of X Y dy now what are these things this is what we figured out in the first video this is what we figured out just now in this video this thing right here is the exact same thing as that over there so this is going to be equal to this is going to be equal to the double integral over the region over this region right here of the - partial of P with respect to Y the - partial of P with respect to Y instead of a dydx we could say just you know over the differential of area and then + this one this result Q this thing right here is exactly what we just proved is exactly what we just showed in this video so that's + I'll leave it up there to be able do it in the yellow + the double integral over the same region of the partial of Q with respect to X da where that's just dy DX or DX dy you can switch the order it's the differential of area and now we can add these two integrals what do we get so this is equal to and this is kind of our big grand conclusion maybe magenta is called for here the double integral of over the region of I'll write this one first says it's positive that one's negative of the partial of Q with respect to X minus the partial of P with respect to Y D the differential of area this is our big takeaway this is our big takeaway let me write here over the close a line integral of the closed loop of f dot d R is equal to the double integral of this expression and it's something just to remember we're taking the the the function that was associated with the X component or the I component we're taking the partial with respect to Y and the function that was associated with the Y component we're taking the partial with respect to X and that first one we're taking the negative of that's a good way to remember it but this result right here this is maybe I should write it in green this is greens this is greens theorem greens theorem and it's a neat way to relate a line integral of a vector field that has these partial derivatives assuming has these partial derivatives to the region to a double integral of the region now and this is a little bit of a side note we've seen several videos before if f is conservative we've learned that if F is conservative conservative which means it's the gradient of some function that it's path independent that the closed integral around any path is equal to zero and that's still true so that tells us that if F is conservative this thing right here must be equal to zero that's the only way that you're always going to enforce that this whole integral is going to be equal is going to be equal to zero over any any any region I mean you could think of situations where they cancel each other out but really over any region that's the only way that this is going to be true that these two things are going to be equal to zero so then you could say partial of Q with respect to X minus the partial of P with respect to Y has to be equal to zero or these two things have to equal each other or this kind of a corollary to greens theorem kind of a low-hanging fruit you could have figured out the partial of Q with respect to X is equal to the partial of P with respect to Y and when you study exact equations in differential equation you'll see this a lot more and actually well I won't go into too much but conservative fields you're actually the differential form of what you see in the line integral is actually if it's conservative would be an exact equation but we are going to go into that too much but hopefully you might see the parallels if you've already run into exact equations in differential equations but this is the big takeaway and let's maybe do some examples using this takeaway in the next video