 This principle, this general principle looks like this. We have two integrals, one on the left-hand side and one on the right-hand side. On the right-hand side, we have an integral of some quantity, which is to be decided and which depends on the, on circumstances, which dimension we are, what we'd like to integrate and so on. And on the left-hand side, we integrate some sort of derivative of that object, okay? Some sort of derivative, which I just schematically denote by d omega. Now, the next question is, what do we integrate over? And we are integrating on the left over some domain d and on the right, we integrate over the boundary of this domain. This is a boundary of d, okay? So, this formula represents a kind of trade-off between an algebraic operation, namely, taking the derivative going from right to left and a geometric operation, which is taking the boundary of your region. So now, we'd like to see how this general principle plays out in different contexts, okay? And what we discussed last week was one particular, in some sense, the simplest possible context of this. In this context, we took d to be one-dimensional, one-dimensional. And so, its boundary, b of d, is zero-dimensional. It consists of two points. So, here is a picture. Here is d, it's a curve in this case. And being a curve, we usually denote it by c. C for curve, okay? And the boundary has two points, like this. And also, there is an important aspect of this picture, which is that we have to choose our orientation. We have to say which is the initial point and which is the end point. We travel from here to here or from there to this point. So, let's choose our orientation like this. So, we travel from here to here. Let's call this point a and this point b, okay? So, the formula which we learned last time looks as follows. We have, on the left, we have the line integral of a vector field f. So, f is a vector field. Actually, let me rephrase that. So, let me start from the right-hand side. Because, see, the point is that the algebraic object, omega, actually is found on the right-hand side, not on the left-hand side. So, I should really start on the right-hand side. And on the right-hand side, I have the values. I have, first of all, a function f on the plane. And what I have on the right-hand side is just I evaluate my function at the end points. At the point b. And I also evaluate it at the point a. And I take the difference. So, this one comes with coefficient plus one. And this one with coefficient minus one. And that is a testimony to the fact that this is the end point and this is the initial point, which is due to the choice of orientation that we've made. So, this is the right-hand side. And I can really think of this as kind of integration, integration of the function f over the boundary, over the boundary, right? Even though it's a zero-dimensional boundary. So, integration doesn't really make sense. What I'm talking about is really evaluating the function at these two end points. But it does fit the general formula, right? Because I have a function, that's my omega, in this case. And I have the boundary, which is two points b and a. And integration over the boundary is simply represented by the difference of the values of f at b and a. Okay, what about the left-hand side? On the left-hand side, I'm supposed to take some sort of derivative of omega. So, d omega. And I have to integrate it over my curve c. Right? So, my domain now is the curve c. And what I'm going to integrate is the, I'm going to take the line integral of the gradient vector field of my function. So, this is this derivative that, this is the real incarnation of the derivative of my function in this particular context. In different contexts, this will mean different things. In this particular context, the derivative simply means taking the line, taking the gradient vector field. And then I simply integrate that gradient vector field over the curve, okay? So, what I get is the fundamental theorem for line integrals, which we talked about last time. And my point now was to emphasize how it fits this general principle. So, you can think of this as the first application of this general principle, which is explained on this board. All right, so, what we're going to do now is we're going to make the next step. But you have a question. d omega. Why did I, well, I'm just saying that if you don't like an upper or arrow, I will make it a sideways arrow, like this. Is that better? Okay. Don't read too much into the arrow. It's just to say what d omega, what that d omega is in this context, okay? Omega in this case is a function f, that's right, which is what I wrote. That's omega in this case. All right, any other questions? All right. By the way, for our worldwide viewing audience, and we have a pretty wide worldwide audience, as it turns out, the lectures, all the previous lectures starting from lecture one are now being uploaded on YouTube. And so, everybody can now learn what happened previously on math 53. You know, the struggle, the intrigue, the romance, the broken hearts. It's all there for our viewing audience. Season one and two, the unforgettable. Season one and two. Okay, let's go back to season three, though. I have to say I saw the lecture one today and it sort of made me a little nostalgic about it. I remember the good old days when I first came to this classroom and saw all of you, you know. That was a priceless moment. Anyway, so what we'd like to do now is we want to make the next step. We would like to generalize it when the domain is two-dimensional instead of being one-dimensional. So now we generalize domain D is two-dimensional. And so what I'd like to do is actually kind of make a chart and make a comparison between the one-dimensional and the two-dimensional cases. So this will be the one-dimensional case and this will be the two-dimensional case. So in the one-dimensional case, first of all, what is D? Okay, D is a curve, which I denote it now by C because it's a curve. So let me draw another picture like this. And we have, so we have these two boundary points, right? And so what should be the analog of this when we generalize it to the two-dimensional case? It should be a two-dimensional region, obviously, right? This is one-dimensional. And the analog of this in two-dimensions would have to be just a two-dimensional region. So what is it going to look like? Well, it's going to be some region on the plane, like this. When I draw this curve, I really mean the interior of this curve, right? This will be D. This will be D. And the curve will actually show up now as the boundary. So see, this is a very important point that now the domain is two-dimensional, but the boundary is one-dimensional. And so now the boundary is an object which kind of looks like the original domain in the previous example. Okay, so that's the, that's the domain. So it is two-dimensional region. And this is the region, two-dimensional region. And I will just keep denoting it by D. It will not change the notation. Okay, what about, what about the boundary? So the boundary here, as we discussed, just consists of two points, two points. So I'll just repeat. I'll just draw them again. There's two points, right? What about here? Here the boundary, as I said, is this yellow curve. So let me draw it one more time, approximately. So see, now I'm not shading the interior of this picture. I'm really focusing on the curve. The curve is one-dimensional, and the curve shows up as the boundary of this two-dimensional region. So this is a curve B of D. Okay, next, what is omega? And I, again, I am, I would like to think in terms of this formula. So I would like to define all of this data and then see what kind of equality I get from this, right? So what is omega? Here, omega was F, was the function F. And the derivative of omega was just a gradient of F, vector field, which is a gradient of F. And so what did I do? I integrated, I integrated F over the boundary. So I evaluated my function at the point B and A. But on the left, I was taking the line integral over the curve itself, right? That is the line over the curve itself over the gradient of the function F. So now the question is, what should I do? What should I take in this two-dimensional generalization now, okay? So I have to take something as omega. I have to take something which I can pair with this curve because omega is supposed to be integrated on the right-hand side, right? Omega is supposed to be integrated on the right-hand side of the formula. And on the right-hand side, I integrate over the boundary, over the boundary, not over the region itself, which is two-dimensional, but over the boundary. And that's this curve. So I have to think of the most general integral that I can do over a curve. And we've learned such integrals, these are line integrals, right? So I should be integrating a vector field. That's a natural thing to do because in this analysis so far, we're integrating over curves, we're integrating vector fields. Functions, we also integrate, but in this case, it seems more reasonable to integrate a vector field. So what I'm going to take is a vector field F. And this vector field is going to have two components, P and Q, and that's going to be a vector field. And so I will integrate it on the right-hand side over this boundary. But I also have to say, what is the derivative of this? What do I mean by the derivative of this vector field? That should be some expression involving derivatives of P and Q, which I would then integrate over the two-dimensional region itself. And that would give me the left-hand side, right? So I have to decide what this is. I have to find out what this is, and then I can build the formula. And it's not obvious from the beginning. It's not obvious what that derivative should be. Well, likewise, it was not so obvious that this should be the gradient. So let me just give you an answer and then we'll see what is the meaning of this, okay? So the answer for this derivative is given by the following formula, which at first may look kind of strange to you, but then I will give you kind of an explanation how this could be derived. And the expression is the following. It's actually going to be a function, whereas omega here is a vector field. The derivative here is actually going to be a function. So it's kind of opposite. The operation of differentiation in this context does something opposite to what it did in the one-dimensional case. In the one-dimensional case, the original object was a function, but its derivative turned out to be the gradient vector field. And now we start with a vector field, but we end up with a function. That function is a combination of partial derivatives of p and q, namely dq dx minus dp dy, okay? So in some sense it is like a derivative. It's not a derivative in any obvious sense because we haven't talked really about derivatives of vector fields, right? All we know that it should be some combination of derivatives of the two components of the vector field. And there it is. It is a combination of partial derivatives of the two components. Namely, I take the second component, q, differentiate with respect to x. I take the first component, p, differentiate with respect to y, and I take the difference. So it is, and of course in this discussion, it is assumed that all my functions and my vector fields are differentiable and in fact have continuous partial derivatives. So this actually makes sense. Okay, so let's accept this. Let's accept this formula for now. I'll get back in a minute to the meaning of this formula and how to derive it, but let's accept it for now. And so let's now see at our brand new example of this general formula in the case when the domain is two dimensional. Let's see what it's going to look like. Well, again, on the right hand side, we know already what it is on the right hand side. On the right hand side, we integrate over this curve. Let me draw another replica of this curve. That's the boundary of my two dimensional domain. And I'm taking the line integral of this vector field, which I would like, I could write it as f dot dr, if you want. But because I have already represented f as a vector field with two components, p and q, I might as well just write it in terms of p and q. So I will write it as p dx plus q dy. And this is certainly just this expression, it's the same. This is the same expression. So that's what I integrate on the right hand side. What about the left hand side? On the left hand side now, I should have integrate over my domain D. And actually, someone pointed out last time, at the last lecture, that in this formula, I didn't keep track of how many times I put the integration sign. It was an approximate formula. I just wanted to use a sign of integration, just one sign regardless of how many times I have to integrate. But now I'm writing a precise formula. This was just a guiding principle. Now I'm writing a precise formula. So I would like to keep track of the number of integrals, which is the dimension of the domain over which I'm integrating. So now my domain is two dimensional, right? And therefore, it should really be a double integral. It should really be a double integral. And it should be a double integral of the derivative of omega. About the derivative of omega, I have given by this formula. So what I get is dq dx minus dp dy dA. And that's the formula I get. So now, if you've never seen this before, it looks kind of surprising because so it's a formula which equates a line integral of a vector field to a certain double integral of a function over the domain. Here over the domain, here over the boundary, of this domain. Perhaps let me emphasize this by drawing again the domain. This is the domain. Right, this is domain and this is its boundary. So that is actually the theorem which is the analog of the fundamental theorem for line integrals in the two dimensional case. And it is called Green's theorem, okay? So question. That's right, very good. So there are still quite a few loose ends here, which I'm going to now unravel, okay? So then the first question is a very legitimate one, which is, which orientation do I take on the right-hand side? This is a very good question because if I just write this formula, it actually does not make sense, right? Because a line integral of a vector field is only well-defined when I prescribe the orientation of the curve over which I integrate, right? And if I change the, well, there are two possible orientations. In this particular case, I'm dealing with a closed curve, a curve which does not have a boundary as you see, right? So the way I will be referring to the orientation is, I'll be saying it's clockwise or counterclockwise. So there are two possible orientations. In other words, I can go like this and go around. Or I can go like this. There are only two possible choices. So you can say, well, what's the big deal? It doesn't matter which choice I take. There are not so many choices. Well, it does matter because I get two different answers this way. Granted, the two answers only differ by a sign. But of course, sign is important because, you know, if I want to say that this is actually cool to this, I have to say under what assumptions and under, in particular, under what choice of orientation? And so the point here is that the orientation has to be chosen counterclockwise. For this formula to be correct, I should take it counterclockwise. This is not to say that the integral with the opposite orientation is not well-defined. It is also well-defined. It's just not going to be equal to this. It will be equal to minus this. You see? So this is a very important point. The second point that I would like to make is this distinction between between closed curves and open curves. Or curves with boundary. You see, there is a difference. This is a curve. And there is also a curve in our previous example. These are both curves. And the word curve here refers to the fact that they are one-dimensional options. But they are very different. This one has boundary, has non-empty boundary. It has a boundary which consists of two points. But this one does not have a boundary or more properly, it has an empty boundary. Such curves are called closed curves. And it is interesting that it is exactly such a curve which shows up as a boundary over two-dimensional region. So a curve like this is called a closed curve. It has empty boundary. And you should contrast that to a curve with a non-empty boundary. This is not closed curve. This is a curve. Well, terminology here is slightly misleading because a mathematician would also say that this is closed because it does include the boundary points. But when we say closed in this class we will mean a curve like this. So it closes on itself, so to speak, right? This one, strictly speaking, is also closed because I cannot say it's an open curve because I do include the boundary points. So let's just say it's a curve with boundary points. So curve with two boundary points. And of course that immediately begs the question, is it possible to have such a curve, not a curve like this, but such a curve as a boundary of something two-dimensional? What do you think? Yes. Well, any other answers? Not that there are too many answers here, right? But if you think about it, you will see that this, it looks like I'm giving you an example of this curve being a boundary of the shaded region. But that's not true. What about this part? What about this part? Now granted, both if I include this and this, this would be the boundary of something but then it would have to go to infinity or it would have to close somewhere, you see? So the regions which we will consider in this case, the regions like this, the regions to which we will apply Green's theorem, these regions are going to be bounded. They don't go to infinity. Yes? Yes? That's right. Let's go back to this. So the question is defined to be for counterclockwise, defined as positive, right? So the terminology, you know, we are free to make any terminology we like. We can call it positive, negative, whatever we want, right? Some people like counterclockwise, some people like clockwise. The standard terminology for this is counterclockwise, orientation, counterclockwise. And the standard methodology is that it's also called positive, but it's a matter of choice. Again, I am free to take line integrals over any curve I want. I could take it over a curve oriented this way or oriented that way. My interest here is to choose orientation for which this theorem is correct. And I know that if one of them is correct, it cannot be correct for the other one. So I have to make a choice. And the way, actually, the better way to think about it is as follows, which we will actually, we'll need to slide generalization of this for which this rule that I will explain now will make more sense or will be more sort of easy to remember. Right now it looks kind of ad hoc. Why do I choose it counterclockwise? So the actual rule is the following. That if you imagine somebody walking on the, it's just a sketch. Somebody is walking on the boundary of this domain. So it's like, more realistically, I would have to draw this domain on this stage and I would be walking on the boundary. So the rule is that my orientation of my path should be such that the domain is to my left always. It's to my left. Well, it's, you have to choose it some way, right? So this is well defined. If you walk this way, it's to your left. If you walk this way, it's to your right. So it has to be to your left, orientation to the left, to the left, domain to the left. It does sound a little ad hoc, but actually it all makes sense in the end. So believe me, it is a good rule. And anyway, it's a rule for which everything works. If you choose a different one, you can do that, but then please put a sign. Change the sign on either side, on one side, not on both sides, okay? So let me go back to the issue of boundaries. First of all, the domains that we will consider, domains like this, will have to be bounded. Bounded is something else. Bounded is something which doesn't go away to infinity, off to infinity, you see what I mean? So the entire plane is not bounded, and this is bounded. This is something we talked about when we talked about double integrals. When we do double integrals, we usually do them over bounded domains. If you have a bounded domain, its boundary would have to be closed, a closed curve, you see? So this is important. This is a very important observation, which is how maybe it's not so obvious at first, but it is correct. Now, there is one other subtle point, which I guess I should mention, which is that it looks like this curve is slightly, this is not really a curve, you can say it's not really a curve, it's a combination of four different curves, right? Whereas this one looks kind of smooth, although I kind of messed it up a little bit, so it's not so smooth, but I was trying to draw a smooth curve. This certainly is not smooth, it has four corners, at least, right? But in fact, for all intents and purposes, these are both curves. In other words, I will think of this entire thing as one curve. This curve will have four different segments, which are smooth, and such curves will call them piecewise smooth. In other words, they're built from pieces which are smooth. The curve itself, the entire curve is not smooth, but it's a union of pieces which are smooth, in this case, four. So when I talk about the boundary of this region, I can say that it's a combination of these four pieces, but I will usually just say it's this curve. I will ignore the fact that there are some corners in there because after all, I can draw it in one stroke like this, right? So it's not really, it doesn't have much value to say that, well, how many corners it has. It's not going to matter for us in the end. When we calculate integrals over a curve like this, of course, we will do it as a sum of integrals over each of the smooth parts, but that's sort of a calculational matter. It's a technical issue. It's not an issue, it's not a fundamental theoretical issue. The theoretical issue is that the boundary of this domain is this curve, this closed curve. A boundary of a domain like this on the plane cannot consist of a single curve like this, which has a boundary of its own. In other words, boundary of boundary is empty, if you know what I mean. If you take boundary of something and then you take a boundary again, this is empty. I'll use the standard notation for empty, empty. A boundary can be non-empty. This curve has a boundary of its own, but then it itself cannot appear as boundary of a two-dimensional domain. If a curve appears as a boundary of two-dimensional domain, it has to be closed. Its own boundary should be empty. Do you see what I mean? Are there any questions about this? Yes. It should be, that's right. So the question is, how does this apply to Green's theorem? Well, here's how it applies to Green's theorem. In the case of Green's theorem, the curve I have is a boundary of something. So this principle which I explained, the boundary of a boundary is empty, means that this one will have empty boundary, will have no boundary. So that means that in Green's theorem, I will, on the right-hand side, I will always integrate over closed curves. You see what I mean? Yes? No, we could apply Green's theorem to this one as well. Let's look at this example. Let's suppose that my domain here is a square. So this is my domain. I can apply Green's theorem to this domain, but then on the right-hand side, let me use the same chalk. On the right-hand side, I will have the square as my boundary, okay? This boundary is slightly different than the boundary in the previous example in the sense that it's a curve which has corners, but it is a curve and it is a closed curve. This curve does not have any end points. You see? This is a closed curve, closed curve, empty boundary. These points, it has the special points. Let's, I grant you that. It has some special points which are different from the rest because these are the singular points, so to speak. They are not smooth, but these are not boundary points. This is not the boundary. It's in the middle, right? If I had something like this, those would be boundary points, right? But this is still not the boundary point. If I have something like this, these are the boundary points. You have a question? It's empty. It's not, no, this is not a definition of a closed curve. This is a journal principle which is that if I start with a two-dimensional object and take its boundary and then take the boundary of that boundary, this will be empty. This is illustrated here. I start with a two-dimensional domain. I take its boundary. I get a curve which already has no boundary. Bless you. I know it's a little confusing, so I'll just let this sink in, as I say. And this is an important point. This is an important distinction. So from this point of view, Green's theorem can be, if you have a line integral and you would like to use Green's theorem to understand that integral, you have to make sure that you are integrating over a closed curve. In Green's theorem, on the right-hand side, you will never have a curve like this or like this. You see what I mean? You will never have a curve which has boundary because the curve in Green's theorem is not independent object. It is a boundary of something on the left-hand side. And because it is a boundary of something on the left-hand side, it has no boundary of its own. Yes. What's wrong with integrating over an infinite two-dimensional domain? What's wrong is that the integral may diverge, right? It might actually be infinite. It may not be well-defined. If we integrate over a bounded domain like this, which sort of doesn't go off to infinity, we are assured that if our function is defined everywhere on this domain, we are assured to get a finite answer. Whereas if we integrate over something infinite, we are not assured of that. For instance, take function one. If you take function one, then the integral over any domain is going to be just the area of that domain. If the domain is bounded, the area is finite. If it's not bounded, it's infinite, so it's not well-defined. But it could be well-defined, that's right. It could be well-defined. And so oftentimes we are forced to consider integrals over infinite domains. And the way we do it is quite obvious. We approximate the infinite domain by finite, bigger and bigger finite domains, right? And then we integrate over those finite domains and we take the limit. You're familiar with the notion of limits. So you see roughly how it would have to be defined. But we are not assured in general that the answer, that the limit will exist and so the integral would be well-defined. Here we don't want to deal with these issues. So from the beginning, we restrict ourselves to domains like this, which are finite. Any other questions? Yes. Okay. Let me, let's talk about this. Let's talk about this. So the question is, suppose you have two curves which connect. How do we calculate the boundary of the sum of these two curves, so to speak, right? So for example, let's suppose that we are, let's suppose that we have, let's suppose that we have a curve like this. So what is the boundary of this curve? Just from what I said, just from intuitive description of what boundary means, you can see that it's going to be this and depending on what orientation is, let me put orientation like this. So the boundary is this minus this. So that's the boundary of this total curve. Let's call this points. I want to call it ABC, but I don't want to do that because I have C for this, right? So let me call them alpha, beta and gamma. I think as long as it's not epsilon and delta, you are fine with that, right? So the boundary of C is gamma minus alpha, right? But on the other hand, you can think of this curve C as the sum of this curve and this curve. So it's, well, let me just draw it. I mean, they're not shifted, it's just two different sheets of paper. I can think of this curve as a sum of this one and this one. So let me see what will I get if I treat it like this. There will be C1 and C2. The boundary of C1 would be beta minus alpha, right? And the boundary of C2 would be gamma minus beta, right? So if I take the sum, the boundary of C1 plus C2 is going to be beta minus alpha plus gamma minus beta. So you see beta's cancel out and I get the same result. So it's all consistent. The boundary of this curve is gamma minus alpha, no matter how you look at it. You look at it as just boundary of this curve or you look at it as a boundary of the sum of these two curves because the point is that you have to keep track of the signs. If you keep track of the signs, actually this comes with minus, this comes with plus, this comes with minus, this comes with plus. And so these two guys annihilate each other. They kill each other. So they disappear. Does this answer your question? Okay. All right, so let's move on and let's talk more about this Green's formula. Okay, so now that we have figured out more or less what the meaning of the boundary is and what kind of curves we can get on the boundary of a two-dimensional region. Let's talk about this very strange, let's talk about this very strange looking expression that we integrate on the left-hand side. So on the left-hand side, again, we are integrating this d dq dx minus dp dy. And I kind of want to write it as dx dy. This is what we call dA. This is what we call dA when we talked about double integrals. I'm just rewriting the formula, but it's just in a slightly different way. I want to focus now on the integrants, on the algebraic objects. We have discussed the geometric objects. Now let's discuss the algebraic objects. So in what sense is this a derivative? Why should we think that this is a derivative? So let me give you kind of an intuitive explanation of this. But let's do it again by analogy. By analogy with the one-dimensional case. In the one-dimensional case, we were integrating the function on one side and we were integrating the, it's gradient on the other side, right? So in the one-dimensional case, this is a two-dimensional case. In the one-dimensional case again, I just want to put them next to each other, is nabla f dr equals, well, let's write it like this, f of b minus f of a. But what is nabla f dot dr? We can write it in coordinates. It is dp dx dx plus dq dy dy. That's what we call nabla f dot. So you see, this is a derivative. This is what we call the derivative. And in fact, if you look at this formula, what do you see? What do you recognize? Sorry? Do I should change this? No, I shouldn't. Here I have the sign, but these are two different formulas, right? Why do you think that I should change sign? Nabla f, nabla f is a vector field which has two components, dp dx i plus dq dyj, right? When I take line integral of this vector field, I have to take the first component, multiply by dx, the second component, multiply by dy. That's what I integrate. But my question is, have you ever seen an expression like this before? In season two, it made its appearance, right? You know the answer. What is it? Somebody say it. Differential, thank you. So, this is a differential. So in fact, this is df. This is df, the differential of f. So you can think that you are doing the line integral of the gradient of the function, but actually a better way to think about it is the integral of the differential. The differential is an example of what mathematicians call differential forms. And you see that every time we differentiate, we have to have some ds, dx, or dy here, only one because we're doing a single integral. Or dx, dy, both of them because we're doing a double integral. And sometimes we have dx, dy, dz, right? So these objects, all of those objects, are mathematically called differential forms. And actually a slightly more consistent theory of integration can be developed in which the objects that we integrate are differential forms. We take a slightly different point of view, which for the, because we want to think about these integrals more in terms of applications, like work done by force and other things like a flux of a vector field that we will talk about later. And this is the reason why we are thinking about integrating vector fields rather than differential forms. But algebraically, it's easier to think of this as integrating this differential form, which is just the differential of the function. So in the case, in the one-dimensional case, in the case of the fundamental theorem for line integrals, what we are doing passing from this, from the right-hand side, which is a, which we have a function. Right-hand side, we have a function. And on the left-hand side, we have what I call the derivative. This derivative actually is just the differential of f. You can think of integrating just the differential of this function. Yes. Shouldn't be a derivative of what? I'm sorry, that's right. I made a mistake. I'm sorry, it's df, that's right. Sorry. And that's why you didn't answer, that's why you didn't answer my question. Sorry, it was my fault. Yeah, I was, I got, I got carried away with all this df, dp and df myself. Yes. Of course, there is no p and q in this formula, right? p and q shows up here, because we have p and q here. Here, there is just, what we are given is just the function f. And what I should have written is just the derivative of f in both cases, right? Okay, now it's much better. Do you see now what I mean? Okay, is it better? Okay, so, so what did we do to get this formula from f? What did we do? We took the function f and we, we took two possible derivatives of f with respect to x and respect to y, right? So we took f and we replaced it by df dx plus df dy dy, right? So what is this operation? I differentiate, I take partial derivative, partial derivative, and at the same time, multiply by dx, partial, x partial derivative and multiply by dx plus y partial derivative and multiply by dy. That's what it is, right? So there are two possible partial derivatives. I take partial derivative multiplied by dx, take second partial derivative multiplied by dy. So let's take this as a guiding principle. Now I would like to apply it here. So now my starting point will be p dx plus q dy. And I want to get, to see what I can get from this by applying the same idea, the same procedure, okay? So let me start with the first term, p dx. When I start with p dx, I apply the same, the same rule. So I take the derivative of p with respect to x and I multiply by dx and I keep the old dx which I had before, right? So I just treat it like this. I apply my differential to p, but I keep this, this extra factor as well. So I get this term plus dp dy and then I have dy dx. Now, have you ever seen dx dx? No, and for a very good reason, right? Because we see dx dy because this represents the area of an elementary parallelogram with sides delta x delta y. The area is delta x delta y. And this is what eventually gives us this term dx dy which we call dA in the double integral. But delta x, delta x makes no sense, right? Delta x, delta x does not represent any area whatsoever. So get an area, you have to have increments in two transversal coordinates like x and y, or maybe r and theta if you switch to polar coordinates if you like, okay? Or dx dy, dz if you're doing triple integrals. But delta x, delta x makes no sense and neither does this. So whenever you have dx dx, this just is prohibited. So you have to cross it out or more properly you should say it is equal to zero. It's like, it sort of represents the area of the segment which is zero, right? To have an area you have to have something two dimensional. Now this on the other hand is fine, it's dy dx. The only problem is that we switch the order in which we normally take the area which is usually dx dy. So the rule in fact is that this should be replaced by minus dp dy dx dy. This is an issue of orientation. Remember we talked about different types of orientation and we know that if we, when we talked about general changes of coordinates under double integrals, we saw that if you switch to coordinates, the Jacobian that you get under the integral gets a minus sign. We kind of skirted that issue by saying, well we will just take the absolute value of the Jacobian. But at this point actually we can't avoid this issue and the correct solution to this issue is to say that dy dx is equal to minus negative dx dy. So when we apply this procedure to p dx, this is what we get. But I also have the second term q dy. So let me calculate what I get for q dy. For q dy, I get dq dx dx dy. This is good. First of all there are two different variables and second of all they are already in the right order. I do nothing. Then I also have dq dy, dy dy. Again I got dy dy. Makes no sense. Cross it out, okay? You see what I'm doing, right? I'm just taking q and q gives me this plus this. But I also have the second factor dy which I keep. And it is the second factor which tells me that the second term should disappear. So let me collect all the terms that I got. If I start with p dx plus q dy, I end up with minus dp dy plus dq dx dx dy, you see. And that's exactly the formula that I get on the left. Well, I write it as dq dx minus dp dy, but I just switched the order. Yes. So the rule is the quantity function that you have, each of the functions that you have, you should take partial with respect to x and multiply by dx plus partial with respect to y, multiply by dy, right? So this here is how it works for this term. For this term I take partial with respect to x and multiply by dx. But don't forget I had dx from the beginning also. So I just keep it, right? Plus partial with respect to y, multiply by dy and keep the dx. That's what I have. But in addition, I have this term. For this term I get this answer. So I get a priori four terms. But out of these four terms, only two are non-zero. This one and this one. I put them together and I get this answer. And that's exactly the answer which Green's theorem tells me I should put here. So this is the explanation of this expression. This expression at first glance looks unnatural. But now I have shown that you can obtain it in exactly the same way as the expression for the differential of a function which certainly is quite natural. We know that the differential of a function is actually very meaningful, okay? Yes, that's right. So this is a subtle point. Question is why do I put the minus sign? This is a subtle point because when we did double integrals we actually kind of swept this under the rug because when we oftentimes would write dx dy and oftentimes we would also write dy dx. Depending on which order of integration we choose, right? Because what we did, we represented double integrals as iterated integrals. We would first integrate over x and then over y or first over y and then over x, right? Depending on that we would just write dx dy or dy dx. But in fact, the correct meaning of this expression should involve also the choice of orientation. And the orientation, if you do that, the DA, the measure of integration for double integrals should really be dx dy and dy dx should be minus that. The reason why we didn't do it because essentially we were actually always integrating with respect to the same measure. Just the fact that we do the integral in one order or the other order does not change, does not introduce the sign. But if we do the integration over surfaces keeping track of the orientation then this would actually become much more essential. Our next topic will be more general surface integrals where we will talk about orientation and at that moment this will become more clear. Right now it looks a little strange that why we didn't care about it before and now suddenly I care about it. Well because we didn't have to explain it before but now we have to so that's roughly the answer. That's right. Well the question is about line integrals. In the case of a line integral, right, we were supposed to take this vector field df dx i plus df dyj. And we were supposed to take the dot of this with dr. When you take the dot with dr you get this expression. I have explained now this expression as one obtained from F by this rule, by this procedure. The reason why I explain this is I wanted to show there is some uniformity between these two formulas. Without this explanation, these two formulas look completely disjoint, right? In this case I take this expression, in this case I take this, this looks very arbitrary, very ad hoc, right? So it looks like you have to simply memorize this expression. There's no other way to see like why is it dq dx? Why is it not dq dy minus dp dx, for example, right? That's a natural issue, natural question that arises here. What I try to explain is that actually this is not an arbitrary expression. This is an expression which can be obtained by some universal procedure. That procedure can also be used to explain the formula from last week, right? So you don't have to, this is something which is not in the book. And you don't have to memorize it. I think it's a good idea to memorize it because then you, if you're not memorized, to understand it because then you don't have to memorize this formula. If you understand this procedure this just falls out as I have just explained. But you don't have to do it. This was just an explanation. I wanted to give you an intuitive explanation of what this formula is about and the connection between the two formulas. But if you like, you may just memorize these two formulas separately and not necessarily think of them as two special cases of something general. Yes. Here, that's right. You can think of this as an integral of a vector field. That's right. Oh, you're asking what is an integral of a function f over a curve? Okay, so that's a very good question too. So remember, we also talked about, when we talked about line integrals, there were two types of line integrals, right? There were line integrals of vector fields and line integrals of functions. And then line integrals of functions was kind of, it's like one of those cousins nobody wants to talk about, right? So it's like, it's there somewhere, but we suddenly, we don't talk about them anymore, right? But actually, because the reason is, this is, it is very useful. Those integrals are very useful, but those integrals are not as easy to fit in this formula, so to speak. But since you've asked me, those were, strictly speaking, those were not exactly integrals of functions. We kind of thought of them as integrals of functions. But we never wrote integral of f, right? We never wrote integral of f. And this doesn't make sense unless you integrate over points. For example, here, you can think of this as an integral over two points of a function, which would be just evaluating one point minus evaluating the other point. We never write this as an integral over a curve or over a two-dimensional domain for that matter. Over a curve, we can only integrate fds. Or fd, you have to put some measure here. You have to put ds or dx or dy, something like this, right? So ds is actually also a differential form. It's a kind of object like dx and dy. And I spent some time about a week ago talking about this, right? So you have to keep, you have to remember this. And if you treat this not as an integral of a function but integral of this expression, right? Then it actually fits into this general theory, okay? Yes? Where did it come from? It was just, well, for example, when we go from a function to its differential, right? It's just, that's the way it is. That's the formula, which gives you the, formula for the differential involves partial derivative with respect to x dx plus the same for y, right? The formula was explained before because this was the formula for the linear, for the linear approximation of your function, right? Now we are developing a formalism which allows us to relate integrals over domains and their boundaries, right? And it turns out that within that formalism there is a certain procedure which you have to make starting with an object on the right-hand side to get the object on the left-hand side. And we have now seen two examples of this in action. This is the first example, which was last week. Here you actually have a function and on the left then you have to take its differential. And now is a new example. Here we start with a vector field with components p and q and then that's this expression. But if I had just written this expression without any explanation, you know, you would say, what does it mean? How did you get this formula? I have tried to give you an explanation of that formula in a way which would also make it parallel and similar and analogous to this formula, okay? But let's move on now and do a few examples of this. As you can imagine, this actually has some very, has important applications because you can now trade an integral on one side of this formula for an integral on the other side. And oftentimes one of these integrals is very complicated and maybe not even possible to evaluate easily. And the other one may be very easy to evaluate, okay? And so that's what we're going to try. That's what we're going to do and this is how we're going to apply this material, apply this formula. So here is an example. Suppose that your curve C is a boundary over a parallelogram like this. This is your curve C and it will be positively oriented, which means that you go counterclockwise, okay? This is your curve C. And suppose you're asked to evaluate the line integral over C of the following expression, x e to the minus 2x x e to the minus 2x dx plus x cube plus 2x squared y squared dy. So this is exactly the kind of expression that I'm talking about in this formula, in Green's theorem. You have general expression is p dx plus q dy. But as I already explained, in order to apply Green's theorem, you have to make sure that the curve over which you integrate here is closed, has no boundary, right? So let's see, can we actually have a chance to apply Green's theorem to this line integral? For this, we have to make sure that the boundary of this curve is empty. Is it empty? Yes, it is empty, it is a closed curve. So we are in good shape. We can actually replace it by the left-hand side of Green's theorem. Why would you want to do this? Well, someone might say, let's just, I'll just calculate this like this. And to which I would say, good luck. Because see, first of all, you'll have to break it into four parts. You'll have to integrate over each of the segments. Each of the segments, you'll have to parameterize. Once you parameterize it, you'll have to substitute parameterization here. And then go and find the antiderivatives of those functions. Okay, maybe in this case, it's not so bad, but I could have written something much more complicated, obviously, for which you wouldn't be able to find the antiderivatives, which I will on the exam, right? And that would be the first indication for you to know that you shouldn't do it in a straightforward way. You should not do it directly. It's just way too complicated. Okay, so you think of other ways to evaluate this. And here it is beneficial to use Green's theorem. What does Green's theorem tell me? It tells me that this is also, this is equal to the double integral over the interior of this curve, which is the parallelogram itself. This is D. Of what? I have to take, let me write it here. I have to take DQ DX minus DP DY. And what are the P and Q? This is P, right? And this is Q. So let's take the derivative of Q with respect to X. I get three X squared, right? Plus four X Y squared. And now I have to take also the DP DY. And P is this. And low and behold, P actually does not depend on Y at all. It only depends on X. You see, so we kind of lucked out here. This complicated looking function will actually disappear when we go to the other side of Green's theorem. Because this actually will be zero. This, this is zero. So I don't have to add anything to this. In general, I have to take DP DY. But DP DY is zero, right? Everybody agrees with that? DP DY is zero. It does not depend on Y. It's partially derivative with respect to Y is zero. So this is already simplification. So you can imagine, here is the quickest exercise that you can give to somebody. But kind of, which looks complicated, but actually the easiest to calculate. That would be if I actually wrote here something which only depends on Y. Let's suppose that I give you a formula here where I wrote here a very, very complicated function but only depending on Y. I would write some square root of one plus E to the sine Y or something, you know? And then you look at it so it really, it looks horrible. But then you apply Green's theorem and you see that DQ DX actually is zero. No matter how complicated this function could be in Y, if it doesn't depend on X, this would be zero. And then this would be zero and this would be zero and the answer would be zero. So you don't have to calculate anything. So you already see the power of this method that sometimes you get the answer right away even though the original integral looks very complicated. Well, this would be sort of a gimmick. If I gave you this, it would be sort of a gimmick. You'd be saying, okay, so it's just playing with us. But this is sort of a more serious example, right? Which is kind of intermediate level where one of the functions disappears and the other one does not disappear. So you actually have to calculate something meaningful. You have to calculate this. Of course, DX DY or DA if you want. So now we have to remember what we learned about double integrals. You have a question, right? So the question is, do I have to put DX DY or I can put, there are two issues here with DX DY, okay? So in this discussion of this procedure, you have to keep track of this. And that is a discussion which goes into the issue of how, of the meaning of the formula. So it's a kind of a fundamental stuff about deriving this formulas, right? This is one aspect of it. Another aspect of this is when we do double integrals, we treat them as iterated integrals where we first integrate over one of the two variables and then we integrate over the second variable. Both of them are okay. They don't introduce any sign. The switching that I'm talking about is a switching in the formula when you pass from here to here. But once you are doing a double integral, you can do it in either way. You see what I mean? So don't worry about it. The sign only happens in this discussion so as to explain why there is a minus here and there is no minus here. But when you're actually doing the double integral of an expression like this, the rules which we have used to evaluate these integrals and those rules involve integrating over X and then over Y or over Y and then over X. Both of these rules will give you the same answer. Okay. So let's actually evaluate. So you have, so let's say, so this is going to be, now we do it as an iterated integral. And you choose, as I said, you can choose whichever you like as the first variable and whichever you like as the second variable. So let me see, let me do, let me say do here Y from zero to two and there here the X will be from zero to five and then I will have three X squared plus four X Y squared. Here I have DX and then I'll have D1, right? So here you will have X cubed plus two X squared, Y squared from five to zero, which is, what's five cubed? 125, right? Plus 50 Y squared, right? Tell me if I'm making a mistake somewhere. And then you do the outer integral. I'm doing this because I'm trying to remind you because this is something we haven't done since I guess, since the exam. So you, I hope you haven't forgotten, but we certainly have to use this knowledge, right? In this discussion. So 125 plus 50 Y, and that's going to be 125 Y plus 25 plus 50 over three Y cubed, two to two and zero. So you have to 50 plus 50 over three times eight. Okay, whatever. So this is how you, this is how you compute it, yes? That's right. So this is a very easy way to calculate what could have, would have been a nightmare calculation if you had approached it in the most, in the most direct way, right? This is the answer for the integral of this vector field over this curve, over this curve. Now, we are lucky here. I want to emphasize it one more time. This is something which we cannot apply to a general line integral. We can only apply it to a line integral over a curve like this, which has no boundary. If my question was to calculate over this segment or even over this or even over this, you know, something which has a boundary, I would not be able to do this. Of course, what I could have done, say, if the integral was like this, this, this, I would say, okay, I'm missing this part. But if this part were there, it would be this. So let me say that this then is equal to this part minus this integral. So I would be able to trade this integral over three segments for a double integral plus or minus the integral over one segment. So in other words, not everything is lost even if your curve is not bounded. It is not closed. You could trade a complicated curve for an easy curve also. But here we're actually lucky because the curve is closed. So we just trade the single integral, line integral for a double integral, which is much simpler to calculate. Yes. I'm sorry? Does what look like a square? Does this curve look like a square? Does it have to look like a square? First of all, it looks like a parallel graph, not a square, right? At least I meant it to be, to have an equal size. But a curve, to answer your question, let's look, let's do another example where the curve is not a square or a parallel graph for that matter. Let's do another example. Suppose that your domain, you have to evaluate a line integral of something which is not, which is not, which is not, which has two components. So I would like to draw a picture like this. So I would like to say that here, again, I can use Green's theorem. This is equal to p dx plus q dy. But where the region, my two-dimensional region will be the annulus, which is confined between these two circles. The first circle, let's say, would have radius one and the second circle will have radius two. In all the previous discussion, I assumed that my region, so to speak, didn't have holes. In other words, the boundary had only, consists of only one, a single curve. And now this region sort of has a hole in here, right? So the boundary now consists of two circles. There is an outer circle and there is an inner circle. But it turns out that Green's theorem is applicable to such regions as well. But you have to be very careful with the way you put orientation on the boundary. The orientation on the exterior circle we have already figured out is counterclockwise. And that's going to stay the same in all cases. But now the question is, which orientation you should put on the inner circle? Clockwise, exactly. Because see, this is exactly a situation where the counterclockwise rule doesn't apply. Counterclockwise rule only applies to the exterior boundary. But for the inner boundaries, it's going to be the opposite. It's going to be clockwise. And the reason is that the universal rule involves this. You have to think of a person walking here. And if you walk, and the domain has to be always on your left. So that means that you have to walk counterclockwise outside. But if you go counterclockwise on the inner circle, on your left will not be the domain, but will be there, something else, right? So you have to walk like this. So like this here, I will not draw the other way. But like this here to have the domain on your left. So, which brings us to the second example in which let's say you have some function. Oh yes, here it is. Bring me to the second example where let's say you have the following concrete functions. So let's say this C now represents the union of both of these boundary curves, right? Taking with the appropriate orientation. And so let's say you're evaluating the following thing. You have cosine x dx plus x square sine y dy. So this is P and this is Q, right? So we use Green's theorem. And Green's theorem tells us that this is going to be an integral over this analysis of dQ, again. dQ dx minus dP dy, which is 2x sine y minus dP dy. So again, I chose a function to be such that actually this disappears to make it easier, okay. And now I need to evaluate this double integral. And of course, here you have to remember all the tricks which you've learned in dealing with double integrals. When you do double integrals, you have this entire toolbox available to you. You don't have to do it in a straightforward way, first text and why. You can, for example, use polar coordinates. And surely this region calls for polar coordinates, right? So what you need to do is you need to switch to polar coordinates and the integral that you will get will be as follows. You will have zero to two pi and over theta. And then from one to two, I recall that this has radius one, this has radius two. So theta you integrate over from zero to two pi, but r you integrate from r to two. And then you substitute the polar coordinates. So you get two r cosine theta, right? And then you have, you have sine of r sine theta, right? Then you have r dr d theta q dx. Maybe I made it too complicated. Let me just get rid of this. It's just for the sake of the, for the sake of practice. So let's just get it to, I think it's good enough, okay? And then, oh, but don't forget to put an extra r in da. So then you have this extra r dr d theta. And then of course this you can already do very easily. Okay, so we'll continue on Thursday.