 Okay, how are you guys doing? Did you have a good Thanksgiving break? Yes? Are you ready for the last week of classes? Yeah, okay. It's hard to believe it is the last week. But actually next week I will give a bonus lecture. I will give a review lecture on Tuesday next week. And also I remind you that there is a lot of materials for the final exam, which are available on this page and also on the class home page, which I hope by now you know how to find. All right? And now we go back to the study of Stokes' theorem. So we've been slowly laying the foundations, the necessary foundations for Stokes' theorem. And finally it's time to sort of reap the benefits of all of that. And so finally today we'll be able to state the result and to see what it means and what the applications of this result are. Okay? And but first of all I would like to revisit the material which we talked about last week. Okay? A week ago we talked about surface integrals. Integrals over the surfaces. Okay? And the point was that actually we studied two types of integrals. Two types of integrals over general surfaces in the three dimensional space. As a good working example of such a surface we'll look at the upper hemisphere. It's a surface with a boundary which is a circle. So you can actually, you can think that there is a, it's actually a part of a sphere which is centered at the origin. So you could also imagine that there is a coordinate system here, x, y, z. And so the surface, what I'm talking about is this upper hemisphere, the surface of the upper part of the surface of the sphere. Okay? Which has a boundary which is a circle. So what we're trying to understand is how to integrate over such surfaces, surface of this type and relate integrals over such surfaces to integrals over their boundaries. That's the, that is the point of Stokes' theorem. Okay? So let's denote this by M. And M stands for membrane because I explained last time that integrals over surfaces of the second type can be viewed as, as fluxes over fluid through membrane. But here I want to recall that there are two types of such integrals that we have looked at. The first type, the first type you pick a function and then you integrate over M. This function with a measure which we denote by ds. We talked about the interpretation of this. This has to do with the surface area and more generally with the mass. So this represents the mass of, of the surface. Think of the surface as an aluminum foil, say. And f is a mass density function, density function. And when you integrate the density function, you get the total mass. It can also be a charge density function. Then you would be integrating total charge. So that's the first type. The important thing here is you integrate a function. And the second type is when you start with a vector field. So it's not a function. It's much more than a function. Actually, we represent it by a triple of functions. It has components p, q, where each of these three is a function of its own, on its own right. In this case, we also have a double integral, which we discussed last time. And that's what's called a flux, also, a flux of a vector field, or a surface integral of a vector field, which we denote like this. E dot ds, ds. So this is sometimes called flux of E through this membrane. And it represents the rate of flow of a fluid through a membrane M if f is the velocity vector field for this fluid. I explained last time. And now, of course, even if you were not here, you can actually watch the lecture on multiple places. I think it's a good resource. It makes it easier for me to just say, I did it last time. If it was not recorded, that would be a little unfair. But now I know that if you don't remember, you can always go back and watch it again. Now, how to compute this? So the key to computing this is to, as well as the other one, is to parameterize the surface. So parameterize it, which means that we write x as x of some u, v, and y as some y of u, v, and z as some function z of u, v, where u and v belong to a certain domain now on the plane. Well, when I say parameterize it, I mean parameterize m. And so when you parameterize it, you actually get a very explicit formula for this integral as just a double integral over d. So we get an integral which is now in the realm of the previous chapter of this course, something which we studied before. Now, the key to this is the following. The way we introduce this integral, we actually wrote it as an integral of the first type, wrote it as an integral of the first type over m, where what we do is we take the dot product between e and n, where n is the unit normal vector field to the surface. When we do that, when we take the dot product, we convert our vector field into a function. This becomes a function, and that function we can now integrate in the sense of the integral of the first type. So this is how we introduced it. Today I will explain a different way to think about this, where you don't have to use the concept of the integral of the first type to understand the integral of the second type. But this actually brings up the question as to what this n is, this normal vector. So in my picture, let me draw this picture again. In my picture, this vector field n would look like this. At each point of the surface, we would have a vector which sticks out and which is normal, or normal means perpendicular, perpendicular to the surface. If you have a surface and you have a point on the surface, you can talk about the tangent plane. We talked about this, tangent planes. We even know equations for tangent plane. How to find equations for tangent planes? So we also have the notion of the normal line. A normal line is perpendicular to the tangent plane. So at each point you have a tangent plane. I'm not going to draw this, not to mess up the picture, not to make it too heavy. But there is a tangent plane here. But there is also a normal line. So it's perpendicular to the tangent plane. When something is perpendicular to the tangent plane at the point, we might as well say it's perpendicular to the surface. What else can it mean to be perpendicular to the surface? It just means being perpendicular to the tangent plane. Now, on this line, there are many vectors. If you have one vector on this line, you can take any multiple of this vector. So what we'd like to do is we'd like to specify a normal vector at each point. And we want to specify it with as little ambiguity as possible. So the way to get rid of the ambiguity, get rid of the redundancy is to normalize the vector, to make it a unit vector. If you have any vector, it has a certain magnitude. If it's non-zero, you can divide by the magnitude and you get a vector of length one or magnitude one. Such a vector is called unit vector. So out of all the vectors along this normal line, we can say let's take a unit vector. The problem is that when you do that, you still have a small ambiguity. It is a much smaller ambiguity. Instead of infinitely many vectors, you end up with two. There are two possibilities. There is this one, but there is also one which goes in the opposite direction. So it's kind of a mirror image of this one. And just by saying it's a unit vector, we can't kill this redundancy. Both of these are bona fide unit normal vectors to the surface. If you think about this upper hemisphere, one of them sticks out, goes outside, and the other one goes inside. So there are these two choices at each point. At each point, there are two choices. And what we'd like to do is we'd like to make a choice at each point along this entire surface so that this choice is continuous, smooth. In other words, we don't want to be in a situation where we choose a unit vector at this point to go outside, outward, and then at the nearby point to make it go inside. That wouldn't be smooth. So when we talk about a unit normal vector field, we mean that we want it to vary in a smooth way. So that's what this n is, n of this formula. This n is the unit normal vector field. And then I will emphasize varying smoothly, smoothly along m. Such a vector field is called orientation, orientation. I would like to contrast this to the notion of orientation of a curve. In the case of a curve, if you have a curve going from point A to point B, we talked about its orientation as a direction along the curve. And if the curve has boundary points, specifying orientation is the same as specifying which one is the initial point and which one is the end point. So this would be the initial point, this would be the end point. If the curve does not have a boundary, for example, a circle, we could say it's counterclockwise or it's clockwise. But let's look at this one which has end points. It has a boundary. In this case, by analogy to this, we observe that actually in this case there are also two possible orientations. We can go this way or we can go that way. So there's a lot of similarity with the notion of orientation for curves and surfaces. The main difference is that orientation for curve goes along the curve. You kind of flow, you go with the flow, so to speak. And here you also go with the flow, but the point is that the flow goes across the surface usually. You don't want to flow, you can't flow along the surface in some sense. I suppose you can in some, maybe, but it makes much more sense to flow across the surface. For example, as I explained, you could have a pipe, you know, and this would be a membrane which kind of covers this pipe and you would be measuring the amount of fluid which passes through it. So it's much more natural to talk about orientation across rather than along. In the case of surfaces, unlike curves for which going along means a lot more sense. Now, so this is one important difference between orientations for curves and surfaces. There is a similarity, right? There's a similarity that in both cases there are two choices. So here there are two choices. Here there are two choices. In this case, there are also two choices. One indicated in white and one indicated in red. And if you think about it, once you choose it at one point, then basically there is no choice at nearby points because you have to wear it smoothly so you know it that nearby it will be like this, nearby it will be like this, like this, like this. Because you don't allow jumps. If you don't allow jumps, once you know it at one point, once you specify it at one point of your surface, you know it everywhere. But then the question is, can we actually define it everywhere in a consistent way? And this is, here we kind of get surprised because in the case of a curve, there are always two orientations. For any curve you can always choose two orientations. One goes this way and one goes that way, right? But in the case of surfaces, there actually exist surfaces which have none, which don't have any orientation. And this is actually kind of cool. You must have heard about Möbius Strip, right? Have you heard about Möbius Strip? Yes. Okay. So here is a little demonstration for you. So we need to close up. Sometimes I delay, so I'm saying it and I'm waiting for a few seconds. Okay. So here is a surface. I mean this is a surface. I colored it in two ways. One side is white and the other side is pink. So now what I'm going to do is I'm going to make a surface. I can glue a surface out of it in different ways. So for example, I can do glue it in this way, right? If I glue it in this way, it's like a cylinder. It's like a little cylinder, not so, I mean, kind of a low height. And in this case, if you look, you can see that actually it clearly has two sides, right? Clearly has two sides. There's an inner side and there's an outer side. One is red, one is pink and one is white. So when you talk about orientation, you can also think in terms of choosing the side of the surface. I was talking about vectors sticking out, but once you have a vector sticking out, you have the notion of what's outside and what's inside. So let's say that if it sticks out this way, then that's the outside of the surface. And if it sticks out this way, then that would be the outside and that would be the inside. So it's like, you know, if you put your t-shirt inside out, this kind of stuff, right? So that's a clear notion, inside or outside of the surface, right? So that's what we're talking about, orientation is saying which side we consider as inside and which one you consider as outside. Because of course, our priority is not clear. You can choose, you can make your choice, right? And sometimes you can have a t-shirt which actually looks even better if you wear it inside out. It kind of looks cool, right? But imagine you have a shirt which only has one side, a t-shirt which only has one side, right? So that would be difficult to wear. Now, but actually you can create such a, not necessarily a t-shirt, but you can create such a surface very easily by taking this same strip and gluing the ends in the opposite, kind of by twisting by 180 degrees, okay? If you do that, you say, okay, so let's say that this is the outside, right? But then you trace it, if you trace it, the outside, you trace it, you trace it, and then suddenly it becomes white. And when you make the full circle, you come back on the opposite side, right? So if this was outside, outside, it should be the notion that if we follow it, you know, on my shirt, if I follow it everywhere, it's going to be on the same side always. I will never be able to jump on the other side. One would hope anyway. And that's the way it's supposed to be, right? But here you go, you make a full circle and you come back on the opposite side. So that means this is outside, but also this is outside, right? So that means, but then which one is it? So it means at this point it's not defined. There is no global notion, there's no global notion of inside and outside. You see what I mean? So in this case, we will say that the surface is not orientable. And this is a really remarkable phenomenon. Again, in the case of curves, all curves are orientable. It's just that there is a redundancy, there are two possible, or ambiguity, there are two possible orientations, you have to make a choice. For surfaces, actually, we don't even have that luxury. There are many surfaces which are not orientable, for which you cannot consistently say which is inside and which is outside. And that's the example, okay? Any questions? So in this case, if the surface is like this, there is no notion, there is no orientation, there is no unit vector that you can make which varies smoothly and gives it a normal vector at each point. Because, you know, if I were to give such a vector, let's say, again, here the vector would be sticking out this way, but if I varied smoothly, I would be forced to come out here on this side. So that means it becomes ambiguous. And at this point, there is no, it has to be simultaneously this and that. And that's not allowed, right? So it does not have a consistent orientation. So in the case of such a surface, this integral is actually not well defined. You cannot set up an integral of the vector field over a surface which is not orientable, which does not have this type of orientation. In other words, the consistent notion of inside and outside. So we have to be aware of this. Now, the good news is that most surfaces are orientable. More surfaces that are of interest to us. For example, if we are thinking about really a fluid flowing in the pipe and you're thinking of something which is like a cover at the end of the pipe, it is going to be orientable. So it's not really going, we're not going to be too handicapped by the fact that, by the fact that for non-orientable surfaces, this is not well defined. Most of this, all the surfaces we will consider in this class, for sure, will be orientable, okay? But we have to be aware of the fact that actually it's not, we should not take this for granted. A surface that may not have a particular, even one orientation. Of course, if it has one, it will have to have two, right? The point is that if n is one of them, the negative n will also such an object, also such an orientation. So it can't be that you have just one orientation. You either have none or you have exactly two always, okay? And in all of the cases which we will consider here, our surfaces will have two orientations. And this will give us a little bit of a headache, because when I write now the Stokes formula, I will have to be very precise in specifying how the orientations on the left and right-hand sides, so as to make the equation correct. So let's go back to this integral. So the integral is actually, let's suppose that it is orientable. So this is important. It's not a given that the surface is orientable, but let us assume from now on m is orientable. That is to say m has an orientation vector field, n, and then it has necessarily two of them. It has two orientations, n and negative n, okay? Let's assume that. And so pick one, pick one. We will pick one. It's pick one orientation, which is what I call n. In this case, we have this integral. We have this double integral. Let me give myself more space like this. So you have E dot ds, which can be defined by the formula here. I just re-write it. E dot n ds. Now, there are two choices. So if you take the second choice, the second choice means instead of n, you take negative n. If we choose negative n, we just get minus of what we had before. If we choose the other orientation, this will just result in the overall sign. So in that sense, you can say, well, it's not such a huge ambiguity. You don't get sort of a totally different answer. You get something up to a sign. But, you know, it wouldn't be serious if we, in this course, we said all our results were defined up to a sign. We have to be precise. We have to be able to say, is it 10 or negative 10? Because oftentimes, you know, your life depends on it to know it's positive or negative. So that's why we take this seriously and we say the definition of such an integral has to include the orientation. So if you do a homework problem, it has to say the integral of the vector field over the surface with respect to which orientation. And then there is standard terminology. Standard terminology is you can say orientation is upward. Upward means that the vertical component of the vector field is positive. So the white orientation on that picture is upward. Or it could be downward. That would be the red one. And oftentimes, if we have a closed surface, for example, not the entire sphere, but say, sorry, not the upper hemisphere, but the entire sphere, you could say outward or inward, which also has a clear meaning. Okay, so let's pick one. How do we actually compute this? We compute it by parameterizing m by choosing two auxiliary parameters, u and v. Right? And then when we parameterize it, so this integral becomes, we can write it very concretely as follows. It's e dot r u cross r v over d. Where d is a parameterizing domain on the plane, on the u v plane. Parameterizing domain. On the plane of this coordinates u and v. What is r? So r is a vector which we get by combining the three functions which parameterize our curve. x of u v plus y of u v j plus z of u v. So these are the three functions. You put them together into a vector field. r sub u simply means that you take derivatives with respect to u. Maybe you write dx du plus and so on. And likewise r sub v means that you take dx dv i plus and so on. So once you have your parameterization, you have this r and you have r sub u by taking simply partial derivatives of these three functions with respect to u and you have r sub v. So then you can just take the cross product. And when you take the cross product, you get another vector field which you can then dot with your original vector field. This will give you a function which you then integrate over the parameterizing domain. And of course the point here is that both the original integral and this integral are integrals over surfaces, but there is a huge difference. This surface is actually on the plane. It's flat. So by this formula we express a complicated integral over a very complicated surface which could be very curved like a sphere or part of a sphere as an integral over a domain on the plane which is flat. So that's the definition which we talked about last time. And then in some special cases where, for example, your surface is a graph of a function, we have some, the formula simplifies and we have various formulas which allow, which kind of give us a shortcut to the answer. So finally we are ready to state the Stokes theorem by using this notion of a surface integral. So let me draw this one more time. So you have our surface m and we have the boundary, the circle, the circle boundary which I'll call B of m. So on the right hand side of the Stokes formula we will have a line integral over vector field over this boundary. So we'll have a vector field f which will be defined in a three-dimensional space. In particular it will be defined on this entire surface. And on the right hand side we'll simply take the line integral of this vector field over the boundary. So this is something we learned before. So I don't have to say much about it. This is just what has become by now the usual thing, the line integral of a vector field. Which is again, to calculate it, we need to parameterize the curve and then there's a very simple formula which allows you to take this integral. And now on the left hand side we are going to have an integral which is a double integral and it's going to be an integral over m. And then the integral, it will be the integral of another vector field and that vector field will be nothing but the curl of f. So you have your vector field and it appears on the right hand side. But on the left hand side you have another vector field. This is something which I was denoting by e in all my previous formulas. So we have double integrals of a vector field e over surface. But now take as e the curl of your original vector field which is given to you from the beginning, f. f is given, take its curl, that's another vector field and take the integral of the curl. A remarkable statement is that you always have this equality. You have this equality of a double integral and a single integral over surface and over the boundary of the surface. Now what I have said is still incomplete. And the reason it's incomplete is that both of these integrals depend on the choice of orientation. Did you have a question about this? Well, curl of f would be what I call here e. So I purposefully called this vector field e and I called that vector field f because on the left hand side when we do the double integral we are actually integrating not f but curl f. So that's why I'm using a different notation here. So it's not to confuse you. We are given a vector field f but we are sticking that vector field on the right hand side into a single integral, into a line integral, right? Whereas on the left hand side we do curl. And curl is another vector field which we know how to calculate. And today we'll actually see what the meaning of that curl is and we'll get a better explanation of what the meaning of this vector field is. Okay, so that's the formula. But as I said, on both sides we have two integrals which as I explained requires the choice of orientation, right? So in other words, depending on how I choose orientation on m I will either get plus minus some answer. I will get some answer or it's negative. And if I want to pin it down I have to say which orientation I should take. Both integrals are well defined. There's no canonical choice of orientation, right? There's no canonical choice of orientation. So there are two possibilities and there are two possible answers. They differ by sign. But I have to say which one I should take for this to make this well defined. Likewise, here I also have two different answers. In the case at hand I have this circle. I can choose orientation which goes, you know, counter-clockwise or clockwise if you look from above. Which differ by sign. If I'm saying that these two things are equal to each other I better be precise and say how the orientations should be defined so as to make this into an equality. Because if I make the wrong choices instead of getting this formula I'll get this equal to negative of that. In other words, if I wanted to be careless about orientation I would not be able to state this formula at all. The best I could do I would be able to say this is equal to plus minus this. And then if I did that I should be fired. So we don't want this to happen. Well, maybe you do but I don't. So that's why I'm going to explain how to choose orientation so that we don't have this ambiguity. And of course the point is that you probably guessed already because this is very similar to the way, to the problem which already arose when we talked about the Green's theorem which is actually kind of a little brother of this talk's theorem we'll talk about in a minute. In Green's theorem there is also this issue of orientation. And the way we handled that was by saying that was the following. Let me do it here. So in Green's theorem we also have a two-dimensional region but it was flat. This is flat, this is not flat. Well, I have drawn it on a flat blackboard but 3D, you know, I was just reading about this new movie, Avatar, you know, it's going to be in 3D. So I was thinking, I was thinking, you know, maybe some years from now these lectures like calculus will be given also in 3D. So you will all be wearing these cool glasses and you will actually see this three-dimensional, I will be able to draw three-dimensional surfaces. I should talk to Cameron about this. I'm sure he's a huge fan of multivariable calculus. Or if he isn't, he should be because he uses a lot of special effects and of course how can you do special effects if you don't use calculus? Anyway, so in Green's theorem you have a region like this and the way we handle this, as we said, once you have this region, the orientation you define on the boundary is such that when you walk on the boundary the region is to your left. Okay? If you walk on the boundary, the region is to your left. So it would be like this. It would be counterclockwise in this picture. So I would like to say the same. I would like to say that if I walk on this boundary, if somebody walks on this boundary, then the region should be along the orientation, with respect to the orientation. The relation should be such that as somebody walks along this boundary, the region itself, the surface, is to the left. But the question is how do you expect that person to walk on this? There are two possibilities, right? It could be that they walk like this or they could be walking from the other side of the blackboard. You see? A priori there is no choice. I mean, there are two choices, a priori, right? There is no canonical choice. There is no given choice. In the case of a plane, we do have an obvious orientation. We do have a special orientation. And the reason is the following. Because there is an orientation which kind of sticks out towards us. Another way to say it is that there is an orientation which we call K. So we draw this coordinate system, which is I, J, and then there is a K which is obtained by the core screw rule. The orientation of the right hand rule, whatever you call it. Remember this picture? This was one of my fondest memories of this class. This discovery of this visual aid. This is how you show the normal. So the point is that you see, you can't just say, since I can actually move this, you can already appreciate the problem in giving it orientation. It could, orientation could be like this or it could be like this. But the point is when we actually draw something on a two-dimensional plane, we look at it from a certain direction. We have to be looking at it from a certain direction. We either are on this side and we're looking on this side or we are there. And let me tell you, there's not a whole lot of space behind that board. So we should be on this side, right? Once we are looking at it from this side, we can give it any coordinate system, any orientable coordinate system. In fact, this is not even canonical either, but I can rotate it and so on. But whichever coordinate system X, Y I choose, the cross product of the I and J, we'll be looking in this direction. We'll be sticking this way. So that's why when we have a region on the plane, we actually have a canonical orientation. That's what we call vector K. K being the cross product of I and J. In this order, first X, this is the unit vector in the X direction. This is the unit vector in the Y direction. We take the cross product in this order, right? And so that's what gives us this Z axis, the unit vector along the Z axis. So we're kind of lucky in this case, we don't have to make a choice. The choice is made by the fact that we are actually observing it in a certain way. We are looking at this blackboard in a certain way. But if we have now a surface in a three-dimensional space, for example, I could just put this blackboard somewhere in the middle of this class, and some of you will be on one side and some of you will be on the other side. So for some of you will be looking from one direction, some of them will be looking from the opposite direction. So for some of you will be one orientation, for some of them it will be natural to take the other orientation. So in general, there is a choice of orientation. But once you make that choice, like this, you can say if a person walks so that their head is, you know, points in the right direction, up, then the domain is on the left. Then it actually makes sense. So somebody could be walking like this, their head would be this way, right? And so they will have the region on their left. So that would be, we get the old rule. If on the other hand I would just say let's choose the other orientation which goes in that direction, I would still have the notion of somebody walking along the boundary, but that person would be on this side, right? And then it would still make sense to say that the region is to their left, which would then mean the opposite direction. Well, you would have to believe, trust me, this is how it works, but you can just imagine it if it were made of glass, you will see it, right? So the rule in general is like this. If a person is walking along this, pointing in the direction of the orientation, if this is your orientation, and you're not upside down, you have to be, you know, oriented in the right way, your head should be above your feet, basically. With respect to this orientation, I'll try to, not to make too many jokes about this, but I'm sure you can do them yourself. So if you walk this way, the surface should be to your left, so this is how it works. And if I chose a different orientation which would be inward, then I would have to draw this little fellow pointing in this direction, and then it would have to walk so that the region is to the left, okay? So that's the rule. Is that clear? Okay, so this is pretty clear. So that's the rule I choose. If I choose that rule, which it will give me a consistent formula, it will give me a choice. Whatever choice I take here, once I make the choice of orientation, and I get orientation on the boundary as well. And it is for these two choices that these two integrals will be equal to each other. So that's the statement of Stokes' theorem. Now, the next thing I would like to do is I would like to kind of try to demystify this formula because at the moment it looks pretty bizarre in a way because you have this strange, you have two different types of integrals. So one of them is of the vector field. The other one is of this curl, okay? So I would like, what I would like to do is I would like to explain how this formula actually fits in this general principle, which I talked about earlier, okay? Of which we have now seen several examples. So I would like to convince you that this formula is actually a special case of this general principle, just like the fundamental theorem of four-line integrals and the Green's theorem. So that's the next thing that we're going to do. So in order to explain this, I would like to actually give a slightly more precise formula for a surface integral. And so this is going to be a slightly long and somewhat boring calculation, TDS calculation. But bear with me because at the end of the calculation, I hope that I will be able to demystify this formula and you will see that really it is part of a pattern which is much more, how should I say? Much more convincing, okay? So the first thing I'd like to do is I want to go back to a general surface integral, like this, a general. So here E doesn't have to be the curl. It doesn't have to be the curl. It could be some vector field. And I would like to write a slightly more explicit formula for it, assuming that I have a parameterization. So let me recall what I wrote on the top right corner there. I wrote that this is equal to the integral over D, the parameterizing domain of E dot r u cross r v. Where r u and r v, you have, well, the beginnings of r u and r v, you have on that blackboard. So now I want to actually calculate r u cross r v for you, okay? So I'm going to calculate what r u cross r v is. So this is just the usual cross product. This is just the usual cross product where I put d x, d y, d u and d z, d u in the first row. And I put d x, d v, d y, d v and d z, d v in the second row, okay? So now I just open, I just write down what it means. So let me just start with i. So this is going to be d y, d u, d z, d v minus d y, d v, d z, d u. This is times i, okay? Maybe here plus. So now j. So I have to look at this and I have to remember the sign. So it's going to be d z, d u. It's going to take a little time. But you will like it when it's over. Trust me. So it's worthwhile to do this calculation. d z, d v, d x, d u, j. You have to do it once to believe it. And then after this, a lot of things become simpler. Plus, finally, I have to do the k. And the k is d x, d u, d y, d v minus d x, d v, d y, d u, okay? Check this, okay? Are you with me on this? This is good? You can only, if you agree then, you know, you will now have a chance to re-disagree with me later. You have to, this is your last chance. Okay? So if we agree on this, have you ever seen this formula before? You're under oath. So have you ever seen this formula before? What is it called, this kind of expression? Come on. It starts with a j. It's a Jacobian, okay? It is a Jacobian for the coordinates y and z with respect to u and v. This is a Jacobian for the coordinates z and x with respect to u and v. And this is a Jacobian for x and y with respect to u and v. Is that, everybody agrees with that? So I actually can rewrite this in a compact, in a more compact way by using the notation which we introduced when we talked about Jacobians. Okay? So this is a notation for the Jacobians. You put a del like this and then you put y and z over del of x, y, u, v. This is a notation for the Jacobian, which we have used extensively about months ago. Okay? So this is times i plus we have the Jacobian for z, x over u, v, j, plus the Jacobian over x, y, and then you put k. So that's, this is actually the explicit formula for this cross product. And this cross product enters in the definition. Now let me actually take the dot product so that I get the entire expression. For this I have to also say what is e. So let's say that e is ai plus bj plus ck. Then what do I get? I get e dot r u cross rv, which is my integrand in the surface integral, is a times d times this Jacobian plus b. Okay? So that's the formula. Here is actually explicit formula for what you should integrate when you take double integral of a vector field. Explicit formula, which is written entirely in terms of the components of your original vector field and the Jacobians of the coordinate change from u, v to all possible pairs of coordinates amongst x, y, z. Now, how can you remember this without memorizing this formula? It's very easy. So first of all you have a, b, c, right? These are the components of the original vector field. And then you also have x, y, and z, the three coordinates. So you have to remember the structure of the formula, that there are three terms. Each term corresponds to one of the three components of the vector field. And then you have to remember which Jacobian to put. You see there are three different Jacobians. But there is also a subtle point, which is in the Jacobian you also have to remember in which order you put the coordinates. You may remember from doing double integrals by using general coordinates changes that if you switch the coordinates the Jacobian gets a minus sign, right? The Jacobian gets a minus sign. So here it's important that I put y, z over u, v and not z, y over u, v, right? So first thing you notice that if you take a, the variables which get involved, so a corresponds to x. So to get the variables involved you have to cross it out. So you see you have a, y, z here. Then you have b with x, z. And c with x, y. So that the two coordinates which you take, which appear together with the component of the vector field are the remaining two components, right? So a is the x component. So the coordinates which are relevant to this term of the formula are y and z. And then you have to ask yourself in which order. And the order should be what mathematicians call a cyclic order. Which means this is first, this is second, this is third, this is fourth. And so on. So it kind of goes in a circle. So for example, y and z have to appear in this order. Y, z. But z and x should appear, you are at z. What's the next one? The next one is the first one. It's a cyclic order. So it's like z, x. And then like x, y is clear. So x, y, z, z, x. And then you put the remaining component of the vector field. So that's what the meaning of this formula is, okay? So that's the first thing we should agree on. The double integral can be written in this form, okay? But now let's go back to Stokes formula. In Stokes formula, there is an additional complication, as if this formula was not complicated enough, right? We actually have to take the surface integral, not of the vector field itself, but of its curl. So gee, we have to do one more cross product, right? But actually the funny thing is that when you do that cross product, and you combine it with this stuff, which you can imagine is going to look horrible, right? Actually it will look better, in some sense. Or you will get something which will have a much more conceptual explanation. So let's do that. Let's do that because I promise that I will explain it to you. So let me explain this to you. So we're halfway there. What we need now is we need to take the, supply this formula to the left-hand side of Stokes' theorem, where we are just doing it to the curl of the vector field, and not to the vector field as before. So if E is curl of F, then what does it look like? We have to again do this. i, j, k. Then we put d dx, d dy, d dz, and then we put pqr. So what do we get? We get dr dy minus dq dz times i, plus dp dz minus dr dx, plus dr dx minus dp dz. But please check that I did it correctly. dr dz, because if I mess up one term, then it will not come out, right? So dr dx. Okay, good. That's right. And in fact, for good reason, because this is expression in Green's theorem, right? So we know that Green's theorem actually is a special case of this when only this term shows up. That's right, dq dx. And here dy, exactly. Good. I think now it's correct. So now I want to put this in here. In other words, this will be my a, and this will be my b, and this will be my c, okay? So you would think it will look awful. And in some sense, in some sense it will. But then I will explain another way to get that same expression in a much more conceptual way, so to speak. So that, and this will, at least for me, and I hope for you, this will demystify this expression in this result. Okay, so let me put this into this formula. And I really want to do it in real time instead of showing it to you already written before, because if I show you, you'll be like, okay, whatever, there's some derivative, so he says it works and it works, but I want you to really follow me and to really see how it works. So dr dy minus, so please keep track of what I'm writing, because as you see, I can make mistakes. But now I want to do one more thing, actually to simplify, to kind of accelerate this slightly. At the end of the day, I'm going to integrate this, right? I'm going to integrate this. I'm not just taking this expression, I'm actually integrating this. And I'm integrating with respect to du, dv, right? So let me actually then multiply this by du, dv. But if I do that, then this becomes a dy, dz, right? Because remember, let me write it here. When we did double integrals, this was the formula which we used. If we integrate, well, the way we did it was for x, y. We did it for dx dy. But the same would apply if I do it for xz or yz, right? In other words, when we integrate a function in x and y, we said it can be written as an integral over unv, but you have to insert this factor. I'm sorry. This doesn't look right, okay? What I mean is of course this, right? So the way we wrote it is we said that when you have a double integral of some function f, this is equal to... But what I'm doing is just stripping it off the integral and remove this function. I'm just saying that this is really the identity which we used and we proved it, right? So what I'm going to do is that whenever I have this ratio, because I now multiply overall by du, dv, I'm just going to replace this by dy, dz. I'll replace this by dz, dx. And I'll replace this by dx, dy. Can I do that? Are you agree that this is okay, legitimate? Yeah? Okay. So this was the purpose. I'm just trying to remind you that this was the purpose of introducing these decobins in first place because we wanted to replace things like dx, dy by du, dv. And we knew that this was the price to pay. We had to insert this factor. So I might as well replace this expression by this expression. And likewise for the remaining two summits. So this is a formula I'm going to use and I'm going to now take this formula and say that A is... A, B, C are those components which I got from the curl. So now here's what it's going to look like. dy, dz. So B now dp, dz, rdx. dq, dx minus dp, dy. Check that I didn't shorthand you. Shortchange you. So this is the formula, right? I just take each of these three components. qd, d. So the easy way to check is that it should be yz, yz. This should be zx, zx. This should be xy, xy. And then it should be dr, qp, rq. So this is what I get. So this is what this expression stands for. When I integrate this, I have to take this... I'm integrating this expression. This is the left-hand side. This is the left-hand side. More precisely, if I put now the double integral over m, this is the left-hand side of stocks. Of stocks there. And now I want to write the right-hand side. And the right-hand side, of course, we've known for a while. The right-hand side, I recall that dr is dx i plus dy j plus dzk. And so f, f dr. And f is pq. I should have said it, but f is pqr. But this is what I was using. So f is pi plus qj plus rk. So f dot dr is, of course, then just pdx plus qdy plus rdz. So the right-hand side is the integral of the boundary of m of pdx plus qdy plus rdz. This is actually a familiar expression for line integrals. For line integrals, we've known this all along that line integrals could be written in this way. If we did it often in two variables, then it would be just pdx plus qdy. For three variables, it's pdx plus qdy plus rdz. So what we are saying now, what folks and others have taught us is that this integral over the boundary is equal to that integral over the surface. So how can we relate these two expressions? So there is actually this very simple way, which I already explained in the context of Green's theorem. And this involves the notion of differential. Remember the notion of differential of a function? The notion of differential of a function was like this. If f is a function, then its differential df is df dx x plus df dy dy plus df dz. You remember this? This is differential. So I would like to apply the same operation d2 through this expression. So now let's apply... So maybe I should have written this. This is the right answer. So now what I want to do is I want to apply d2 pdx plus qdy plus rdz. What am I going to get? The rule is very simple. Just apply it to the function which stands in front the same way you would apply it here when you can calculate the differential and just multiply by whatever it remains. For example, dx here. So let's apply it to the first one. What are we going to get? Let's apply it to this. So it means that I take the differential of p. Let me do it slowly. I take dx plus dq dy plus dr dz. So what is dp? What is dp? dp is dp dx dx plus dp dy dy plus dp dz dz. And now I have to multiply this by dx. So let me just put it here. So I have dx dx dy dx. Right? This is what the first term will give me. I just take the differential dp and I multiply by dx. So there is always this dx. But otherwise it's exactly the terms that I get in the differential. Is that clear? Any questions? That's the expression. Let me do the second time. dq. So it's going to be dp dx dx plus dq dy dy plus dq dz. Let me open the brackets. When I open the brackets I have to put dz here. And here I have two dz's actually. dy. Right. Sorry. I'm already jumping to the last line. dy. So now the last line. dr dx dx plus dr dy. Let me leave some space for it. dr dy dy plus dr dz. Let me open the brackets. dz dz. Okay. Now we're almost there. So first of all if you have the same variable twice you have to get rid of it. I already explained it once. Because if you have two coordinates which are not independent from each other you cannot think of them as two new coordinates. In other words the area of the parallelogram which they span is zero. So this has to go. This has to go. And this has to go. So I had nine terms of which only six terms will survive. And I claim that these six terms will be equal to this. For the six terms here. Okay. So let's check that. Let's start with x, y. dx dy. Where do I have dx dy? I have dx dy here. And I have dq dx. And I have dy dx. But remember as we discussed before if you have dy dx it's the same as dx dy but you have to put a minus sign. So you get minus dp dy. dq dx minus dp dy. Voila. Okay. I hope the rest works as well. I'll return to this issue of switching and getting the sign in a minute. But I just want to convince you that we get the right formula. Okay. Next is dy dz. So here's dz dy. Here's dy dz. Dr dy. dq dz. But we have to switch them here so we get a minus sign. Which is correct. And finally, dz dx. I'm not denying the right one. This one. dz dx. So this is dz dx. And this is dz dx. dz dx you get dp dz. And here you get them in the wrong order. So you have to put a minus sign. Minus dr dx. It works. Right? So this very complicated formula. This very complicated formula, which I got by this very long and tedious calculation is nothing but d of p dx plus q dy plus r dz. Right? This is what it is. And now I can rewrite Stokes formula in a much nicer way, in a much nicer way by using this operation. Namely, the left-hand side is just p dx plus q dy plus r dz over the boundary of m. And the left-hand side is the integral of d of this p dx plus q dy plus r dz over m. Where this differential, d, is understood in the way I explained. In the way I have explained. Right? So this is what it boils down to. And certainly now it looks much more beautiful. There are no curls. There are no ad hoc definitions of integrals and stuff like this. You see a very clear pattern. Let me call this omega. And this is the integral of omega over B of m. And this is the integral, a double integral of d omega over m. Where d now acquires very concrete meaning as this operation. So this is the general principle which I have been invoking throughout these lectures. And I told you that I will explain that all of these formulas that we are studying in this course, in this part of this course can be thought of as just special cases of this general formula. And now I can make it much more precise in this particular case. And you can see that this is how it works. Okay? Any questions about this? Yes. Do we have to worry about simply connected regions? At this moment, no. So the simply connectedness shows up when we try to determine whether a given vector field is conservative. Right? Whether on the plane, given a vector field, you try to see whether it is conservative and then you want to find the potential function. So at that moment we have to worry about this. But this issue of simply connectedness is sort of one dimension lower. It has to do, rather with fundamental theorem for line integrals than stock's theorem. You see, because in that issue there was no... It was about finding line integrals. So here, no. The short answer is no. We don't have to worry about this. Any orientable? Smooth orientable surface. Smooth orientable surface with a boundary. That's right. So this is a very powerful, this is a very powerful equation. Now, this derivation that I have given you, you're not responsible for it. In other words, it's not going to be an exam and so on. I just wanted to explain for you so that you can understand a more conceptual way of sort of viewing and appreciating this formulas. And I think it really clarifies a great deal. Now, going back to what you have to know and what you are responsible for. I want to give you an example of the application of this stock's formula which is very similar to what you will have on your homework for this lecture. Okay? So example is actually from the book. It's 16.8.8. So you are given a vector field f which is e to the negative xi plus e to the xj plus e to the zk and you are supposed, you are asked to evaluate the line integral of this vector field over a triangle which is given by the, which is a part of the plane 2x plus y plus 2z equals which is contained in the first octant for positive x, y and z. So what it looks like is like this. Here you have one, here you have two and here you have one and it's this triangle. This is a triangle of intersection of this plane with three coordinate planes. You see because for example if you put y and z equals 0 you get x equals 1. If you put x and z equals 0 you get y equals 2 and then z equals 1. So this is x, y, z. Okay? So you are supposed to evaluate this integral which is fine. It's not too complicated. What you need to do is you have to say which orientation. Okay? So I have to be careful, right? So orientation is counterclockwise like this. So you can evaluate by breaking into three pieces by parameterizing each of this line segments, right? And then just doing this line integrals. But it's going to take a while because you have to do three different integrals. You have to parameterize so that, you know, there's some work involved. So what Stokes theorem does it gives you a shortcut to this. It allows you to evaluate this integral in a much more direct way you realize that this curve is actually the boundary of this surface. This will be our M. And so by using Stokes theorem you can actually rewrite this as a as an integral over M of curl of f dot ds. So I'm not going to use any of this fancy machinery. I'm just going to use straightforward the formula of the way it is written in the book. Curl of f dot ds. So instead of doing this integral we might as well do this integral, okay? And in fact it is a very good idea because curl f is not exactly zero. So that would be like the, maybe the first example for you would be to, I could write this in such a way that the curl is zero, right? And then you actually say okay so you don't have to calculate anything, right? So in this particular case it's not, it's almost zero. It's one component. If I calculate it correctly it is e to the x times k, right? Because this one you take derivative back to y and z here you take respect to x and z so this is what gives you that and here you take derivative to x and y so they disappear. So it's almost zero. If it were zero you just get zero, right? So that's already that no matter how complicated this curve is if the curl of vector field is zero then you have a solution of this result where you can calculate the integral very quickly without doing any calculation. Here it's almost zero. So what we need to do is we need to evaluate the surface integral of this vector field, okay? So for this we have to choose, we have to use the formula and the formula is that in this case you can say that this is this plane is a graph of a function you can write it as z equals 1 minus x minus 1 half y. Let's call this g of x, y and then we know that if you have a vector field e which is abc or maybe okay let's call it pqr because I think this is the way I wrote it before then the surface integral of e dot ds in such a case is equal to double integral over the base over the domain in x and y where here you put p minus pdg dx minus qdg dy plus r, right? Remember this formula? So in this case you only have p and q are zero so you only have r to the e to the x and so you end up with a double integral of e to the x dx dy over the triangle on the plane which is the the projection of this triangle onto the x, y plane so it has x coordinate 1 at this end point and y coordinate 2 at this end and this is your d okay? And this is very easy to calculate so I'm actually out of time so I stop here and we'll continue on Thursday