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# Green's theorem proof (part 1)

## Video transcript

let's say we have a path in the XY plane that's my y-axis that is my x-axis and my path will look like this let's say it looks like that just try and draw a bit of an arbitrary path and let's say we go in a counterclockwise direction like that along our path and we could call this path so I'm going to counterclockwise direction we could call that path C and let's say we also have a vector field and our vector field going to be a little unusual I call it P P of X Y it only has an eye component or all of its vectors are only multiples of the I unit vector so it's P capital P of X Y times the unit vector I there is no J component so if you have to visualize this vector field all of the vectors only go they're all multiples of the I unit vector or they could be negative multiples so they could also go in that direction but they don't go diagonal or they don't go up they all go left to right or right to left that's what this vector field would look like now what I'm interested in doing is figuring out the line integral over a closed loop the closed loop C or the closed path C of of P dot d R which just our standard kind of way of solving for a line integral and we've seen what dr is in the past dr is equal to DX times I plus dy times the J unit vector and we you might say hey isn't it DX DT times DT let me write that isn't it you know can't dr be written as DX DT times DT I plus dy DT times DT J and it could but if you imagine these differentials could cancel out and you'll just slip to the DX and dy and we've seen that multiple times and I'm going to leave it in this form because hopefully if we if we if we're careful we won't have to deal with the third parameter T so let's just look at it in this form right here with just the D X's and the dys so this integral can be re-written as the line integral the curve C X let me just do it over down here the line integral over the path or the curve C of PDR so we take the product of each of the coefficients let's say the coefficient of the I component so we get P I'll do it in green actually let me do it in that purple color so we get P of X Y times DX plus well there's no zero times J times dy zero times dy is just going to be zero so this our line integral simplified to this right here this is equal to this original integral up here so we're literally just taking the line integral around this path now I said that we if we're if we play our cards right we're not going to have to deal with the third variable T that we might be able to just solve this integral only in terms of X and so let's see if we can do that so let's look at our minimum and maximum X points that looks like our minimum x point let's call that a and let's call that our maximum x point let's call that B what we could do is we can break up this curve into two functions of XY as functions of X so this bottom one right here we could call as y1 of X right this is just a standard curve and you know we're just start dealing with a standard calculus this is just you can kamancheh this is f of X and it's a function of X and this is y 2 of X Y 2 of X just like that so you can imagine two paths one one path defined by y1 of X let me do that in a different color little magenta one path defined by one y1 of X as we go from X is equal to a to X is equal to B and then another path defined by y2 of X as we go from X is equal to B to X is equal to a that is our curve so what we could do is we could rewrite this integral which is the same thing as that integral as this is equal to the integral will first do this first path of X going from A to B of P of X and I could just say P of X and y but we know along this path Y is a function of X its y1 of X so we say X and y1 of X wherever we see a Y we substitute it with y1 of X DX so that covers that first path I'll do it in the same color so this is that is we could imagine this is c1 this is kind of the first half of our curve well it's not exactly necessarily the half but that takes us right from that point to that point and then we want to complete the circle maybe I'll do that and that is going to be I'll do that in yellow that's going to be equal to sorry we're going to have to add these two plus the integral from X is equal to be X is equal to B 2x is equal to a X is equal to B 2x is equal to a of I'll do it in that same color of P of X and now Y is going to be y2 of X wherever you see a Y you can substitute with y2 of X along this curve y2 of X DX this is already getting interesting and you might already see where I'm going with this so this is the curve C - C - I think you appreciate if you take the union of c1 and c2 we've got our whole curve so let's see we can simplify this integral a little bit well one thing we want to do we might want to make their their endpoints the same so if you swap a and B here you just turns the integral negative so you make this into a B that into an A and then make that plus sign into a minus sign and now we can rewrite this whole thing as being equal to the integral from A to B of this thing of P of X and y1 of X - this thing - P of X and y2 of X and then all of that times DX maybe I'll write it in third color all of that times DX now I'm going to do something a little bit arbitrary but I think you'll appreciate why I did this by the end of this video and it's just a very simple operation what I'm going to do is I'm going to swap these two so I'm essentially going to multiply this whole thing by negative 1 or essentially multiply and divide by negative 1 so I can multiply this by negative 1 and then multiply the outside by negative 1 and I will not have changed the integral I'm multiplying by negative 1 twice so if I swap these two things if I multiply the inside times negative 1 so this is going to be equal to we do the outside of the integral A to B if I multiply the inside I'll do a DX out here if I multiply the inside of the integral by negative 1 these two guys switch so it becomes P of X of y2 of X and then you're going to have minus minus P of X and y1 of X my handwriting is getting a little messy but I can't just multiply just the inside by minus 1 I can only I don't want to change the integral so I multiply the inside by minus 1 let me multiply the outside by minus 1 and since I multiplied by minus 1 twice these two things are equivalent or you could say this is that or this is the negative of that either way I think you appreciate that I haven't changed the integral at all numerically I multiply the inside and the outside by minus 1 and now this the next step I'm going to do it might look a little bit foreign to you but I think you'll appreciate it if you might want to I mean act well you know it might be obvious to you if you've recently done some double integrals so this thing can be re-written as minus the integral from A to B of and I'll do let me do it in a new color of the function P of X Y evaluated at Y 2 of X y2 of X minus it valuated and let me make it very clear this is y is equal to y2 of X - this function evaluated at Y is equal to y1 of X and then of course all of that times DX that's this statement and what we saw right here and this statement right here are completely identical completely identical and then if we assume that this that a partial derivative of P with of capital P with respect to Y exists you can you hopefully you'll realize and I'll focus on this a little bit because I don't want to confuse in this step let me write the outside of the integral so this is going to be equal to and this is kind of a neat outcome and we're starting to build up to a very neat outcome which we'll probably have to take the next video to do so let me do the outside DX this right here if we assume that capital P has a partial derivative this right here is the exact same thing this right here is the exact same thing as the partial derivative of P with respect to Y dy the antiderivative of that from y1 of X y1 of x2 y2 of X now I want to make you I want to make you feel comfortable that these two things are equivalent and to realize they're equivalent you probably just have to start here and then go to that we're used to seeing this we're so used to seeing a double integral like this and then the very first step we say ok to solve this double integral we start on the inside integral right there and we say ok let's take the antiderivative of this with respect to Y so if you take the antiderivative of the partial of P with respect to Y with respect to Y you're going to end up with P and since this is a definite integral the boundaries are going to be in terms of X you're going to evaluate that from Y is equal to y2 of X and you're going to subtract from that Y is equal to y1 of X I want to normally we start with something like this and we go to something like this this is kind of unusual that we started we kind of solved we started with the solution of the definite integral and then we slowly built back to the definite integral so hopefully you realize that this is true that this is just we're kind of going in the reverse direction that we normally do and if you do realize that then we've just established a pretty neat outcome because what is this right here what is this right here let me go back let me see if I can fit everything this is I have some function I have some function and I'm assuming that the partial of P with respect to Y exists but I have some function defined over the XY plane you know you can imagine if we're dealing in three dimensions now let me draw a little bit neater so that's why that's X that's Z so that's why that's X so this you can imagine is some surface it just happens to be the partial of P with respect to X so it's some surface on the XY plane like that and what are we doing we're taking the double integral under that surface right around this region the region's boundaries in terms of Y are defined by y2 and y1 of X by y2 and y1 of X so you literally have that curve that's Y 2 on top Y 1 on the bottom Y 1 on the bottom and so we're essentially taking the volume above we're taking the volume above this so if you imagine with the base is the whole the whole floor of this is going to be the area inside of this inside of this curve and then the height is going to be the function partial of P with respect to Y so the height is going to be this function partial of P with respect to Y and it's going to be a little hard for me to draw but this is essentially some type of a volume if you want to visualize it that way but the really neat outcome here is the really neat outcome here is if you call if you call this region R we've just simplified this line integral and this was a special one it only had an X component the vector field but we've just simplified this line integral to being equivalent to to being equivalent to and maybe I should write this line integral because that's what's the really neat outcome we've just established that this thing right here which is the same as our original one so let me write that the close line integral around the curve C of P of why DX we've just established that that's the same thing as the double integral as the double integral over the region R this is the region R over the region R over the region R of the partial of P with respect to Y the partial of P the partial of P with respect to Y and we could write dy DX or we could write da or D or whatever you want to write but this is the double integral over that region so the neat thing here is using a vector field it only had an X component we were able to connect its line integral to the double integral over region oh and I forgot something very important we had a negative sign out here so this was a minus sign out here so the minus or we could even put the minus in here but I think you get the general idea in the next video I'm going to do the same exact thing with a vector field that only has vectors in the Y direction and then we'll connect the two and we'll end up with greens theorem