 Have you recovered after the meter? No? Yes? Good, good. You actually did pretty well. I was happy about it. I think, good job. So keep on doing it. And you know, these lectures are on webcast.brookly.edu, right? They're also on iTunes. In the iTunes version, I sync all the formulas. So check it out. And now, our lectures are on YouTube. So I'd like to welcome our new addition, new addition to our audience, our YouTube audience. I think this is great because, actually, I try to search for math courses on YouTube. And I was surprised that, actually, there are very few of those. So if you people want to learn about multivariable calculus, you've come to the right place. And even if you don't want to learn about multivariable calculus, you've come to the right place. So, you know, unfortunately, there is only one lecture available for now. And it's like when you start watching a television series, and it's like in its third season, and you don't know what happened in the first season and the second season. And it's exactly the same here because, you know, I would say first season is like from the beginning till the first midterm. And second season, it's from first midterm to second midterm. And now we start our third season. I personally think the first season was the best one. But anyway, I think the third one will be the best. But, well, of course, all the lectures are available on webcast.brookly.edu. And I was told that they will all be available on YouTube also, perhaps this week. There were some problems with uploading. Anyway, so if you want to know what happened on Math, previously on Math 53, you know, should watch out. Stay tuned, right? OK. Having said that, what is this third season, so to speak? What is it about? And I personally think this is the best one. It's the most interesting one because that's where we will actually be able to apply all the stuff that we've learned up to now. And this is actually, this has been sort of a recurring theme in this course. I've said it many times that calculus, you can basically break into two parts, differential calculus and integration calculus. And in some sense, integration and differentiation are two operations which are inverse to each other, OK? So in some sense, you can write, you can say very roughly that integration is like differentiation inverse. Sort of a, this is a thesis which, you know, I will have to follow up on this. And I will have to explain what exactly this means in the context of multivariable calculus. We all know what it means in the context of single variable calculus. In one variable calculus, we have the beautiful formula which relates integration and differentiation, which is called the fundamental theorem of calculus or single variable calculus or the Newton-Leibniz formula. And this formula looks as follows. On the left-hand side, you have an integral. And to make this formula to kind of support this thesis, I would like to write it in this way. Integral of a function which itself is a derivative of another function. So you put a function here which is derivative of some function f. So you put f prime. And you integrate this function from a to b. And on the right-hand side, we simply have the values of this function at the end points. So we have f of b minus f of a. So I would like to think about this formula in a following way. On the left, we integrate our function over an interval from a to b. So on the left, I have an integral over this domain which is just an interval. Well, in a single variable. And by the way, no walking in front of the front row. It looks really bad on YouTube. I'm just kidding, it's OK. So on the left, we have an integral over this domain. And because we are in a single variable calculus now, there are not so many domains to go around. The only domains you can think of are intervals, or perhaps unions of intervals. So that's what we have on the left-hand side. And on the right-hand side, we have these two points which I would like to think of as the boundary of this interval. So this is an interval. And this is the boundary of this interval. Two points, maybe. Let me put them like this. Dots, really, and not this. There is something interesting that happens here. f of b enters with a plus sign. And f of a enters with a minus sign. So actually, this point comes with a plus sign. And this point comes with a minus sign. And we can actually read this off from the orientation of this interval. This interval has an orientation. We integrate from a to b. So there is an orientation. And so the arrow points from a to b. And therefore, the end point comes with a plus. And the initial point, a, comes with a minus. So you see this already indicates that orientation is going to play an important role. But for now, the essential point is the following. On the left, you have a domain interval. On the right, you have the boundary of this domain. That's about the region of integration. In some sense, you can think of the right-hand side also as integration. It's just that you are integrating over zero-dimensional domains, which means simply evaluating your function. What does, what do negative and positive mean here? Well, for now, it's just a notation that I put. I put, I assign plus to this point, which is point b. And I assign minus to this point, which is point a. And for practical, the practical meaning of this is the following, that when I evaluate the integral, so to speak, I evaluate at this point with assign plus and at this point with assign minus, which is what's written here, right? Okay, so there is some rule here that some part of the boundary comes with plus, with coefficient plus one, and some part of the boundary comes with coefficient minus one. All right, but integral is, in a way, integral is sort of a union of two things. The integrand, the object you integrate, right? And the domain of integration. The domain over which you integrate. So you cannot make an integral with just one of the two. You know, you need both the integrand and the domain of integration. So in this formula, there is a trade-off between something that happens to the integrand and something that happens to the domain. We see now what happens to the domain. When we go from left, from the left-hand side to the right-hand side, we take the boundary of the domain. On the other hand, let's look at the integrand. On the right, the integrand is F, and on the left, the integrand is F prime. So, what we do when we go from right to left is we differentiate, we differentiate. And here we take the boundary. And that is the structure, this is the basic structure of this formula. So what we would like to do is to find analogs of this formula in multivariable calculus. And that's what we're going to do in the remaining part of the semester, okay? So in other words, what we would like to do is to prove formulas of the following type. On the left, we have an integral of some quantity. And on the right, we'll have an integral of some quantity. On the right, it will be integral of some quantity, let's say omega. I will explain what kind of object this will be. For in some cases, it will be a function like here. But we will also have more general objects integrated here. And on the left, we will have some sort of derivative of this object. This is just a schematic notation for now. Don't read too much into it. It's something like taking differential or taking the derivative. For example, here it's just taking f prime, okay? So that's as far as the integrand goes. And as far as the integration domain goes, here we will have some domain D. And here we will have the boundary of D. You had a question. So here I'm talking about the domain of integration. We have a formula, right? We have a formula. Formal has a left-hand side and a right-hand side. And each of them is an integral of something, right, a function over a domain. So I'd like to trace here on this blackboard what happens to the domain and what happens to the boundary as we go from left to right or from right to left. And I observe that for the domain, on the left, we start with an interval, but on the right, we take the boundary of this interval. So it's like going from the left-hand side to the right-hand side for domains corresponds to taking the boundary. But as far as the integrand is concerned, on the right, I have some function f and on the left, I have f prime. So going from right to left, I differentiate. So this formula could be thought of as a trade-off between taking the boundary and differentiation, taking the boundary at the level of domain, taking the derivative or differentiating the integrand. Is that clear? So this is the basic structure that I would, this is how I would like to think about the fundamental theorem of calculus, which is of the one variable calculus, which is written here. I would like to, first of all, think of taking an integral as combining or pairing two different types of objects. One is algebraic, say a function. The other one is geometric, a domain. Okay, that's the first thing. And if I think about integration in this way, then I clearly see what is the structure of this formula. At the level of domains, it's going to the boundary. At the level of functions, it's taking the derivative. So this suggests that perhaps, if I have, if I am not on the line, but I am on the plane or in three-dimensional space, if I'm doing multivariable calculus now, I should also have a formula like this, where on the right, I would have some object integrated, and on the left, I'll have some object integrated. But the correspondence would be like in the fundamental theorem of the one variable calculus. That is to say, at the level of domains, we go from a domain to its boundary. And at the level of integrants, we go from omega, whatever it is. This we will have to discuss in more detail, it's derivative of some sort. You see, that will be the basic structure of the formulas that we will derive. And they will be the analogs of the fundamental theorem of calculus, and they will allow us to learn a lot of things about integrals. And we will also talk about various sort of real life applications of these integrals. What do these integrals mean? And various applications of those integrals. And such formulas will allow us to express integrals of one kind in terms of integrals of the other kind. That's basically the outline of what we're going to do next. Okay, so now of course the question is to discuss what kind of objects we should be integrating and over what? In other words, the integrants and the domains of integration. What are the most general objects that we should introduce which we could then use so as to obtain a formula like this? And what is the meaning of the corresponding integrals? These are the questions that we will have to ask now. So let's remember first of all, what kind of integrals we have learned up to now in the multivariable calculus. We have learned in the second season, so to speak. In other words, in the last months or so, we have learned how to integrate functions. So integrals, double integrals of functions and triple integrals of functions over some regions in the two dimensional space on the plane and over solids, three dimensional regions in the three dimensional space. But in both cases, we have integrated functions. So it turns out that this type of integrals are not sufficient to really do justice to the kind of formulas which I promised you. We will have to generalize these integrals first. And there are two types of limitations that we have now. There are two types of things that we'll have to relax and consider more general integrals. So we'll generalize these integrals in two ways. In two ways. First of all, we will learn how to integrate more general objects and functions. We will integrate vector fields. We will learn how to integrate vector fields. We will talk about what they are, vector fields, and what is the meaning of the corresponding integral. And that will be the first kind of generalization that we'll make. And the second generalization is that we will consider more general domains. More general domains. They will be, I will say roughly, the difference is that they will be curved. We'll consider most general curved domains. Whereas the domains we have looked at up to now are flat domains. So let me explain what I mean by this. Flat versus curved. So let me just illustrate this. The kind of regions that we consider here, the two-dimensional regions are, it could be some region on the plane like this. And someone might say, well, this is curved, right? It kind of looks curved. But that would be misleading, because what is curved is not the domain itself, which is the interior of this curve, but the boundary. In other words, in this subject, one of the most important things is to distinguish domains and their boundaries, okay? So that's why I would like to emphasize it, and I would like to trace this boundary in a different color. So this is the boundary, and this is the domain. These are two different things. This is the domain. For example, the domain is two-dimensional, right? It's two-dimensional, it's on the plane. But the boundary is one-dimensional. And this is always the case. The boundary of an n-dimensional domain is going to be n minus one-dimensional. You lose one-dimensional when you go, when you pass to the boundary. Same thing happened here. Here, the domain was an interval. It was one-dimensional, and the boundary was zero-dimensional. It consisted of two points. Maybe I should make them yellow also, to keep the analogy, okay? So the boundary here is curved. It is curved, I wouldn't disagree with that. But the domain itself is flat. It's flat in the sense that if you look at any point here, and you look at a small neighborhood of this point, if you look at a small neighborhood of this point, you don't know what the boundary looks like. So you don't have any illusions about the domain. You just look at a very small part of this domain here. And you see that it's just the same as a small part of the plane around this point. And the plane is flat. The blackboard is flat. So locally, this domain everywhere looks like the blackboard itself, so it is flat. You should contrast that with the boundary itself. The boundary is not flat. A flat one-dimensional domain would be an interval, or a line segment. So this would be flat. For example, this is flat. This is also flat. So you could have, of course, you could have a domain which is flat, and its boundary is also flat. You might have some corners. So this clearly, you see what I mean for the one-dimensional boundaries. Here it's curved, here it's flat, okay? Then you can ask, well, what's, give me an example of a two-dimensional curved domain. And that's very easy. Think of a sphere. A sphere is curved, right? Because a sphere cannot be, is two-dimensional, the sphere is two-dimensional. It's the same dimension as the plane. Same dimension as the blackboard. But you can't put the sphere inside the blackboard. I mean, you can project it, but you will have to make some violin transformation to the object. You will change its geometry, right? So that's the point. A curved two-dimensional object cannot be fit on a two-dimensional plane. Anything that is two-dimensional and that fits in the two-dimensional plane is flat. But there are a lot of two-dimensional objects like a sphere or a torus, right? And many other things, if you can imagine, like a surface of the ocean, many examples you can make, those are also two-dimensional, but they are not like the plane. So they are curved. And so far, we have not really talked about integrating over such domains. And we will need to do that here, okay? Is that clear? What's the boundary of a sphere? That's a very good question. So, before answering this question, let's talk about a simpler question, right? But analogous question in dimension one lower. You see, there will be many analogies between objects of different dimensions, but always it's easier to look at the situation where dimension is lower than higher. So oftentimes it's better to start with lower dimensions and then go proceed to higher dimensions. So, a good question here would be, what about boundaries of, what about boundaries of one-dimensional domains? So, if you just have an interval like this, we know what the boundary is. It's very similar to what we discussed in one variable calculus. So, if your domain is like this, it's one-dimensional domains like this, then the boundary consists of two points. But here's an example of a domain which does not have a boundary or more proper way to say it, it has empty boundary, a circle, right? So that's a question very similar to your question. You ask me, what is the boundary of a sphere? So the simpler question would be, what is the boundary of a circle? Boundary of a circle is empty, right? So you see the difference between this and this. In some sense, in a way you can think of, you can obtain this by combining these two endpoints, and closing these two endpoints. And you see, it actually all makes sense because the boundary, as we discussed, in the boundary, these points will enter with different signs. This one would say, if we orient it like this, this one with sign plus and this one with sign minus. When we close them, they will sort of cancel each other. So that's consistent with our natural observation that this circle has no boundary. Likewise, a sphere also has no boundary or has empty boundary. But if you take the upper hemisphere, for example, take the upper hemisphere. The upper hemisphere has a boundary which is a circle. So it's like upper hemisphere is like a dome, and the dome has a boundary which is a circle. So there are many things that you need to be able to visualize here and to understand and to see. There are certain, there are objects which do have boundaries, there are objects which don't have boundary. And you should always keep track of where you are. You talk about the domain, the boundary, inside three dimensional space, three dimensional space and so on. So there are many different options. Okay, so this I have explained. So this I have now explained. The second aspect is that we will integrate over more general domains, which in practical terms means that when we talk about single integrals, up to now we would only consider integrals like this. In other words, we just integrate over an interval. But now we will learn how to integrate over a curve of domains like this. Curved one dimensional, so that means this yellow things. And it could be closed like this. By the way, this has no boundary either, just like the circle. Neither of them has a boundary. This one has a boundary. And of course, you could have a curved curve, which, okay, it sounds stupid, but a general curve, a curve which has a boundary too, right? We'll have this minus this as a boundary. So there is a whole variety. There are closed curves without boundary. There are curves with the boundary. And the curves could be flat like this, or they could be curved, okay, right? So that's, but we will have to learn how to integrate over all of them. And that's just for single integrals, for one dimensional integrals. For two dimensional integrals, we will have to learn how to integrate over general surfaces in R3. For example, over a sphere, over a hemisphere, or over a torus. And in case you get worried, that's it. Because next, when we come to three dimensional, right, to have a curved three dimensional domain, we'd have to go to four dimensional space, right? And that's a subject for the next course. So season four, so we don't do it here, right? So we stop at three dimensional. We only consider in this class objects which can be embedded, immersed in a two dimensional or three dimensional space. And therefore, as far as the three dimensional domains are concerned, they're all going to be flat. They're going to be, for example, the interior of the sphere, right? The interior of the sphere, or a ball, is three dimensional. And unlike the sphere, which is two dimensional and curved, the ball is three dimensional and flat because it just sits in a three dimensional space which is flat, okay? To have something curved three dimensional, you'd have to go introduce one more dimension. So you'd have to think of time, but it will be very difficult to visualize. So we're not going to do that. Okay. So now we need to talk about the integrants. We need to talk about the objects that we will integrate over those domains. Question? Yes? That's right. That's right. I would take this. So the question is what would be the analog of passing to the boundary on this picture, right? So on this picture, on the left-hand side, we would have a double integral over this domain. And on the right-hand side of the formula, we would have a single integral over the boundary of this domain. No, inside will be on the left, right? When I say domain, I mean the interior. That's the shaded area here, right? That's the domain, two dimensional. So an example of a kind of formula which we will prove will be, we'll look as follows. On the left, we'll have a double integral over this, let's call this R of something which I'm not going to specify now. And on the right-hand side, we'll have the integral over this, over the boundary, boundary of R of something else. You see, so that's, is that clear? Okay, any other questions? Yes? Right, what did I mean when I said that the upper hemisphere has a boundary which is a circle? I'm kind of, I don't have good visual aids, unfortunately, so I have to just use my hands like this. But if you think of an upper hemisphere, let me try to draw this. Here. So it's going to look something like this, right? That's the dome. So it's like this, right? But it does have a boundary which is this. The circle on the bottom is its boundary, you see? The boundary means the following, that these are the points, the difference between a point here and a point in the interior is the following. If you have a point in the interior, you can go in any direction you like. There are two sort of, two degrees of freedom. There are two possible, two independent directions in which you can go. By the way, the same here. If you take a point in the interior, you can go in any direction, right? But if you are on the boundary, I mean, in the boundary, you can also go in all directions, but when I say all directions, I mean to go in some directions and stay within the domain. If you want to stay within the domain, you can go inside, you can go along the boundary, but you're prohibited from going outside. So some directions are prohibited. That's what it means to be on the boundary. Likewise, if you look at the circle at the base of this dome, you see that you cannot go down. This is prohibited. You can only go sort of up on the dome or you can go along the boundary, right? That's the difference between these points. But even if you have a point, even very close to the boundary, you can still go down a little bit. So that's the difference. Any other questions? Yes. Right, right, not every, because see, I have not put the integrand yet. We'll have to learn that. See, the trick is that at the level of domains, you take the boundary. But at the level of the integrand, you take the derivative this way. So in other words, what you will have here will schematically, I write it as a derivative of something, which is on the right-hand side. So it's not the most general integral you see here. It's the integral of something which is a derivative of something from the right. So this is not a formula which allows you to express any double integral as a single integral, but only allows you to express a double integral of something which is a derivative as a single integral of the quantity of which it is a derivative. That's the idea. But we'll talk about this more later, so I don't want to spend too much time making it precise at the moment. Okay. So what we need to do now is to describe the kind of objects that we'll have to integrate to really do justice to this kind of formula. And the point is that it's not enough to just integrate functions. We have to integrate more general objects, which are called vector fields. And of course, everybody knows what this concept is, maybe not the term. So it's actually very intuitive concept, which I'm going to explain. The way to think about it is as follows. First of all, let's remember what is a vector to begin with? What is a vector? A vector, we could work in on, everything that I will say will be, can be applied in R2 and R3, on the plane and in space. But it will be easier to say in R2, there's just fewer components and so on. So let me say it first in R2, and then we'll generalize it to R3. R2, on the plane. What is a vector in R2? A vector in R2, as we know, first of all, we can draw it like this. It's an object which has not only a magnitude, but also direction. So this is a vector. But algebraically, we represent it by a pair of numbers. We write it as PQ, say. So it's not just a single number. It can be represented by a pair of numbers once you choose a coordinate system, PQ, and we also use a different notation, PI plus QJ. Remember, we've learned, this was, I and J was our notation for the unit vectors along the X axis and along the Y axis. So that's a single vector. This is just one vector. But suppose you would like to attach a vector to each point on the plane. Suppose we have attached a vector, a quantity like this, to each point. On the plane. So what is this going to look like? Say this point will have this vector and then this point will have that vector and then this point will have this vector and this point will have this vector. So a priority, they can all be different. They can all be different. There should be no connection. Of course, the simplest situation would be just assigning the same one to all points, but it's kind of boring. So you can actually, you can just assign them in a very complicated way. So this actually occurs in practice, in reality, in many different ways. And the simplest way to think about it is wind maps. Wind maps. Actually, it's funny because in the book, the picture which illustrates vector fields is the picture of the wind map of San Francisco Bay Area. So I guess for a good reason because our winds are changed and kind of have an interesting pattern. Some years ago, I used to do a lot of windsurfing and I remember there was this website where you could go and you'd get a beautiful picture of the winds real time, the winds real time all across the San Francisco Bay. So if you want to windsurf, so let's say that this is a bay, roughly. This is San Francisco. This is a Golden Gate Bridge. This is San Francisco. So there are, it's very schematic, okay. And this is Berkeley. I'm looking from the top. Oh yeah, this is important to say that, right. I'm looking from the air. So there are a few, let's say there are many spots across the bay for windsurfing. So let's say here in the Berkeley marina, there's a Emmerville and here you have a treasure island. And so on. And so if you look at this map, you know right away that the wind here is say east points in the east, eastward winds and then like here will be north and here will be south, whatever, right. So that's the meaning of, you've all seen such maps. You can, or you can imagine such maps because the thing about wind is that the wind does not just have magnitude, the force of the wind, but also has direction. It could point in the northern direction east, west, south, and so on, right. So wind at a given point is a vector. It's not a number, it's a vector. But wind is everywhere. So therefore, at each point you have a vector, the wind vector, right. So wind represents a vector field. So this is a vector field. So a vector field is an assignment of a vector to each point on the plane or maybe some given domain on the plane. Maybe not everywhere, but I mean of course, say here I talk about a particular domain which is on the map say, which is just San Francisco Bay Area. I'm not talking about the entire United States or entire state of California. So this is a vector field. Is this clear? Okay, so now let me contrast that to a function. Up to now we've talked about functions and we've been integrating functions. A function would be just a function means assigning to each point, not a vector, but a number. For example, think of a parametric pressure. At each point you have a parametric pressure. And parametric pressure is not a vector, it's a number. So at each point you have a parametric pressure. At each point you have a number, so you get a function. What we normally call the function is assignment of a number to each point on the plane. And vector field is almost like this. The only difference is that we assign to each point not a number, but a vector. But in some sense, why not, right? Okay, so how would we represent such a vector field algebraically? Algebraically, well, each of those guys, each of those vectors would have two components, but now these components depend on the point. So a point x, y, so we have a point x, y on the plane would have a vector which will have components p of x, y and q of x, y. So we will write p of x, y, q of x, y. And we will put the square brackets as we usually do. Or we can also write it as p of x, y times i plus q of x, y times j, just like we did before. Just like we did before, pq or pi plus qj. The only difference is that now p and q themselves become functions, right? So it's p of x, y, q of x, y. And we will call this vector field f of x, y. So oftentimes, you know, for practical purposes, you can say that a vector field is nothing but a pair of functions. So it's not a single function, it's two functions because the vector has two components and now each component becomes a function of x, y. But really reducing it to a pair of functions is not a good idea because it kind of strips the vector of its geometric meaning because the vector is much more than just a pair of numbers. First of all, this pair of numbers can be given only once we choose a coordinate system, right? I mean, after all, what are these numbers, p and q? I have drawn them here. Those numbers are the projections of the vector onto the x and y axis, right? But if I choose a different coordinate system, the vector will stay the same. The vector will stay the same. But the projections will be different. So you have to remember this algebraic representation really depends on the choice of coordinates, right? But the vector itself doesn't. In other words, let's say here is a map of San Francisco Bay Area. San Francisco Bay Area doesn't know anything about coordinates. It exists without any coordinates. Somebody will come and make coordinate system like this, but somebody else will come and make it like this. Who knows, maybe for some purposes this will be more convenient, okay? The wind doesn't know about the coordinate system. This vector will stay, the wind will blow in this direction no matter what coordinate system you'll prescribe, right? So the wind is for real. It's a real geometric object. And the two numbers that we assign, or two numbers which we use to represent this vector is an abstraction, is a way to represent them algebraically. But this way of representing it algebraically depends on the choice of coordinates. So that's why I am saying that reducing a vector to a pair of numbers or reducing a vector field to a pair of functions somehow doesn't do justice to the object. It sort of reduces, strips it of its geometric meaning. Nevertheless, in our calculations oftentimes, we will be working with vector fields as though they were just represented by these two functions. So for all practical purposes, we just have to integrate those functions, say, or differentiate those functions independently, okay? Yes, that's right. So let's say at this point, this is a point x, y. So you have this vector. So here is another point which will say, we'll have say x1 and y1 coordinates. And at this point, let me use a different chalk to represent vectors. So let's say this vector goes like this. The vector assigned, this is the vector assigned to the point xy, to the point xy. And say that this vector assigned to this point will be this vector. So this vector will have, will be p of x1, y1, comma q of x1, y1. Chapriory has nothing to do with this. You see what I mean? But each point, you have a vector. Well, a function represents a bunch of numbers, right? So in other words, say this function, what is a function? Function is a number attached to each point xy. For any xy, you have some number. And now you have these two functions assembled into a vector, right? So you have a vector for each xy. The vector attached to xy, say, would be this vector. And a vector attached to x1, y1 will be this vector. A vector attached to x2, y2 will be this vector, and so on, represented by a formula. So let's see, let's look at an example of a vector field. Okay, so the simplest example would be, let's say that f of xy is just i. What does it mean? It means that p in this case is 1 and q is 0, right? From the point of view of this general formula, p times i plus q times j, p is 1 and q is 0. So I just write it as i. What is the geometric meaning of this formula? Well, it says that the vector attached to any point is one and the same vector i, right? So what does it look like? Well, at each point I just have vector i. This is vector i. This is a vector which is parallel to the x axis and has magnitude one. So here I have this vector and here I have this vector and so on. I cannot draw them all because I cannot account for all the points on this blackboard or even on the small part of the blackboard because there are infinitely many of them. So I'm just drawing a few of this. And this is what's called the diagram. The vector field diagram. And that's what is meant to, it is not meant to represent the entire vector field because, of course, inevitably, no matter how many vectors I will draw, I will miss some points. There are infinitely many of those. This is just meant to give you an idea as to what the vector field looks like because, of course, the point is that in these studies, in this course, we are going to consider kind of nice, the vector fields will consider will be nice in the sense that the vector field, in general, the vectors will change from point to point but they will change in a continuous manner, in a smooth manner. So even if you know what the vectors look like at a certain number of points, this should give you an idea as to what vectors look like in between. So this is a simplest vector field. It's called a constant vector field. Okay, so let's do something more complicated. Something more complicated. Question, yes, shoot. That's right. If we say a vector field on the plane, that's right, very good question. So the question is, do we have to assign a vector to every point on the plane or not? Okay, and it's the same kind of question as asking, suppose you have a function. Does it mean that we have assigned a number to each point? And the answer is, it depends. By default, if I just say this is a function on the line or on the plane, I'm saying that it is defined everywhere but we know that there are many interesting functions which are not defined everywhere. For example, even in one dimensional context, say function f of x equals one over x. This is an important function and it's not defined at zero. So this function assigns a number to every point x on the line except x equals zero. Sometimes we only want to know the values of the function on a certain domain. Let's say on an interval from zero to one and we don't care what happens at other points. Same goes for the vector fields. But if we say vector fields on the plane, we kind of implicitly saying that there is a vector attached to every point. But in many situations, first of all, a formula that we write will not be well defined at some points on the plane, right? So then it means that the vector field is not defined there or it could be that we're just only care about the values across a certain domain. So we only define it on that domain and we don't care what happens outside of it. Any other questions? All right, so next example, let's say f of x, y equals x times i. No, let me use the y, x, i plus y, g. So this is slightly more interesting. This is certainly not a constant vector field because now the function p of x, y is not a constant function but it is function x and q of x, y is a function y, right? Let's see, so what does it look like? So for example, let's say we take the point one, zero. If we take the point one, zero, I should put an arrow, of course. When I talk about vectors, I have to put arrows on top. That's the usual notation. So let's say f at one, at one, zero. This is a point one, zero. If I substitute one, zero, I get i. So here it's going to be vector i just like in the previous example. But this is really the only point where it is going to be vector i. Let's take, for example, the point zero, one. Zero, one, we get j. So this is a point zero, one, and we get this vector. And likewise, for example, you can take negative one. Say you can put plus minus one here and this will be plus minus i. So this will go in this direction. This will be negative five at this point. And likewise, here, plus minus one will be plus minus j. So this will be this vector. So this already kind of tells you what the pattern is. The pattern is that at each point, in fact, the vector is going to go in the direction of the position vector of that point. For example, this is a point one, zero. The position vector of this point is this vector. And the value of our vector field is going to be just that vector. But we draw it. See, it would not be right to draw this vector starting at this point because we are drawing this picture with the understanding that the vector that starts at this point is the vector assigned at this point. But of course, as we discussed many times, a vector, the same vector can be represented in many different ways. We can start with any point. Here, we start with the point to which it is attached. This vector is attached to the point one, zero. So we draw it here. This vector is attached to the point zero, one. It's written here. So we draw it here and so on. And more generally, if we have a point with coordinates x, y, coordinates x, y, the vector attached to this point is going to be xi plus yj. But that just happens to be in this example. It just happens to be the position vector of this point, which we normally draw like this. So what should I draw then for this point? I have to take this vector and transport it here so that the initial point is here and not here. So then I end up with this. I just transport it here. So that's the vector ij. This is vector xi plus yj, which is prescribed to this point. And it's the same as this vector. So I can now kind of draw more. So you see, they all move away from the center. They all move away from the origin. At each point, the direction is exactly away from the origin, but the magnitude actually changes. The magnitude changes. The magnitude is equal to the distance from the origin. So if I go far away, the magnitude will be even higher. If I am very close, the magnitude is very small. So that's the vector field that I get. It's kind of an explosion. It's kind of an explosion. Well, it's not quite an explosion because if you have an explosion, it actually decays as you go farther away. But here, actually, the force increases. The magnitude increases. It's like a universe expanding, that's right. Or it's one of the scenarios. Of course, we would like to think that it's expanding, and it is expanding now, but we don't know what will happen eventually. And it might all collapse. But not before the end of the semester, don't worry. So unless some of those formulas are incorrect. So okay, but let's suppose that you wanted the same kind of deal. In other words, you wanted the vector field which also goes away from the origin, but you didn't want the magnitude to be higher and greater and greater. Let's say you wanted the vector field where each vector has magnitude one. This is very easy to do. You can just convert each of those vectors into a unit vector pointing in the same direction. So let's say the third example would be you would have to take x times i plus y times j and divide by the magnitude. And the magnitude is x squared plus y squared, a square root of x squared plus y squared. So now each of these vectors is a unit vector pointing in the direction of the position vector like before. So in other words, the magnitude is not increasing, but it stays fixed equal to one. Or if you want the magnitude to the vector field to decay, you can divide by a higher power of this. So for instance, you can also take f of x, y equals xi plus yj, but divide by, so actually you could divide by the square of this, but it is a little bit better to divide by the cube. And I was playing y squared plus y squared, three halves. If we do that, then the magnitude of this vector, the magnitude of this vector, or the length if you will, of this vector will be what? When we take the magnitude we get square root of x squared plus y squared in the numerator, right? And we get the, is this the right? No, actually I don't want three over two. I want actually, how do I want it? I want to have twice square of the length. That's right, no, it was correct. That's exactly what I want. So the magnitude would then be one divided by x squared plus y squared, right? Which is the square of the length, which is the square of xy. So that corresponds to vector field where the magnitude is proportional to the, or equal to the inverse square of the length of the distance. And this is very typical for very important forces in nature. So this is going to be the behavior of this vector field is like the behavior of the force of gravity, force of gravity, or the electric force, electromagnetic force, okay? The force of gravity is inverse proportional to the square of the distance. And that's exactly what we get. But because in the numerator we have the vector, position vector itself, we divide by, instead of dividing by the second power, we divide by the third power. So that when we take the length, there will be one power here and three powers here. So only the two powers will be left, okay? Yes? Because I would like to get a vector field such that the length of a vector is equal to the one divided by the square of the length of this. Because this kind of behavior is characteristic of important forces in nature, like the force of gravity. You know that if you have two masses, say sun and earth, two masses, then two objects and then the force of gravity is inverse proportional to the square of the distance between them. And also proportional to the masses and there's some coefficient of proportionality. Do we do what? Do we cube every problem we do? This is just one example, okay? One example. I'm not saying that all vector fields look like this. It's just one vector field, okay? All right. So just trying to illustrate how you can produce new vector fields by starting with some known vector fields. Okay. So the next question is, so if these are the vector fields, the next question is, how do we integrate these vector fields? So what I am going to explain is how to integrate vector fields. That's right. Why would we be interested in this? Right? So that's the question. Wow, yeah. This is a math class. Come on. But no, seriously, of course we wouldn't be doing that if there weren't a good reason for it. And there is a good reason for it, which I'm going to explain now, okay? What I mean by this comment is that in math, I'm not trying to say that we do useless stuff. On the contrary, we do very useful stuff, but sometimes it takes a while to really appreciate what exactly this is good for. And so you have to be patient sometimes, okay? But in this case, actually you don't have to be patient because I'm going to explain it right away. So this is one of those cases. Okay, so I would like to explain the real problem which actually will lead us naturally to the notion of an integral of a vector field, over a curve. And this is actually very easy to explain. And this has to do with calculating the work done by a force. So by the way, one important example of a vector field which I have not, which I kind of alluded to here, when I said force of gravity, I should say it more forcefully probably, which is that forces, forces in general are vector field, are represented by vector fields, right? Because a force also has a magnitude and direction usually. Like force, it pushes you in a certain direction, not only with different levels of strength, but also pushes you in a certain direction, or pulls you in a certain direction. Like force of gravity, force of gravity pulls me down and not, you know, I'm not flying out of this class, which some of you may have wanted, but I'm staying here. And the only thing, and the only thing that happens is that, well, this floor prevents me from falling down, but if suddenly this floor were disappear, I would go down, right? So I would go down, not up, and not right, not left. That means that the force of gravity in this case actually has a very specific direction down, like the apple which fell on Newton's head, right? So force has a certain direction, and likewise electromagnetic force and other forces, they are represented by vector fields. And if you have a force, then oftentimes you may be interested in the amount of work that a force, certain force, first vector field, does in moving an object from one point to another, from point A to point B. And the option is that this kind of quantity, the work done by force, is precisely represented by an integral over curve, where the curve would be the path along which the force moves the object. Let's say particle, if it's electromagnetic field, or if it's a force of gravity, it could be any object that you want. So, but to explain this, to explain why an integral, okay? As always, we have to go back to kind of the elementary situation where everything is simplified, where the force is constant and the displacement is kind of linear, and we look at the formula for the work, and then we kind of average it out over a general curve, okay? And that's how we get to the integral. So the first question I'm going to ask is, what is the amount of work that is needed to move a particle from, say from this point to this point, along a linear segment with some displacement delta R? If you have a constant vector field, if a force is constant, so force is everywhere, constant like in my first example, where it was I, like this. Where is the amount of work needed to move particle from point A to point B, to B in the constant force vector field, like this. So in other words, the value of the force is the same everywhere. And actually it's very easy to illustrate this in the following way. And I will do it, not with a particle, but I will do it with this eraser. You see, this is what I'm talking about. I'm moving this, unfortunately I've erased it, but I hope you've memorized what was here. Let's do it again, no. So, okay, you see, I'm moving this eraser from this point to this point, and I'm applying force. Now, the most efficient way to do it would be to use my force to go exactly parallel to the blackboard, right? But in reality, I don't do this because I also want to, I want to push sort of against the blackboard. So first of all, I don't lose the eraser and also to erase it better, right? So in other words, oftentimes the displacement vector and the force vector are not parallel to each other. That's natural. For example, I can go like this. So see, there is a component of my force, which is moving it this way, but there is also component which goes this way and keeps it close to the blackboard. I can do it like this, or I can do it like this. A stupid thing to do would be to go like this because then I wouldn't be able to move, right? So in order to do this, I have to have a non-zero component in this direction. But then I can have some component in the transverse direction. So the question is, what is the work that has been done? And the point is that what is essential for this process is not the force itself, but only the horizontal component. The component in the direction of displacement, you see. In other words, the work is equal to the displacement, which is this, let's just write it in words, displacement times the component of the force, component of the force along the displacement vector. In other words, what matters is how thoroughly I erase the blackboard. It doesn't matter how much I sweat doing this, you see. So what I'm talking about is not how much work I have done, but the question is how much, what is the result, you see. And for the result, it doesn't matter how much pressure I was applying towards the board. What matters is how much force I applied in the direction of the displacement vector. So that's why I get this formula. But if so, it means that algebraically, algebraically, I draw this, this is a displacement vector. Let me draw it again, right? And so this is the force, the force vector. And this is a displacement vector. Let's call it delta R. And so the work is, the work done moving this particle is, is a, is delta R times F, but not F, but rather this, this is some angle theta. So, so this is actually, the length of this is F, length of F times cosine of this angle. So times cosine of theta. So what I'm trying to say that it is just a dot product between DR, that's what I'm trying to say. The work is a dot product of the displacement vector and the force vector. In this kind of simplified situation where displacement is linear and the vector field is constant. Okay, or if you want, you can write it as FDR, F delta R, doesn't matter. Sorry, I should have written DR, not delta R, this fact. Okay. So that's the idealized situation. And now the general situation will be as follows. I will have a more general curve, not necessarily a line segment. So it will be, let me draw it like this. It will be like this, let's say, and this will be some point A, A and this B. And at each point, I will have some point like this. So you see the vector field changes. Has different values at different points. Okay. So let's suppose I would like to calculate the work done by this force in moving this particle from this point to this point along this curve. So what should I do? I should break it into small pieces. And this is a typical thing. This should be almost a reflex by now because this is the kind of thing we always do when we talk about integrals, right? We break it into small pieces. And each small piece, we approximate by this idealized situation where this will be, displacement will be just linear, right? And the vector field will be constant because if the vector field varies in a continuous way, in a smooth way, more precisely differentiable, if it is differentiable, right? So then it is a good approximation on each, on a small segment, sub-segment here to think that the vector field actually doesn't change so much. So it's a good approximation to replace it by the value at any given point within this small interval, right? And then the work will be equal to, the total work will be the sum of the work done on each of the small intervals. And this will be the sum of f, evaluate at some points along this, within this interval, dot delta Ri. And so you know what will happen, you know what will happen when this partition becomes finer and finer. This sum will converge to an integral. Because this of course is very reminiscent of all kinds of integrals we've done up to now. We've done, you know, double integrals and we have done integrals as a triple integrals and many others. And it's always the same idea that if you have an expression like this where you have a sum over this small, you know, over this kind of partition, then in the limit, you just replace this by an integral. And this is valid where if the quantities involved satisfy some natural conditions like being differentiable. And so the result of this is an integral of f dr. I would like to say from A to B, but it's, it would be a little bit, it would not be very, would not be a good idea to write it like this. You see, when we were in the one variable calculus, there was only one way, connect point A to point B because we were going along the single line that we are given. Here we are on the plane and there are many different ways to connect these points. And of course the answer will depend on which way I'm going, right? The longer the path, the longer the work, the larger the work that will be done, okay? So instead I will write it like this, we're over C and I don't mean like A, B, C, but I just want to denote the curve by C, by letter C. So A and B denote the end points, but C denotes this entire curve. It is important, however, to remember the orientation of this curve. We go from A to B and not from B to A. If we were going from B to A, the calculation would be different, okay? So now, let's talk about this formula. So this was to convince you that an integral of a vector field over a curve is important. It gives you the quantities like the work done by a force on a certain particle, let's say. But now I would like, I have to explain in more detail what exactly this integral stands for. So I would like to explain it in more detail and also give you recipe, how to calculate it, okay? So let me write it down one more time. It's F dr over C. So for the dot product, we have a nice formula in terms of the components of the vector. So let me remind you that F is a vector field which looks like this. It's p times p of x, y times i plus q of x, y times j, right? And what is dr? Dr is a vector which is dx times i plus dy times j. So let's take the dot product of these two. Dot product is going to be p of x, y dx. Plus q of x, y dy. So in retrospect, the result actually looks very natural because you see in the single variable calculus, the only quantities that we were allowed to integrate were integrals like this, you know, let's say F dx, F of x dx. Well, first of all, there was a function which depended only on one variable because it's a single variable calculus. But also there was only one dx. We could only integrate against dx because there were no other variables. Now we are on the plane. So there is dx and there is dy. So we could integrate a quantity of the form p times dx and we could integrate a quantity of the form q times dy. And this is the most general expression which we could actually integrate over a curve in the two dimensional space. Now we got this by thinking in terms of the work done by force and by writing as a dot product of the force vector field and dr. But you see the end result is actually very nice. And we could have actually written this end result from the beginning as the most natural expression really that we could think of integrating over a given curve. Again, because you have dx and dy and then each of them could come with some factor, with some function which would be some function p and some function q, okay? So now the only thing that remains is we need to explain how to actually compute this. And this is where the methods which we developed at the very beginning of this course are going to become very useful. I'm talking about of course a parametric curves. To compute this integral, we need to parametrize our curve, okay? So to compute, we parametrize our curve, our curve c. In other words, the curve was like this, but we parametrize it, which means that we're actually identifying it with a certain interval. But an interval and then some auxiliary line with a coordinate which we usually call t. So we identify this curved object, this curve on the plane with a straight object with a line segment on this auxiliary line. We kind of identify this point with this and we identify this point with this and the points in the middle identified in a certain way. So that there is a one-to-one correspondence between points here and points here. And algebraically, we write it like this. We say x is f of t and y is g of t along this curve and we put some limits on t. It's from a to b. So here the end points were actually points on the plane. I could not have, that's why I call them capital A and capital B because actually capital A was a pair of numbers. It was some, which actually now I can write as f of a and g of a. And this b is f of b and g of b, right? And so let's substitute this in this formula. So what do we get? We will get integral now just over this interval which is great because this is already a kind of integral what we did in single variable calculus. Before it was an integral over this curve but now we write this integral over this curve as an integral over the line segment by using this parameterization. And it looks as follows. We have to substitute instead of x and y, we have to substitute f of t and g of t. And then we have to write dx in terms of t. But what is dx? dx is of course just f prime dt just by usual formula. dx is f prime of t dt. And dy is g prime of t dt. So I replace dx, this dx by f prime of t dt. And then I write the second one which is q of f of t g of t times g prime of t. And I close the bracket. Or if you will, you can write it as a sum of two integrals. This is the first integral and this is the second one. But you see, I have, by using the parametric form for this curve, I have been able to rewrite this integral which involved x and y as an integral in one auxiliary variable, namely t. So let's see, we have actually one minute to do an example of this. So let's say compute the integral of z dx z dx plus x dy plus y dz. See, and on purpose I'm doing it in a three dimensional space just to show you that actually everything works out in the same way in a three dimensional space as in a two dimensional space. The difference is that now you have three variables x, y, z and likewise you have dx, dy and dz. And so let's say you have a curve in which x is t squared. Y is t cubed and z is t squared. And t goes from zero to one. So the first term, so likewise for dz, you will have some h prime of t dt where h is the formula for z, right? So this will be the integral from zero to one. z is equal to t squared, dx is equal to two t times dt. Plus x is equal, sorry, I was z, that's right, z, dx. x is equal to t squared, dy is equal to three t squared dt. And finally here you have t cubed and dz is two t. So you end up with a very simple integral in one variable. Okay, so we stop here and we'll continue on Thursday.