 Right? So, sort of the culmination here in the course is, and I'm hoping I can get enough through stuff that I can use Stokes theorem in general. So, Stokes theorem is sort of like my fundamental theorem of calculus for the entire time. Right? And so, what I'm going to do today and what I'm going to do next time is generalizations of the fundamental theorem of calculus for higher dimensions. So, we have, so, we have already, and if we have, okay, why did I just forget what I'm doing? So, the fundamental theorem of calculus then says that in one dimension, in one dimension being the gradient of some G, or G is a function from R and R, then, and so, and gamma is a piecewise line integral over gamma. A gradient vector field, have some gradient vector field here, that color depends, you know, there's some curve, it goes from here, you have to do, this is the one variable, the calculus, the end of the, the culmination of calculus one. Did I screw it up? Well, okay, whatever variable you want to put here, smiley face. Oh, I see, because, yeah. Yeah, I guess I want D. Why can't we just make a D smiley face on the outside? So, let's just do the small smiley face. Yeah. That's awesome. Let me be a second. Doesn't matter. So, the derivative of anything, right, you integrate the derivative and you get back where you started. Yeah, I'm sorry about the Y. I was thinking that this was the variable, okay, so this is the standard fundamental theorem calculus that says if you integrate the derivative you might as well just evaluate the ends, and then there is this generalization to it, which works for gradient vector fields, which tells us again that we can just evaluate the ends, but not all vector fields are gradient vector fields. So that's, if all vector fields were gradient vector fields and there would be more to do and everything would be fine. So the goal of these next couple lectures is to increase this to deal with the vector fields that aren't gradient vector fields. And at the end of the last, well, I don't know, in one of the previous classes I talked about these two quantities called the divergence of a vector field. I hope I will get to this today, which measures the divergence of the vector field, which is just the sum of the partials with respect to each component of F. This is also right as grad dot or del dot F, which, if you know these words, is the trace, trace DF, and this is, so this is also the sum of the eigenvalues, which is amount, so this is all evaluated at some point. Right? This is the local expansion rate. Once again, I'm having trouble reading some faces. I read that face. Right? So we talked about this before. So this is just telling us how much F spreads stuff out or squishes it up or doesn't. I'm going to save that for a bit, but this is a useful, so I guess there's an end to it now. So for example, let me talk about this a little more. Suppose instead of having some open curve gamma, I have a closed curve gamma. So gamma starts again at the same place, and I have stuff flowing through. Well, maybe I want to know how much stuff is stretched out or not. So this is called the flux for this face. And let's take an extreme example where stuff is created in here. Then certainly you would expect there to be divergence here. Stuff spreads out. So this is a quantity that we want to know about. Okay? So this is not what I'm going to talk about yet. So we talked about this a little bit. Was it last time? Or the time before that? Or sometime? I don't know before I ate much face. So this is something we want to know. I'm going to save that a little bit. So this is related to something called Gauss's Theorem. It tells us about this, which I'm going to try to get to at the end. And another quantity that we talked about was the curl of the vector field, which instead of this quantity was, well, let me first do the plane, was this thing, this mysterious del cross f, which measures the circulation. I want to sort of put that on hold a little bit. What did I say? Okay. So I'm going to, and so in the plane, so mostly today I'm just going to talk about the functions vector field in the plane. And I want to try and maybe motivate a little bit why the curl here, which in the plane is this quantity, so f of x, y will have two components, x, y, q and x, y. Right? So I have a vector field in the x direction and a vector field in the y direction. Something that describes how f varies in both x and y independently. Yeah. And so in this case the curl, or the planar curl, I didn't want to write it up for the case, but I did. The curl of f is going to be the partial with q with respect to x minus the partial with p with respect to y. So if f is gradient, because the gradient flows, so f is the gradient of some function and it's continuously differentiable because the mixed partials would be g x y is. Right? Because that's what this is. This is the difference of the mixed partials. If we have a gradient flow, then there is no curl. But if we don't have a gradient vector field, then this measures something. And let's think about what it measures. So if I, and again, I'm going to think about closed paths. So we know that if we have a gradient vector field, and I, so I'm just going to draw a picture. I have some per gamma here and I have my vector field doing the same. Arrows. Then if I compute, so I'm going to start at some point, how about here? This would be gamma of 0 is gamma of 1. Because I start here and I go around this way. I've got too many arrows. Oh well, it doesn't matter. I start here, I go around my curve, and I come back. Then if I integrate f dot dx over gamma, well I should get 0. Because this is of gamma of 1 minus g of gamma of 0, which is just, these are the same. So if I integrate gradient vector field over some curve like this, I get nothing interesting. But if it isn't a gradient vector field, then I might get something interesting. And so remember, this measures, this measures in some sense the amount of work you want, the effect of the vector field as I cruise around this path. And so if I don't have a, if f isn't a gradient vector field, there is not a conserved quantity that I would expect this to be nonzero. And this sort of measures, so then I might close the curve of gamma. In fact, there's some notation that I guess I would start using. If you put a little circle here, this means that gamma is a closed curve. If we miss, so gamma is closed, which it can be anyway, and it's counterclockwise, and it has an orientation as it counterclockwise. I will try to remember to put the little circle on here. But if I don't, mostly we're talking about these. So these line integrals, and those line integrals are the same, but here I must have gamma of 0 equals gamma of 1, or whatever my beginning has to be. Also notice that in this case, it matters which way I go. So I might have that. In other words, go ahead and point that out. So that, so suppose, so suppose that this, I guess it doesn't matter how I write it. In the same way, so this isn't so wacky. I mean if you think about it, let's do one where it's obviously true. So let's, I'm just going to do it by picture. I can write it down because I'm probably going to screw it up. But the vector field, this vector field, so that is, the one that just corresponds to going around is y and negative x. No, x and negative y, x. Right? So let's think about, and then if I just take the circle, the unit circle, then this vector field is, so this is cosine t, sine t. This vector field is tangent here everywhere. So integrating over the unit circle, integrating over gamma of f dx, well the vector field is one. And so that's just adding up one as you go around the unit circle. So this is 2 pi without even calculating. If you want to parameterize it, take the derivative, you'll still get 2 pi. But just by thinking, I'm taking the unit vector and dragging it all the way around and asking how much did it turn, well it turned 2 pi. So nothing changed. But if I go the other way, if I take, let's call it alpha, really it should be fish. L, L, yeah. So if I take the curly L to be the other way, it's the same as cosine t, negative sine t. So that is, instead of going this way, I'm going to go around this way around the circle. Then I'm going to add a negative one as I go around the circle. Right? Because the tangent vector here is the other way from the vector field. So that integral over the upside down gamma. Is gradient and non-gradient, is that the same, same deteriorated and non- Yeah, yeah, same. So there's not, right? So this guy doesn't correspond to any, it's not possible to have rotation around the origin being the gradient of some function. Right? We did that already. And so, okay, so this is what I'm trying to say. If we go backwards around the path, then we change the sine to the other side. And this is exactly the same, the exact analogy here is that for one variable, we integrate from a to b of f of x dx. This is negative the integral from d to a. Exactly the same thing. Should I prove this? No, do you need to switch the direction that you're traveling with curve? This is the vector field. No, I mean the line integral. This is switching the direction? Yeah, but you're not reflecting that in the circle. No, you're saying that by the sentence. Okay, so instead of writing negative gamma, I'll write the upside down gamma there. That's the curve that goes the other way. Yeah, I know, but in the integral where you have the belute, you have it going counterclockwise. They always go counterclockwise. Isn't that the direction of the belute this trigger going? Isn't that the direction of the curve? So if you make your negative, yeah. So the reason that the arrow is put on there, the arrow is put on there so that if I have a curve, if my curve is not an oriented curve, adjust the words unit circle, then this means go counterclockwise. So usually, so you'll see it sometimes without an arrow and sometimes with an arrow. Let me write it without an arrow to mean it is a closed curve. Or in fact, just whatever the hell we were doing before, it's a line integral and it just happens to be a closed curve. If you go the wrong way, you change the sign. The statement I'm trying to make is if you go the wrong way, you change the sign. And that's really easy to prove because here, I mean I did this example by no calculation, but if we do it by calculation, I make the substitution r equals negative t, then my derivative is negative of my other derivative. The derivative pulls through the integral. There it is. I've changed the sign. So without writing it down, that's the proof. So in general, whenever we have a line integral and we go the wrong way, we get a negative of that. Okay. So I was still trying to motivate why we're looking at curl. I have lost track of where I am in my motivation here. So, I mean at least in this example, this is somehow measuring how much this vector field twists around. And it seems reasonable, even without thinking about anything, that this quantity is something like, and so I want to apologize up front for being very hand wavy here. If you want to read a more careful thing, I'll let people go. I want to give you some idea of what's going on, which you may not get from this book, but what it is measuring. This is in some sense a measurement of how far from gradient f is. Because this would be zero when f is a gradient field. And this is measuring sort of how far away it is from being a gradient. And it's also, you can think of it in terms of physics as measuring sort of how much work it is to go around this path. Okay. Right. So, blah, blah, blah, blah, blah. Okay, so let me tell you Green's theorem. Let me leave this here for a minute. Because Green's theorem is in some sense, so Green's theorem is Stoke's theorem in the plane. So, let me put on hold for a minute. There are three important theorems that I'm going to try and cover today, not today, this week. Green's theorem, Green's theorem, which is also called Gauss's theorem. There's many answers. I'll just call it the Gauss's theorem. And then there's Stoke's theorem. And in some sense, they are all the fundamental theorem of calculus. See, their friend is the fundamental theorem of calculus. And a lot of times, once you get, well, I mean, so these look like three different theorems, but they're really all the same theorem. They look like four different theorems. But they're really sort of go by the name of Stoke's theorem. And roughly they say, if I have, so let me say it not as a thing, I have some region and I don't care how many dimensions we're in. So maybe this is a spherical object, maybe this is a 47-dimensional object, whatever it is. It says that, so this is a rough idea. If I integrate, so I have some blob in some number of dimensions, right, this is a blob, it's a high-dimensional blobby thing. And it says roughly that I can calculate the integral of some function on the inside. I'm going to screw this up, right? So let's say it's a derivative. By instead, this is what, this is what the fundamental theorem of calculus says, right? If, I didn't tell you what any of these theorems say, but I'm sort of telling you morally with it. So I have to make sense of what I mean by that and what I mean by this. And it changes depending on which of these theorems we're talking about. But the fundamental theorem of calculus says that if I want to integrate some function here from A to B, well I find an anti-derivative, G prime is F, right? That's the fundamental theorem of calculus where, so I want to compute the integral of a thing on the inside and instead I find some other function that is an anti-derivative and I just compute it on the edges. Here, there are two edges. So I'm really, my blobby thing here looks like this. It's that line. And so the edges of this are here and here. And so I can push the integration on the inside to what happens on the two ends and then everything's great. So what, so these sort of all collectively go by the name of students. This is, this is sort of a content free version of Stokes theorem. Well I don't know if it's content free, but a very squishy version of Stokes theorem says that knowing a function on the boundary, knowing an anti-derivative, whatever the heck that means, on the boundary tells me the integral of the function on the inside. Okay. So Green's theorem says, let me just write it carefully. So suppose that I have some A, some let A bounded plane. So I need an edge on it. I can't have it stretching off to infinity. If I want to let it stretch off to infinity then I have a problem. And I need a little technical condition. Let's say it's a finite, I guess I have to say what this means of simple pieces. I'll say what that means in a second. In other words, it only has a finite number of holes in it. Let me say this squishily. The boundary is some curve gamma, which is piecewise. In other words, it doesn't have too many corners. The corners don't pilot it. So it's not a fracture. So in terms of a picture, this is just saying A looks something, we can put a few corners in it if we want. It can have corners, it can have straight parts. That's all okay. But it's only got a finite number of holes in it. So this condition means that I can break it up into chunks that don't have holes. If I want to, I don't need to. So I can break it up into chunks that don't have holes in it. And the boundary isn't too nasty. So the boundary doesn't, so it's a holes in it like that, and it doesn't have really squeaky bits like that. There are plenty of things that, so anyway. So don't worry about it if you don't understand what I'm talking about. Okay, so now also suppose that I have this function vector field that is continuously differentiable. So its partials are equal. Actually, no, that's not what I want. Well, P and Q are continuously differentiable. So F, X, Y, and those guys are good. Q continuously differentiable. So P and Q are nice-ish. They have mixed partials that are equal. And I already wrote gamma, right where the camera is. The boundary of A is per gamma. So in this case, gamma consists of these three pieces. Let's call this gamma 1, gamma 2, gamma 3. And so I also need to orient gamma with right so that A is on its left. That means that if I go around, let's do the simple case, A is inside. So if I go counter clockwise, A is inside. So that means in this case where I have three goals, or how about two goals, I'm sorry I'm stating Green's theorem in such a long, long, long, long way. So if I have this case, then gamma needs to be oriented. If you walk along gamma, then A is always on your left. Here when I walk on gamma, A is on the left. Yeah? No? So gamma goes this way here, counter clockwise here. And here when I go this way, I want to keep it on my left. So it goes the other way here. The inside guys go the wrong way. So this says inside, clockwise, and outside. In the usual case like this, but if there's holes, we've got to go wrong. Okay, I'm almost ready to state the result. So with all that crap, it says that if I want to integrate, let me state it backwards, if I want to integrate some kind of a derivative, so if I want to integrate the curl, which is the partial of q with respect to y minus the partial of p with respect to x with y, then this is the same integrating over gamma, y, which is the same as saying f dot dx. Okay? So this says that integrating f is the same as integrating some kind of derivative over the inside. Yeah? What you said over there with the donut. Well, it's not really a donut. It's supposed to be an annulus. Okay. But okay. The donut wouldn't have a fair dimension, okay? Yeah. If you're going to do this as a hole, you must do right. So in this case, if I call this gamma 1 and this gamma 2, then I would have to integrate gamma 1 plus f dot dx plus gamma 2 of f dot dx, where gamma 2 goes together. Right? So this integral over gamma means over all the bits of gamma. But each bit has to be a closed curve. Because if I don't have a closed curve, then A will spill in. And that also means that when you think about it for a second, if I do something silly, imagine that my region has a slit like that. Then my curve gamma really has to go here and back. So gamma involves. In order to get a slit, I cover it both times. In other words, it doesn't matter. Instead of if you have it go the same way on the inside, then you just do minus the integral. Does it work like that? Or you can still take out the negative sign. So when I want to do this, right, if I want to orient gamma the other way, then I would put a minus sign there. I feel like that's more intuitive because I get a little subtraction of holes. Yeah, if you want to think, I mean, so that's why it works. So it's just that you may not be able to integrate in the middle. The function might not be defined here. So it's a little hard to subtract the holes because there's no way. So imagine that your function blows up here. Then if my vector field is really nasty in the middle, I can't even define it. Then it'd be hard to subtract the other way. Well, I mean, so like if you think about this guy, this guy has a problem at the origin. Yeah, so if I want to integrate a circle containing the origin. Well, so notice that what I'm integrating is not, I'm not calculating the volume, I'm calculating the curve. Right, this is the line integral. Right? It's a vector field. Everything here is a vector field. Right? No, no, no, this is a vector field. So this is a volume. Right? I mean, if you want, this is a volume. Yeah, it's like the total force. Right, exactly. It's sort of the volume of force or something. So this is a regular double integral. Right? So this sort of says a double integral. So this is a double integral over the inside is kind of a single integral over the boundary. And that's sort of the analogy here with the, here, I'm going to have a double integral to double integrate. My single integral is a zero integral. So I drop a dimension or I raise a dimension. If I want to calculate this, I could integrate and raise a dimension of something simple. Right? So this, so just as an inside, it's very common in, say, topology and so on. So this symbol means derivative or this symbol means a kind of a derivative. And it's very common in topology that if I want to take the boundary of a shape, let's say a field in circle, that's not the circle. And so it's very common in topology to use this symbol to be in boundary, which is the same. It's not an accident that we use the same symbol for boundary as we do for derivative. And it's precisely because of Stokes theorem relates the boundary, the geometric topological edge thing to a derivative on the inside. And there's all sorts of cool duality between this sort of geometric operation of passing to the edge and this sort of analytic operation of taking a derivative. We won't talk about that. Conceptually, like what measure, what do you measure if you integrate the curl? Is there some kind of collection of forces? You're measuring that, so of course not being a physicist, I'm probably going to screw it up. But you're measuring the circulation. You're measuring sort of, you know, if I have my vector field, let's say my vector field looks like this, but it's not conserved. So there's something funny going on in here. And I want to measure how hard is it to take a particle or a thing and drag it around this curve? Well, that's going to depend on sort of how much this thing wants to twist, although it's not really twisting. And that's what this is measuring. It's measuring sort of the total difference from being gradient. How much work it is to do that. And so here, in this case where it's really clearly twisting, then going around here, I guess somewhere I'm on if I want to, maybe that's negative 2 times minus plus 2 times. I did something wrong because I have a sign wrong. But going around here, let's go this way, is work. Because all the way I'm against the wind. And so this is measuring sort of the circulation or the current or all that stuff. Yeah, I just do the whole class. Oh well. Alright, that's okay. So this, yeah. There it is. So this will be a double integral. Right? You calculate these things. Right? If I have, maybe I should like do an example. Let's do that example. So here, my P, my P minus Y, my Q. Let's just do that. Let's just do my minus. Let's just do this. P of X, Y is this. Q of X, Y, that. Alright, my vector field is this vector field. And so now I want to compute gamma is the unit circle and T sine T. T goes from 0 to pi. And if I want to compute P of X plus Q dy. So this is, this is a line integral. So that is a line integral X, no. Minus Y dx plus X dy. That's what I want to compute over the unit circle. So if I want to do this, then I parameterize gamma as say cosine T sine T. So I can just do this directly. Integrate from 0 to 2 pi. Y is minus sine T. dx is cosine T plus 2 pi X. Am I doing this okay? People follow what I'm doing. Okay. X is, I've lost it. Cosine T dy is sine T dt. So that's the integral I want to compute. Integral from 0 to 2 pi. And so this is integral from 0 to 2 pi of sine square cosine square T, which is 2 pi. Okay. So that's one statement calculating this integral. Somebody asked me this. Somewhere over here somebody asked me and I started doing it. Now I forgot who asked you. But by Green's theorem that also says that this is the integral over the unit disc of, well, the partial of P's, the partial of Q with respect to Y. What did I do wrong here? Q. Oh. Right. It's the mixed partial. Ah. Q and X. That's Q. That's X. Good. Sorry, I wrote it down from the beginning. It's the difference of the mixed partial. If I do this is great. So derivative of Q with respect to X. With respect to X. One. And the derivative of P with respect to Y. Minus one. And so this integral over the unit disc of 2, PX, P1. Yes or no? Well, that's easy. I want to integrate. So this is, if you like a double integral, I'm not even going to write it out over the unit disc of 2, PX, PY. But it's a lot easier to think. I take the disc of radius one and I erect a function of a height 2 over it. And I ask for the volume of that cylinder. The volume of that cylinder, well, the base as radius pi, as the area is pi r squared. The height is 2. So this is 2 pi. So Green's theorem tells me that I can do this either way. At least justify why this makes any sense. I'm not just doing an example. The example I was going to do, but I didn't, would need to take this one. So let me just say, this one is done in the book, so I won't do it. Because I'm running out of time. Take this one. Let me just say some words about that one. Which is the same as this one, except that it has a problem at the origin. So if I take this one and I want to integrate this nasty vector field over some ellipse or something, then when you parameterize it and try and do this integral, it really sucks. You don't want to do that. But if you check, then this quantity, it comes out to be, I change this to a circle, nothing, this quantity doesn't change. And so I can just integrate it over the circle. If I just check. So let me prove Green's theorem before I run out of time. So why does Green's theorem work? Just because it does. Okay, good. We're happy with that? No. So Green's theorem is really just the fundamental theorem of calculus. Because if we look at this quantity of calculus and QDY. So Green's theorem says what? I don't know. I screwed it up. Well let's just, so in the case where is symbol. Okay, so I'm going to say that A looks something like, something like a fish. Can look like a fish. So A is a goldfish. You have to eat it. So let's say A looks like that. And we want to compute, well let's just break it up into two pieces. So I want to compute the integral over my curve gamma of S dx, which is, so the integral over my curve gamma of P dx plus Q dy, I can break this up into the integral over my curve gamma of P dx plus the h integral over my curve gamma of Q dy. You can certainly do that. Again it just means things like this. And now that really means, so I want to look at my fishy curve here. And notice that if I'm just looking at this piece, let me write in some variables. I'm looking at something like that. So I'm going to revise this curve gamma, which goes this way. And I'm going to notice that in X it goes from A to B. And I can write gamma, well it's this, if I start here, let's start here. I go along for a while until I get to here. So there's some curve, let's call it V of X. So let's break it into three pieces in this picture. Here is gamma 1 on the bottom. Here is gamma 2 on the top. And here's gamma 3. So I'm going to break it into three pieces. If you have a more complicated thing you can break it into more pieces and if it's simpler you can break it into less. But break it up into pieces where it's graphs of functions or it's vertical. So more formal in break it into pieces where it's graphs of functions where it's vertical. We'll notice that if I integrate over gamma 3, that is if I parameterize gamma 3, I get a curve whose derivative is 0 in the X direction. So I can ignore this part. Gamma 1, gamma 1 as some curve, it's a graph of a function so it's the curve X, let's call this V of X or bottom if you have trouble distinguishing B's and B's. Gamma 1, I can parameterize it as the graph of X, some function V of X so it's a curve like that. So here X goes from a, is that clear? Those of you that are, yeah. And on M2, U of X, but X goes the wrong way. I guess I'll just start moving this way so I can keep pointing at that same thing. So now let's just calculate what the integral over gamma, let's do say gamma 1 of P of X, Y of DX. Well this is the integral from A to B of P of X, V of X, that is DX. DX here if you remember is gamma prime, DT. I have to adjust by the derivative, right? Yes? No? Yes. But it's really gamma 1 prime. The DX for gamma is whatever speed gamma is traversing this curve in the X direction. But since I parameterize it by X, DX is DX. If I do the change of variables then it comes out that I don't have to adjust anything here. Because DX is just DX, I parameterize this in unit speed in the X direction. Is everybody okay with that? Wait a second. Is the derivative of gamma prime in the first part of the formula? Okay, so here. Let's write it correctly. Let's get about all of that junk. This is integral from 0 to 1 of P, or maybe 1 half, I don't care, P of gamma 1 of T, gamma 2 of T, and then DX is gamma 1 prime of T, DT. Because this is X, this is Y, and this is DX. Oh, that's gamma 1. Yeah, gamma 1, gamma 1. It's just the X part of gamma, right? It's not a vector, it's just the X component of the vector. But if I, okay, I sort of screwed up my notation because I'm using gamma 1 and gamma 2 in two different ways here. Yeah. Okay. Sorry. This means, how about gamma X, gamma Y, gamma X. Thank you. Yes. Sorry for messing up my notation. Okay. So if we parameterize the curve and we want to calculate this over the first curve, then we can do that. But if I choose my parameterization like this, then gamma X prime of T, DT is just DX with this X. So this is the same thing, one integral in one variable. And if I do the other guy, so I take the other piece, where I parameterize the top half of the fish by gamma 2. So if I integrate over gamma 2, I get a similar thing, but it goes from B to A. Now it's the graph of some curve, B of X, and I parameterize the top, and it goes from B to A, so it's backwards. So this is the integral from, negative the integral from A to B. And so what I want, and I already showed that the NEDX is zero, so I have to worry about that one. Almost there. And so now, we put them together. So my integral over gamma, P DX, is going to be, these two things added together, the integral from A to B of P of X, V of X minus DX, integral from A to B of P of X, U of X DX, which is the same, just integrate from A to B, P of X, the bottom, minus P of X, U of X, the whole thing, PX. And now comes the magic observation, that all of this junk is just a function of X. So you can really do the fundamental theorem of calculus here to throw in a Y and not change anything. That is, I can think of this mess, so I can think of, so let me, let me put it in intermediate step here. If I integrate, what do I want? V minus U, so that means V has to be on top, V has to be on the bottom, P of, so if I do a silly thing, I do the fundamental theorem of calculus the wrong way. If I have the derivative of a thing integrated, so I've changed very much to Y, then that's the same P of V minus P of X, U of X. So I mean, this is like magic. It's not magic, you just see this, say, yeah, sure. The thing that makes this theorem is you say, oh, geez, I can do that, cool. So it just seems like magic. Well, that's where this partial business comes from. So this, I want to flip it around, so that's equal to negatives. I can put the U on top because I like the top to be on top and the bottom to be on bottom. Partial P, P, Y, let me leave out the variables. Let me not, I have a Y, let me leave out this part of the variable. That's the inside of this integral. So that is, that means that this equals negative integral from A to B, integral from U to B, right over here. This equals negative of the integral from A to B, negative of the integral from U, from B to U of the partial of P with respect to Y, BY, PX. Because this inside, this bit is that bit. Well, what is this? This is a negative sign here. This is exactly, don't tell me I erased green's theorem. This is exactly the first term or the second term in green's theorem. Now do the same thing laying on your side and looking at this region the other way. And if you check the negative sign, you don't get a negative sign because the top is still the top. And so then you get green's theorem. And I guess you can get anywhere near where I wanted to get. But maybe that's where I should stop. So we get this part, and so that means that we have, but if I do the double integral over the whole thing of Y, Y, B, P, yeah. The mixed partials, Y, DX, that's the same as the integral gamma of F dot N. Okay. So next time I will try, in some way, to move this up a bit. Well, do the diverges theorem and then I probably won't be able to do that. I'll definitely take a look at that because, nope.