 Okay, how are you guys doing? Are you ready for Thanksgiving break? All right, yeah, me too. So, well, we still have one hour and a half to go. One hour, 20 minutes to be precise. So the initial motivation for the material which we are discussing now was our desire to generalize Green's theory so as to apply it to more general regions in the three-dimensional space, okay? And last time, we made the first steps towards that goal. We introduced the curl of a vector field. So now we are almost ready to state Stokes' theorem and it's going to take the following shape. On the right hand side, we're going to have something we've learned before and understand fairly well by now, I hope. And that's a line integral of a vector field over the boundary of a surface in a three-dimensional space. So we are talking about things like spheres or maybe a better example which we discussed before was a hemisphere, an upper hemisphere. Like this. So a sphere, as we discussed already many times, is a two-dimensional object. I'm not talking about the interior of the sphere, I'm talking about just the surface of the sphere. It is a surface hence a two-dimensional object but it's a kind of two-dimensional object which really lives in a three-dimensional space and not on the plane, not in a two-dimensional space because it's curved. So this surface has a boundary. So this surface, M, M will stand for membrane, you'll see why, we'll talk about it later. No cell phones. I'm actually, I'm impressed how rarely we see this, we hear this sound, so keep it up. This surface has a boundary and that's what I call here B of M. So in this case, if it is really an upper hemisphere, if you really think in terms of the upper hemisphere, it is a circle. In any case, it's a curve and over that curve we can take this line integral where F is a vector field. So that's the left-hand side. The left-hand side is kind of easy in the sense that we already know what it means. And the left-hand side is going to involve a double integral over M itself and now we can say the integral of what? It's going to be the integral of what we learned last time, the curl of F. So if you have a vector field F, you can define a new vector field which we call curl of F, okay? So this has already been defined. And now we have to learn how to integrate vector fields over surfaces. So that notation here represents a surface integral, an integral of a vector field over the surface, also known as a flux. And today we will define integrals of this type so that finally we'll be able to say what we mean by the left-hand side and at least we'll be able to make the statement of Stokes theory which we will then discuss next week, okay? So that's the plan. We would like to learn surface integrals, okay? Now this is, by the way, about the final exam which again will be on December 19th at Hearst Gym. No more jokes about it. And now I put the materials for the final on my homepage so you can see the review problems and various other things there, okay? I will put the solutions of the review problems probably sometime during next week. All right, so let me raise it now. So what we'd like to talk about is surface integrals. And as always, when we talk about things like that, it's always good to think in terms of analogies, okay? And an analogy to a surface integral would be an integral over a curve, over one dimensional object. Something which we now understand fairly well. So let's revisit first line integrals. Revisit integrals over curves. One of the aspects of the story, something that I know you have found puzzling at first is the fact that actually the way we thought about it, the way we talked about it, there are two types of integrals over curves. One has, one type is the integral of a function. And this is a kind of, it's arc length type integral. In this case, we have a curve C and we integrate a function. And the expression for this is FDS. So F here is a function on this curve. And this DS represents a very small, the length of a very small piece of that curve, okay? So what is the meaning of this integral? The meaning of this integral is that if F is equal to one, we get arc length of the curve. Arc length of the curve. And if F is not equal to one in general, but it is a non-negative, takes non-negative values, we can interpret this function as a density function. For example, mass density function. We can think of a curve as being a metal wire like this. And that function would then represent the density of mass. Some parts of it could be heavier, some parts lighter. And this integral would correspond to the total mass. Total mass. That's the interpretation of such an integral, okay? So that's the first type. And the second type, which actually we have studied more and which turned out to be more important for our purposes, in particular for formulating things like stock's theorem or Green's theorem, or fundamental theorem of line integrals for that matter, are integrals of vector fields. Line integrals of vector fields. So let me recall that. So in this case, we have a vector field F, we have a vector field, and we are computing its line integral, which we do not like this. And that also had a nice physical interpretation. It's not a mass, but rather it is work done by force. If F represents a force vector field, then this represents the work done by this force. Examples of such vector fields are the force of gravity, which we talked about before. Also the electric field and the magnetic field. When we talked about Maxwell equations last time, you had those two vector fields, E and B, right? So this would be examples of such vector fields. And such a line integral for such a vector field would actually have a very concrete physical interpretation as the work done by this force. For example, electric field, the work done by it in moving an electron from one place to another along this curve C. So those were the two types. The reason I am reminding you of all this is because surface integrals, which we are going to study today, can also be broken into these two different categories. One is the analog of the first type here, and the second one is the analog of the second type here. In other words, the first one, the first type of surface integrals will apply to functions, will integrate functions. And the second type will apply to vector fields. So we'll be integrating vector fields. And in both cases, they will have nice interpretation, very similar to the interpretation we've learned for line integrals. So having said that, having recalled the integrals over curves, I want to now talk about surface integrals. Now for surface integrals. For surface integrals, we'll also have two types. Now, I have to say from the beginning that this looks a little bit disconcerting, the way we talk about this, that there are two types and they kind of seem unrelated. When we talked about these guys, integrals over curves, I actually tried to explain that in a way, these are two special cases of kind of most general integrals over curves. So from the point of view of the big picture, these are actually not so unrelated. They're actually quite close to each other. But for the purposes of this course, for the purposes of the kind of results we want to obtain and kind of applications we have in mind, it is just easier to introduce them separately and kind of think about them as two different types. One applies to functions, one applies to vector fields. And the same actually will be true here for surface integrals. We'll kind of divide surface integrals into these two big groups. But I just want to tell you that actually from the point of view of the big picture, this is unnatural. And there is actually more natural theory. There is a more natural point of view from which both of these integrals appear as just special cases of some general construction. Nonetheless, we'll do fine. We'll do just fine by kind of defining them in two separate ways. Okay. So for surface integrals, we'll also have two types then. Also two types. The way we will introduce them here, two types. And so the first type will be an integral of a function, surface integral of a function. So we will have a surface M again, like in Stoke's theorem on that board. But we will be integrating not a vector field, but a function. And it will be represented in this way. So this is not yet the kind of integral which will show up in Stoke's theorem. This is sort of a preliminary integral, which is like type one integral in the case of curves. And this is something which we will use to define also line, to define surface integrals for vector fields. The reason why we define them first is because they're easier to define than the other guys in some sense. And also we will actually use this theory to define the integrals of the second type. So what does this integral represent? Well, just like in the case of curves, it represents something which is sort of inherent to this curve. The quantity which is to the surface in this case. So here there was a curve. And the inherent quantity for curve is arc length. It's arc length. Or if you have a general density function, the total mass. And likewise here, this will represent, for f equal one, this will represent the surface area. Surface area. And for f, not equal to one, but non-negative or maybe positive is better. This will correspond to total mass. If we interpret this function as a density function, is mass density function. So let's talk about the special case when f is actually equal to one. So in this case, we will denote this integral just like this. And this will simply correspond to the area of f. So let's discuss how to find the area of f. This is not entirely new. It's not an entirely new subject for us because at the very beginning of this course, we actually talked about revolution surfaces. And we talked about the areas of revolution surfaces. So we actually know one example, at least one example of this kind of integral. But now we will talk about areas of more general surfaces. Not just the revolution surfaces. Revolution surface remember was obtained when you rotated the graph of a function in one variable around the x-axis or maybe around the y-axis. But this is only very special kind of surface. It's like a vase, if you know what I mean. Like a vase is a revolution surface if it's a symmetrical vase, you know, like a round vase. Is a revolution surface of a curve around this z-axis in this case, if you think about vertical axis as z-axis. But not every surface is like this. There are even vases which are not exactly symmetrical, right? For example, you could have squares. So we would like to have a formalism which would enable us to calculate the area of any surface. So what are the necessary steps for that? Well, how should we approach this? It's the same drill as before. We have a piece of surface like this. And we would like to compute its area. And as always, what we need to do is we need to break it into small pieces and evaluate the area of each small piece and then take the sum. And then the sum, when the partition becomes finer, will actually give us an integral. But how would we actually break it into small pieces? A priori is not clear, right? So in the case of curves, in the case of curves, if you had a curve like this, you also wanted to break it into small pieces and take the small lengths of each piece and take the sum. And the way we learned to do this was through parameterization. This was in fact the most efficient way to talk about curves in the first place, right? Because a general curve was best represented algebraically if you expressed it in the parametric form. In other words, we would introduce an auxiliary parameter, auxiliary parameter T and identify an interval from, so this is parameter T from A to B. We would identify this curve with this interval, right? And algebraically, we would just write that X is a function of T, Y is a function of T, and well, in two dimensions, that would be enough, but in three dimensions, we would also have Z equals Z of T. And then we'll say T is between A and B, right? So when we write these formulas, what we are doing is we are identifying points on this interval, right? On this straight line interval, which is very simple, was this complicated curve. Once we do that, there is a natural way to break into pieces because we can just break this interval into pieces of equal length, say 1,000 pieces or 100 pieces in the beginning, maybe 10 or 100, 1,000, and then each of them, for example, this one will correspond to this part here, and then we'll try to find the length of this one, and then the next one, and so on, okay? And so the result of this calculation was an integral of the first step. Now, instead of a curve, instead of this curve, we have a surface, right? So we should apply the same approach, and we should try to parameterize this surface. This is something which we have actually managed to avoid up to now because this is something we talked about before, but we never actually got around to trying to write surfaces in the same way, or present surfaces in the same way we've been representing curves, and the reason is very simple because surface is two-dimensional. So surface is two-dimensional, so this M, let's call it M, its dimension is two, and so it requires two independent parameters, two independent parameters, but only one equation, only one equation. So whenever we could refine a surface by equations, we said it's better to define by equations and try to find parameterization because one is better than two in some sense. One is easier to handle than two. One equation is easier to handle than two parameters, okay? Notice that for curves, it's the opposite situation because the curve is one-dimensional, so it requires one auxiliary parameter or two equations. So that's why curves usually show up through parametric representation. On rare occasions, we will talk about representing curves as by two equations or as intersection of two surfaces, but now it's finally time to use the first approach and to parameterize a surface by two parameters now, okay? So how is this going to work? Well, in exactly the same way as in the case of curves, we will choose two auxiliary parameters which will, oftentimes we'll call them UNV, but again, it's just a choice of notation. You can choose whatever letters you like, UNV, and we will try to identify this piece of curved surface in a three-dimensional space like this with a flat domain in a two-dimensional space of these two parameters. What I mean is that somewhere in the background, we will have a two-dimensional coordinate system with coordinates UNV. Note that this is the analog of that line with the coordinate T which I drew on the top blackboard. And the analog of that interval from A to B now will be a two-dimensional domain on this auxiliary plane. So it is going to look something like this, right? See now, of course, there is more choice for such a domain. There are certainly more pictures we can draw like this than pictures of intervals, interval is an interval. I mean, the only things which we can play with is just the endpoints, but it looks, they all look the same. Here the domains look different, right? They have different shapes. Now, you can of course ask, why would be bothered to identify this and this? The reason is that this object is much simpler to handle than this object. You have to realize what this object represents. This object represents something which actually does not fit in the blackboard, right? It is actually, it can only live in three-dimensional space. In other words, without bending and squashing it, I wouldn't be able to fit in the two-dimensional blackboard. Here, I have drawn it on the two-dimensional plane, but that's just a projection, right? So in fact, usually to emphasize the curviness of this object, I would kind of draw a grid, this kind of a curvy grid to kind of indicate that it has this shape. And this is in stark contrast to the structure of this object, you see. So this guy is actually flat. This guy actually fits in a two-dimensional plane. And that's why it's to our advantage to try to identify this curvy object with this flat one. So that's the first idea, okay? And once we do that, we can then try to calculate the surface area. Now, since I drew it like this, it would be more, it kind of, it looks strange that I established one-to-one correspondence between this one and this one, right? So of course, it would be more natural if in fact the flat object, which was on the plane, which corresponds to it, would be something more like a rectangle. Then the picture would make a lot more sense. Here, I drew just a most general region that you could have a priori, but just to simplify our pictures, let's assume that in this particular instance, we will have a rectangular domain. This is not the most general one, but this is just for the sake of the argument, just to do an example. Okay, so if we do that, for this guy, there is a natural grid because just like for the interval, there is a natural grid because we can just break it into equal pieces like this. And once we do that, if we identify this and this, okay, we'll be able to also give a grid to this picture. You see what I mean? Just as before. So when I say identify, it means that I would map, I would try to map this flat rectangle to this curved surface in a three-dimensional space so that for instance, this interval would go here and this one would go here, this one would go here and this one would go here, but not only that, also each of this horizontal lines, horizontal intervals would go to these curves and all the vertical ones would go to those curves. So if I have already a parameterization like this, okay, then I would be able to break it into smaller pieces and then that would be the first step in getting the integral. So let's talk about examples of this kind of parameterization. Let's talk about examples. So the simplest one appears when our surface is graph of a function. If the surface M is part of a graph of a function, E equals, in this case, we can just choose X and Y as the auxiliary parameters. So, or if you will, we can say X is U and Y is V and Z is F of UV. So here is a parameterization. I have represented each of the three variables as a function of my new auxiliary variables, U and V, okay? And so this is just like parameterizing curves, the difference being that now the functions on the right-hand side depend on two variables, U and V, instead of just one variable, namely T in the previous case. So that's the first example. The second example is example of a plane because after all, plane is also a surface. Plane is two-dimensional. In fact, plane is the simplest possible surface, right? So in the case of planes, we are used to representing planes. So this is two. We are used to representing planes by equations. Exactly what I just said now, which is that surfaces oftentimes are easier to represent by a single equation. But now we're talking about parameterization, which is important for integrals. So let's talk about parameterizing a surface, which is very easy because think about, in fact, let me use my favorite surface, this one, right? So in the case of a surface, in the case of a curve, if I had a line, to know a line, I need to know a point and I need to know a direction vector. For a plane, one vector is not enough, right? Because one vector would just give me this line. If I want to get the entire plane, I should also pick a second vector, which is kind of transversal to this one. So if I do that, if I pick two vectors, let's call it R1 and R2, then any other vector on this plane can be obtained as a superposition of some multiple of this one and some multiple of this one, or some combination of this one and this one. And I can use that to find a parametric representation for this plane. So let me draw it on this picture. So let's say I would pick, so this is my coordinate system. So I would have, this would be some vector R0 and then I would have two vectors, R1 and R2. And so a general point on this plane would correspond to some vector R of UV. And it can be found by using this one plus a superposition of R1 and R2. So it will be R0 plus UR1 plus VR2. And then if I write this in coordinates, because as always, R is just X, Y, Z, that's three coordinates. So likewise for R0, R1 and R2. So say R1 is K1, L1, M1, and same for R2. Then I can actually write each of them as a function of U and V. So X would be X0, the first component of R0 plus U, K1 plus V, K2, Y would be Y0 plus U, L1 plus V, L2, and Z would be Z0 plus U, M1 plus V, M2. So here these numbers, K1, L1, M1 and K2, L2, M2 are given and the variables are U and V. So now I have been able to express each point on this plane as a function of two variables, U and V. So that's exactly what I'm talking about in general, except in general these functions will be more complicated. Here they are linear functions in U and V. In other words, the degree is a polynomial of degree one or less than or equal to one. Okay, something slightly more interesting corresponds to a sphere, corresponds to a sphere. For a sphere we could use spherical coordinates. For a sphere we could actually apply spherical coordinates. So we use as auxiliary variable, auxiliary variable will be theta and phi and we would write X sine phi cosine theta, Y is R sine phi, sine theta and Z is R cosine phi, where R is the number, R is the radius of the sphere, it's fixed. So see the difference between the spherical coordinates and this parameterization is, whereas for spherical coordinates we had rho here. So there was three variables, rho, phi and theta and those variables gave us all points on the three dimensional space. And now we are talking about points just on the sphere. So that actually leaves only two independent variables, phi and theta and rho is fixed because rho is equal to a fixed number for all points on the sphere. You have a question. When I drew, the question is what does this plane represent? This picture means like that plane. No, that's a plane in a three dimensional space. But now I am relating it to a UV plane. It's actually easier maybe to think about it in this way that so here is a plane, right? And this plane is in a three dimensional space which is our classroom. And now I would like to parameterize all points on this plane by points of another plane which say will be this table, okay? This table will have coordinates U and V and if I have a point UV here, they will correspond to a point here. Namely, that would be the point for which this vector, that would be this point for which this vector is U R 1 plus V R 2. Meaning that this is U R 1. Actually didn't draw it very well. Let me do, I have to do it in a parallel way, right? Like this, it's moving. That's why I'm not making a good picture. So this is V R 2, okay? Doesn't make sense? So once I fix U and V on an auxiliary plane, right? I can get a point here. Namely I just draw the vector U times R 1, that's this one. Then V times R 2, I take their sum, that's the point. And this way I can get to all points of this plane, of this tilted plane, right? So U and V, this pair of numbers, U and V provide me a unique address system for all points on this green plane. And that's what we are doing in general. It's just that in general, our surface is not flat, it's curved, but we are finding an address system for all points of that surface, just the way I do here for a plane. Okay, so for instance, in the case of a sphere, you know, so here is a sphere. Let me again draw an upper part of the sphere because it's easier to see, to analyze like this. So for instance, so this is the sphere of radius R, some number, let's say R is equal to five. So sphere of radius five. So this number is fixed. But now there are many points here, right? And I would like to address each point on this upper hemisphere by some unique pair of numbers. I call those numbers phi and theta. And I denote them just by using the same rule as I use for spherical coordinates. So for example, if I want the entire, if I want the entire sphere, this would correspond to the limits theta from zero to two pi and phi from zero, sorry, phi from zero to pi. This is a whole sphere. If on the other hand, I want just the upper hemisphere, this would mean that theta is between zero and two pi. And phi is between zero and pi over two, right? I put a limit pi over two because I don't wanna go more than upper hemisphere. So going beyond pi over two would correspond to the lower hemisphere. And then you can also think of, even if you think about this cone, if you just want the part of the cone, for example, the stop part. Think about an ice cream cone. So that would be the ice cream, the part of the ice cream, right? So that would correspond to theta between zero and two pi and phi between zero and pi over four. So by choosing limits in an appropriate way, I can describe different regions on the sphere by using spherical coordinates. And this is the third example. Okay, now once I have a parameterization, I can get to work and try to calculate the surface area. You can calculate the surface area. And now I have to, I just have to look at this picture. And what I need to do is I need to find a way to express the area of this one little curvy rectangle on the surface, which will correspond to a particular rectangle in the parameterizing domain, okay? So here let's say we would be delta U, the length would be delta U, and this would be delta V. And I would like to find the area of this, which is delta S. And this actually is something which we have already done. We have done this when we talked about double integrals in general coordinate systems. When we talked about that, we did exactly the same calculation. The only difference was that actually our surfaces at that time were still part of the plane, but the calculation is exactly the same. If you remember, we took the tangent vectors here, R sub U and R sub V, and we took the cross product, which represented the area of the parallel ground. So the answer is the same, answer is the same as before, as in the old calculation for double integrals is in all calculations. So you can review that if you like. Just to save time, I'll give you the answer. And the answer is that this is the absolute value of R U cross R V delta U delta V. This area of this little piece. So that's the formula. What are the R U, what do I mean by R U and R V? So R is just X of R of U V is just the vector which position vector of the point X, Y, Z. So that's just X of U V I plus Y of U V J plus Z of U V. And if I want to take R sub U, that just means taking the derivative of X with respect to U times I plus derivative of Y with respect to U times J plus the derivative of Z with respect to U times K. So what I mean by X sub U is just DX, DU, but it's just easier to write it. It's faster to write it like this, so that's why I'm writing. And likewise for R sub V is X sub V I. So you see, and now to find the area, the area and of course I would like to emphasize, this is not the exact answer. This is approximate answer, just as before. This is an approximate answer for this little surface area. And what we need to do after this, we have to take the sum of all of this over all pieces of this partition and then take the limit when the partition becomes finer. So that's, and then we get the surface area. So this way we find that the area of this membrane of this two-dimensional surface is a double integral, but now it's a double integral over D, you see. D being this, this is D. D is a domain which is on the flat domain. The whole purpose of this discussion was to replace a curvy surface by a flat surface. And to replace a curvy integral, which we are trying to define by a flat integral. A flat integral we already know. We studied the months ago, it's been awhile, right? So that's something which we understand fairly well. So the whole point of discussion is to define the area as an integral, but over a flat domain which we understand. The question is the integral of what? And now we know the integral of what? Because we know the elementary area. And it's the same game as before. Once you know that the elementary object, you know the formula for the elementary object, which in this case is the area of this little piece and is given in this way in terms of delta U and delta V, then you know that when you take the limit, this will get replaced by DUDV and this will be the function. Yes. Why is the LS equal to this? I would love to explain this, but I'm afraid that if I do that then I won't have time to explain other things. But, so I have to make a choice. The reason why I'm not explaining this is because we actually, it's verbatim the explanation for general coordinate systems. Which actually I didn't explain in great detail either. I have to confess. But I explained enough of it to convince you that that's the right answer. But think about the fact that, remember the fact that the cross product of two vectors always represents the area of the rectangle which is formed by them. So what I do here is that I, the calculation is done in a following way. I approximate this curvy rectangle by a flat rectangle formed by the tangent vectors. And then once I have the two vectors, the area of the rectangle spanned by them is just a cross product. So the only question is what are these two vectors? And you know when you do tangent vectors you take derivatives. So if you actually look, if you look at this question you will see right away. And actually we have all the, you have all the information you need to be able to find this because we talked about tangent vectors two lines. And this actually lines, sorry lines, curves. These are tangent vectors to curves. And you will see that the tangent vectors, these two tangent vectors are r sub u delta u cross. And the second one is r sub v delta v. So actually the answer comes out to this. Comes out to be this. But then I just take the delta u and delta v which are numbers out of the cross product. And that's how I get this formula. Maybe this is an, if you look at this formula you'll see it more, you know, it will be more convincing. Okay? So having said all of that, this is the formula I get. r sub u cross r sub v du dv. Which is actually, you can also write. It's more appropriate to write as dA. You know the little piece of area on this flat region. So that's the formula for the area. And now we can actually look back at all of those examples that we have studied. Here, it tried to figure out what the answer is going to be. And the point is that, well, let's do it in this case. In this case, r sub u, you have to take the derivative of these three functions with respect to u. So this derivative is one. So it's i. This derivative is zero because derivative of v with respect to u is zero. So I'll just skip it. And then the derivative of this one is f sub u times k. And r sub v is actually j plus f sub v k. So now I can actually find this expression by simply taking the cross product of these two vectors. So what do I get? So r sub u cross r sub v. Just use the usual rule. So you'll have i, j, k. And then you'll have one, zero, f sub u, zero, one, f sub v, right? And that's what, that's negative f sub, let me write it here. Negative f sub u, i, negative f sub v, j plus k. And so its norm is the square root of f sub u squared plus f sub v squared plus one. And so that's the function which in this particular case, I would have to put here and integrate it over my flat domain parameterizing my surface. And that would give me the surface area, right? So it's very straightforward. Likewise, you can calculate the answer for sphere. For sphere, you just have to do the same kind of calculation, you have to write, find out what r sub phi is and what r sub theta is, put this together, calculate the cross product. This has already been done. So I'll just give you the answer. And in fact, the answer you can guess because we already know the Jacobian for the spherical coordinates. And this is going to be exactly the same formula as the formula for the Jacobian of spherical coordinates except that now we fix the radius of the sphere. The row is fixed. So the answer is, is r squared times sine phi, right? The answer was r squared times sine phi, but now it's r squared times sine, because it's fixed. So again, if you're working with a sphere, this is a function you'll have to insert in this integral to obtain the surface area, okay? So that's surface areas. So this completes the case when one, your function is one. So this is by definition and the definition goes and usually we write definition from left to right, but now I write definition from right to left, okay? So this is a definition for the integral over m. This was our first task, to calculate the integral, calculate the surface area as an integral over a flat region. So this is the answer, okay? And the next step is very easy. The next step is we want to now calculate not just an integral of ds, not just ds, but we want to integrate an arbitrary density function. For example, think about the mass. If it's an aluminum foil, which has some parts heavier, some parts lighter, so the mass is distributed in an uneven way, described by this function f, so then this would be the mass. How to calculate this integral? Well, very easy, you just insert this f in this integral and that's it. And this completes the definition of the first type of integral. And you know, the reason why I'm kind of going so fast over this is because I would like you to appreciate the fact that this is really parallel to our derivation, for example, of the arc length of a general curve or the formula for the mass of a general curve with a given density function. It's exactly the same derivation. What has changed is that now the surface is parameterized by two parameters, by two variables, whereas before there was only one. But the idea of the calculation is exactly the same. Okay? Any questions about this? So that completes the double integrals of the first kind. And now finally we are getting to, we're getting to something more interesting, which is the double integral, double integrals of the second kind where we will be integrating not over, not, we will not be integrating functions over surfaces, we'll be integrating vector fields, which is what is relevant to the Stokes' theorem in particular. Okay? And now, before I give you a technical definition or I give you the formal definition, I would like to motivate it by considering a real-life problem where this kind of integral shows up. And again, I want you to appreciate the parallel with line integrals of vector fields. For line integrals of vector fields, we talked about work done by force. And then we try to see what this means and how to represent it by an integral. And that's how we came up with the notion of line integral with vector field. So here we're going to do the same thing, except now of course it's not going to be work done by force, but what it's going to be is it's going to be the flux of a vector field or the rate of a flow of a vector field. Okay, this is slightly, this is slightly trickier. I have to say just slightly trickier than the other one, but very transparent nonetheless. Okay? So here is a real-life situation that you can imagine. Imagine that there is a pipe and you have the flow of water through the pipe or any other liquid. I'm not going to make any suggestions. So, and you would like to find the rate of the flow through a given, think about the membrane, that there is some, think of this invisible membrane and you would like to know how much water will flow through this membrane per second. Okay, that's what I mean by rate of flow. So for that, we have to know the density, which would be amount of water per second, or per hour or whatever, per unit of time. And I'm assuming here that my membrane is strictly perpendicular to the flow. So it is perpendicular to this pipe. I'm looking really at the flow through this section of the pipe, which is just obtained by cutting it in a perpendicular way. So what do I need to know? I need to know the density, density function, the density of water, right? And I need to know the velocity, mass, this is mass density, mass density. Then I need to know, so this is a mass per cube of a measure of length, let's say meters. Then I have a velocity vector, right? Whose length is the length of the velocity vector. So this is the velocity vector. And the length of the velocity vector is measured in what, in distance, so it's a meters per second, okay? So what do I need to do? I need to, oh, and I also have the delta S, the area of this section of this membrane. This is the area. So this is a meter squared. So what I need to do is I need to just multiply this. So the rate of flow is equal to rho times V, right? Times delta S, that's pretty clear. Because, well, because over given period of time, it will move, say the water that will move through this membrane will correspond to this part of the pipe, right? Which we find by multiplying, by taking the length of this velocity vector multiplying by time and multiplying by the area of this membrane. So I think this is pretty clear, okay? So now I would like to make it more complicated. So first we have to make things more complicated, then it will become easier. That's always the idea, right? And here's what I want to do. I would like to now understand the flow, not through this one, which is kind of perpendicular, but through another one, which is tilted, through another membrane, which is tilted like this, okay? So what has changed now? Well, the point is what has changed is that what will contribute now to the flow through this membrane is not V. It's the component of V, which is perpendicular to this membrane. Listen. So this little membrane will have a normal vector, N. Let's call it N. And what will matter now is not V, but the component of V along N. Because you can decompose the velocity vector, right? Let me draw this picture one more time. This is my membrane. This is N and this is V. So V has two components now. One is a normal component, which is, let's say, this one, and then it was intangential component. So you can think that the flow of the water is combined, flow of the water through this membrane is combined of two different effects. One is some water flows in a perpendicular way and one sort of water kind of flows is a superposition of flow in a perpendicular direction and the parallel to the membrane. And of course, what is parallel to the membrane is irrelevant, nothing flows. If the flow is just like this, there will be no flow through the membrane. So the only thing that you have that is going to contribute is just the normal component of this vector. And what is this component? So this is V. So let's remember again dot products, right? So the normal component, the length of normal component of V, the length of the normal component of V is just V times the cosine of the angle, which is formed by these two vectors, right? That's this normal component. And that's just V dot N. So my more general formula is going to be like this. So the flow rate is going to be rho times V dot N times delta S, okay? So I hope I convinced you that in this idealized situation where you have the uniform flow of this liquid, that's the velocity vector is the same everywhere. And the normal vector is the same everywhere that the rate of flow is given by this formula. And now I generalize, okay? But before I generalize, there's a question. Right, but N is the unit normal vector. I'm sorry, I should have said that. Right, the question is, I need to put here, I assume that the length of this N is one. So it's a unit normal vector. When I take the dot product, it's going to be the, oh, you're also saying this. Sorry, that's right. You're right, thank you for correcting me. So the, first of all, I should have put the length of V, the magnitude of V. And second, I would normally have to also put the magnitude of this vector, but I assume it to be one, to be equal to what? It's the unit normal vector, if you want. Okay, so now we're ready to generalize it. We generalize it by saying that it could be an arbitrary membrane, and also the velocity could be different in different places. The velocity would just be given by some vector field. So what answer are we going to get now? It's the same idea as in the case of our calculation of the surface area. We break a general membrane, we break a general membrane into small pieces, and for each small piece, we apply the idealized scenario, the idealized scenario, where we assume that approximately the velocity is the same everywhere, okay? And the normal vector is approximately the same everywhere. So then we use this formula. But after this, we have to take the sum over all of those pieces. And when we take the sum, we take the limit when the partition becomes finer. And that's what gives us the integral of the second type. And you can read off the expression for this integral from this formula. As always, just look at the formula for the elementary calculation, which we get from the idealized situation, okay, like this, and just replace delta S by dS, where now we already know what dS is, because we are able to calculate dS by using parameterization, by using this rU cross rV, right? So for general membrane, the flow rate just looking at this formula is, and let me put rho outside, because rho is just a number. It doesn't depend on, actually, might as well, might as well, might as well depend. So let me do something else. So let me say that F is rho times V. The mass density function could, of course, in general be non-constant, that's fine. And V, the velocity vector field is non-constant. But let me just isolate this, the product of the two as a particular vector field. This formula will be nicer this way. So then, well, if I do that, this can be written as F dot N dot times delta S. Times delta S. So what I end up with is an integral over M of F dot N dS, okay? But now, what is N? Now I would like to make sense of this integral by using a parameterization. You know, this is already a definition, right? Because once I take the dot product, I start with a vector field. But my surface has its own vector field attached to it. It's a kind of inherent vector field attached to the surface, namely the normal vector, unit normal vector at each point. And I take the dot product and I get a function. And functions I have already learned how to integrate. You see? Since I've learned how to integrate functions, this actually already makes sense. But let me spell it out in more detail by using parameterization. So suppose now that my surface is actually parameterized in this way, then I can actually find a normal vector and the normal vector is going to be precisely this r u cross r v, which we have used here. This actually, as a vector, it is normal to the surface at this point. It is normal because remember, as I said, in this calculation, you have to take a cross product of two tangent vectors. When you take the cross product of two tangent vectors, you get something which is perpendicular to them and therefore perpendicular to the surface. This is a normal vector. This guy is a normal vector. But in a previous calculation, we just took its norm and we saw that the norm is this. So this gives me a way to actually to get my formula very, obtain a very explicit formula because this is a normal vector. But unfortunately, it's not normalized, which is a horrible pun, right? So it's normal in the sense that it is perpendicular, but it's not normalized in the sense that its norm is not equal to one. But now I will normalize it. So we'll get a normalized normal vector. Okay, I think that this is because our language and our everyday language is too poor to express some mathematical concepts. And that's why it sounds really bad, but so let me just write a formula instead without saying anything more. The formula is that n is r u cross r v divided by its magnitude. If you have a vector and you divide it by its norm or magnitude, you obtain a vector which has norm one. There's always the case. So this is what I should put here instead of n. And also instead of ds, I have to put my old formula has already been erased, but I want to remind you maybe here that the integral over m of f ds was equal to the integral over d, where d is this region of r sub u cross r sub v dA. In other words, sorry, and I forgot the function f. So in other words, ds becomes dA times this guy. So what's going to happen now and it will have a very nice consolation. We will have double integral over d of f dot r u cross r v. Then we'll have to divide by this r u cross r v. And then we'll have to multiply by r u cross r v dA. Do you see? So it's kind of lucky because the same quantity appears in the numerator and denominator and it's non-zero. This is actually something which we have to take care of because it should be non-zero because otherwise it would not be a parameterization. This is like, think about this is like the Jacobian. It is a Jacobian if you think about it. It is a Jacobian. And in fact, not surprisingly, the formulas we get for it, say for spherical coordinates, is a Jacobian, right, it's not a Jacobian. And the Jacobian should be non-zero. For otherwise, this is not going to be a good parameterization. So that's why actually it's non-zero and we have the right to cancel them out. When we cancel them out, we'll finally get a very nice formula. Namely, we get this. And finally, so finally we have come up with the definition of the integral of a vector field, of a vector field across a membrane, across a two-dimensional surface. If the surface is parameterized by a flat domain D, this integral is simply given by this formula, by this double integral over D. What it is, on the other hand, what it is more in a more canonical way, is an integral over M, with respect to the measure of the surface area, of this function, f dot M. But since we don't want to write it each time, we will actually write it. We will actually write it. We'll use a new notation for it. We will call it f dot ds, where we put ds over ds, we put an arrow. In other words, we say that ds as a vector by definition is n times ds. If we do that, then we can combine these two guys into single symbol ds, with a vector over it. And what we're doing is we're taking the dot product of f with ds. So let me summarize. We are trying to understand what to make of an integral over vector field across a membrane. And a good analogy is a line integral over vector field. In the case of a line integral vector field, we had a different interpretation, because in the case of a line integral of a vector field, it was all about pushing something along the curve. So what mattered was the tangential component of the vector field, tangential to the curve. What matters now is the perpendicular component, because we are talking about the flow. The flow through, not along the surface, there's no such thing as a flow along the surface. It could be something along the curve, you could not have something along the surface. It has to be across the surface. If it's across the surface, what matters is the perpendicular component, the normal component, which is what we get by taking the dot product of our vector field with n. So we are trying to find this, and this is called the surface integral of a vector field or a flux of a vector field through or across the surface. And this is very important. It's not along, it's across. This is an important thing, an important distinction between curves and surfaces. For curves, you go along the curve. For surfaces, this is a membrane, you go across the membrane. Okay, that's why it's a dot product with a normal. And so this integral is given by this formula, f is a vector field, and the meaning of this integral is given by this formula. Once you parameterize your membrane m by a domain D, like here, you can actually evaluate this integral in a very straightforward way by simply taking this cross product and taking the dot with f. For example, let's suppose that we are in this situation. We are in this situation where our u cross rv is given by this formula. This, I recall, is the situation where m is part of a graph of a function f, okay? So, and your vector field f, say, is some pi plus qj plus rk. In this case, we can use this formula. So what we have to do is we simply have to take the dot product of this vector and this vector. And so what do we get? That's going to be integral over D of minus pf sub u minus qf sub v plus r times da. Very explicit formula. You have, if you have to do a problem like this, you are given everything you need. You are given a vector field, pqr. You are given your parameterization. In this particular case, although I have to warn you, this is a very special case when the parameterization is like this. This is the case when your surface is part of a graph, right? So you are given function f. All you need to do is to find the partial derivatives of f with respect to the two variables, right? And after this, you assemble all of this data into this formula and you get a double integral. And double integrals, you know how to do. You can use any method you like. In fact, I wanted to do an example, but I'm, okay, well, maybe, let's do a very quick example just to see how this works. And then I actually have something to show you, something fun. Know that this has not been fun. I have to say the way I kind of, the way I present this material is not the most economical and I kind of, I feel a little awkward because this is not the best possible way to present it. It looks a little bit ad hoc again because first you introduce this surface area, that was fine, surface area was fine, and the mass is fine, but defining this flux in terms of the surface area integral of the first type, that's a little bit, that's not the most optimal way. So there is a much nicer way to explain this and I will explain it if not next week than during the review lecture for sure. So I will give you a much nicer perspective on this type of integrals. And then you will see right away where these formulas come from. It will not be, it will not look anymore so ad hoc. But for now, this will suffice. This will give you a reasonable definition which you can work with and you can calculate flux within the surface integrals very easily, like in this case. So here is one example. Find the flux of the vector field F which is minus x i minus y j plus z squared k. And s is part of the cone, part of the cone z equals square root of x squared plus y squared between the planes z equal one and z equal two. So let's do this calculation. First of all, in this formula, in this formula above, let's just say u is x and v is y. I mean, after all, in this particular case, we don't have to pretend that this has some new variables, some auxiliary variables. These are our x and y. So in this case, the function F is this. This is F of x, y. So I'm just substituting in that formula. So what I get is that F dot ds over this membrane. And by the way, it's very easy to draw it. This is a cone. We're talking about the cone like this. And you use the part which is confined between two planes. So it's this one. I mean just the surface of the cone, but between this plane and this plane. It's equal one and it's equal two. Okay, think about ice cream cones. So this is going to be the integral over, under this, we'll have to do a parameterization. And the natural parameterization would be by its projection onto the x, y plane. And under this projection, this will go to an analyst. Somebody said analyst, right? Very good. This goes to, but you don't have to say it's quite so loudly. Anyway, so you get this analyst. And then you have to integrate over this analyst. An analyst of course has a very nice parameterization which is that you have r and theta, polar coordinates r between one and two and theta is between zero and two pi, right? So you can always convert this into an integrated integral very easily, right? But now integral of what? We have to take the dot product of P F sub x. So what is P? P is negative x, right? And some reason, right? So when you do that, what you're going to end up with is negative x. So you first of all have an overall negative sign. Then you have negative x. That's this negative x. Then you have to take the derivative of this, right? That will take the derivative of this. The derivative of this, you know, as well as I do, it's x divided by square root of x squared plus y squared. Then the next one is negative y. Well, first of all, it's minus, overall minus because there is an overall minus here. Then you have to take this negative y and the derivative which is y divided by square root of x squared plus y squared. And finally, you have to take this times just, just take this z squared. I think that, let me make it a little bit easier. So I write it like this. So you have the last one is z squared divided by square root of x squared plus y squared. Actually, no, no, it's okay. Let's just put, let's just do it this way. So you get plus z squared dA, okay? And after this, you have to substitute, you have to substitute the square of this in here. So that would be just x squared plus y squared. So here, you got a very simple formula, which you can now, a very simple integral. You can of course bring this, take the sum and then you cancel out. You get square root of x squared plus y squared and numerator. Here you get just x squared plus y squared. So you just calculate it by using the polar coordinates very easily. So this is how you do it, okay? Now, there is one issue which is left unanswered here. And that is the issue of the choice of this vector n. And let me illustrate it in this example. In this example, we are talking about this surface, right? This part. But now the point is which n, which n am I talking about? Am I talking about this n which goes downward? Or am I talking about this n which goes upward? Okay, so there is this one issue which we have not yet talked about. And to make everything precise, we have to, in other words, to make the definition work, we have to say which normal vector we choose. And the proper terminology for this is orientation. This unit normal vector to our surface or unit normal vector field to our surface is called orientation. It tells us sort of which side is the front and which side is the back, okay? And as always, there are two different choices. It's kind of parallel to the choice of orientation of a curve. But again, for a curve, orientation goes along the curve. And for a surface, it goes across, it's perpendicular to the surface. But there are two different orientations in this case. And we have to say which one. So always on the problems which you will solve on this homework, it will have to say which orientation you have to choose. When it says downward, it means that the z component of the vector is negative. When it says upward, it means that the z component of the vector is positive. In this particular case, we use the normal vector for which the z component is one. So that means that this formula refers to the upward orientation, okay? And then, of course, the really fun part, which unfortunately I don't have time to talk about, is about surfaces which have no orientation. And this is something really cool which we'll have to talk about next week. But for now, happy Thanksgiving and see you next week.