 So in terms of integration, again, you know, following the theme, if we have an integral like the integral from 1 to infinity of 1 over dx over x squared, something like this, so maybe you vaguely remember that this actually makes sense. We have a graph like that, and we're asking about the area under that curve as all the way out to infinity. And if we make sense of this by writing it as the limit, as m goes to infinity, it'll be integral from 1 to m of dx over x squared, which this is all fine. So that will be the integral at 1 over x squared is minus 1 over x, evaluated from 1 to m. That'll be minus 1 over m minus the minus 1 over 1. Now we take the limit as m goes to infinity, and we get 0 plus 1, so that's 1. So the area under this curve, no matter how far we go out, it gets closer and closer and closer. So this says that as we go out further and further and further, the area under this curve tends to 1. So it makes sense to say that the value of this is 1. Right? And similarly, if we try the other end, dx over x squared, let me just use the same integral, then this would be the limit as, let's call it L for lower. So this blows up when x is 0, so we can't just evaluate it directly, but we write it as a limit. So this is the integral from L to 1 of dx over x squared, which is 0 of minus, as L goes to 0, this blows up. So this is infinity, which is saying that this part of the area here there's an infinite amount of area here, whereas here it shrinks very fast. Naturally if we change this 2 to a half and this 2 to a half the rolls will be reversed, this will be infinite and this will be finite. Right? So everybody kind of sort of remembers that from when you did calculus 2 or whatever name it had when you did it. So now we want to make sense of the same process in integrals with more than one dimension. So say we have something like the integral, let me write it symbolically over the unit disk of x squared plus y squared to the 1 half power. So what does that mean? So here the unit disk is the set of x, y such that x squared plus y squared is less than 1. That's the open disk. And so I'm not parameterizing this as a specific thing. So this surface f of x equals 1 over x squared plus y squared square root is something, the graph of this is going to be, well I'll cut it off, something like that. This is sitting under the unit surface. Right? So it really goes up to infinity. I cut it off. So it will be some horn shape where the place you would blow the horn is very far. So this is, I guess this is a trumpet for an angel to blow. It's way up here. So the question is does it make sense to describe the volume here? And the process is exactly the same but we have to take a little bit more care because really the way we're going to do this is let's let d sub delta be the set of points x squared plus y squared is between delta and 1. So in the plane, this region, this angular region here, this ring, so if you speak Spanish then it makes perfect sense because in Spanish this is a kind of needle. We don't use that word so much in English. Anyway, so we're going to integrate over this and then we're going to let delta go to zero. Right? So we would define, if this makes sense then this should be limit as delta goes to zero of this function. It makes sense of that. So we can either, well since this is nice and circular and everything it would make sense to integrate this in polar coordinates. I think so anyway. Of course you can do this. You can parameterize this as what? y equals to square root of 1 minus x squared here and y equals minus square root of 1 minus x squared here. It's just horrible, right? Because you'd have to split it up into several little bits. But in polar coordinates it's very easy. So let's just do it in polar coordinates. This region is just r is between delta and theta. And so the region we want to integrate over is just that. And then we have to write this in polar coordinates. So this would be, if we parameterize this in polar coordinates, let's let theta go from zero to 2 pi and r go from delta to 1. Then this delta goes to zero. Da is r dr d theta. This is r squared. r is positive so the square root of r squared is just r. Yeah, so this becomes really easy. So this is r. That's r. It's gone. So then that integral, when we integrate with r we get a factor of delta. And then delta d theta integrated d theta is 2 pi. So we get the limit. Something's wrong. Sorry about the 1. Okay, good. This is 2 pi 1 minus delta. There, that's better. Somehow I was just thinking it was 2 pi delta and I was shocked that it was coming out to be zero. But it's not. Which is 2 pi. So the area, the volume of this form. So the analogy between the one variable improper initials, just these are called improper initials. Between the one variable case and the multivariable case is exactly the same. You just have to take care to realize that your limit has to be coming in from all directions. So you have to take care that your limit is as a little disk or a sphere, if you're in higher dimensions, which shrinks down to the point where you blow up. What if I should shrink the other way, say constant, but not also pi at volume? Say that again? Wait, what if you have one direction and say constant and one of them goes zero and not also pi at volume? Yeah. So you don't have to keep it up to zero to set it up, right? No. No, just this happens to go to zero. But it could go to some constant. That's my point too. Right? I mean, I did just. I'm sorry? It goes to constant to constant to constant to constant. Before or after anything? Just a little bit. You mean the integrand? This part? No. If, like, how do you understand it? It has to close in on that. Well, it doesn't have to close in on the origin. It has to close in on the place where it blows up. Couldn't one of them stay constant? One of them could stay constant and you could close in on that. Sure. So let me do one more example. So, I mean, you just have to take care that you're taking the appropriate limit. You have to think about where it's blowing up and you have to take the limit as it blows up in the right way. Right? So you could just as well take, you know, we could certainly do something like, it's 10 to 3 in the morning. Okay. So if we had something like, something this is y. So if we had something that was, I don't know if you can see this picture. I know what I'm drawing. This is the function f of x, y is just 1 over y squared. Well, actually, yeah, sure. And let's just go off to infinity. Then, certainly we can just take the limit as y squared goes to infinity. So this doesn't come down to a point, but okay. Or I could try and find, you know, this other one where I get a whole line and that would be fine too. So, yeah, you could do that. You just have to think about what's blowing up and what's the shape of what's blowing up and that you have to take the appropriate limit. So let me do one other example. I don't know if you know that this is true. It's a true fact. Which, as opposed to a false fact. It's a true thing. Which is saying that if we take the bell curve in statistics and we don't normalize it, and the area under it from plus infinity to minus infinity is this square root of 5. We do this integral in one variable because you can't make a substitution to turn this into something in a nice form. But we can be tricky. Instead of doing this, so how do we do this? Risingly, this is hard to calculate, but this is easy to calculate. Which is someone surprising, but have to be a little bit tricky. So we have to view this as saying if we look on any square in the xy plane we put a finite bump. So this bumps both ways. So I'm taking the square and I'm lifting it up. I'm taking the square and I'm pinching it up in the middle. I don't know from minus m to m in all directions. Then the volume of this is that. Well, so the volume of this guy equal to this is the same as if I integrate let's just say over the reals over the whole plane of e to the minus x squared minus y squared dx. This thing is perfectly symmetrical so that if I cut this in half that'll give me the area in one direction and the square will give me the area in the other direction so the product of those two areas will give me the area of this volume. Because of the xy symmetry in this. So that's why the square is the same as if I write this as a double integral over x and y. And I have to integrate it over the whole plane because it's going off all the way out. But this integral is easy. Maybe it doesn't look so easy right here. But this is the same thing as if I integrate in polar coordinates over the whole plane e to the minus r squared r dr d theta. Polar coordinates are magic. In this case, instead of having this square thing I have half for somebody who really needs a lot of shape. This is a really big sombrero. So, yeah. That's a square for y squared minus y squared. Yeah. It's negative x squared plus y squared. Good to see you. Yeah. If I make this, this is a really big sombrero. The rim goes out to infinity in both cases. But now I'm integrating over a circle or a disc. So then this becomes a lot easier. This is an easy integral. So if we write it in polar coordinates, theta goes from zero to two pi, r goes from zero to some big number to the minus r squared r dr d theta. And then we take the limit as r goes to infinity. That's a big r and that's a little r. In this integral, we make the substitution u equals u equals negative r squared. So du is let's do plus r. Is two r dr. And so this is integral from zero to two pi. And as u equals zero means that r equals zero, r equals big r means that u equals big big. What is that square root? So this is the same as the integral limit as r goes to infinity. I guess what is that? So u r is zero to the minus u du theta. One half. Okay, now I'm following this thing. This is then two pi limit as r goes to infinity of minus e to the minus u from zero to the square root of r which is well it doesn't matter. That goes to infinity so that's one. And that's one. Right? Yes. So this is two pi times one minus zero minus one which is two pi to half. So the square root of the integral is pi. The integral is square root of pi. Yeah? I don't think it's important but I just want to get why e to the negative x squared integral square is equal to the cross section of e to the negative x squared plus y squared. But you don't understand why that integral is the square root. Yeah. Okay. No, I need symmetry. Or not, I want to try symmetrical. Does that what we... Okay, so it's true. Yeah, so he's questioning why that's true. E to the negative x squared times e to the negative x squared. We're right. Okay, so this is integral... So why are these integrals multiplicative? That's what you're going to say. Right? This is integral for minus infinity to infinity of e to the minus x squared x minus infinity to infinity e to the minus x squared dx. That's the square. But we can let that be a y. Yeah. So that's really true. And now this if I integrate... So this is some number. Let's call it square root pi just for the hell of it. Square root pi times the integral of e to the negative x squared dx is the integral of square root pi e to the negative x squared dx. So this is, because these variables have separated, I can integrate minus infinity to infinity minus my hat's in the way. It's too late. Minus infinity to infinity e to the minus x squared e to the minus y squared dx dy. Because these are two independent variables. Don't depend on one on the other. So if I had this one, then I could go that way. Because I integrate in order. And since I'm integrating in order, that gives me that. So I can play a trick with squaring to turn it into something bigger. And then I can change and since the limit is going to infinity, I can have a hat instead of a... I have no idea what that is. A square square root. It's not the same. Yeah. Let it dot. I was just changing my limit to do the great stuff. So if r is the wrong... That's basically a square root. Maybe it should be a square root. You have to say it in Spanish. In Spanish. Square root. Square root. Square root. Okay. What? It's a cuadrero. It's a square somewhere. Okay. So I'm done with that. So now I want to move on from this stuff. Are we okay with...so incorporated variables are exactly the same. But you have a few more tricks you can play because you have more dimensions to play with. So in some... In multiple dimensions that you can't do in it. Okay. So now we move on to the next... The section of the class. The fourth third of the class. Okay. So we talked a little while ago. So this is a big changing topic now. We have some function from rn to rn. It's in two ways. Let me do it in the case where n is 2 because it's easier to draw. We can think of...take this as taking something in the x, y plane. Flying up to it and getting...that's supposed to be a... That takes an area in the plane and does something to it. Gives us another area in the plane. There's one way that we can think of a function of two variables. And another way that we saw...these are not going to be the same f because I don't think I can draw that one. Maybe I can. Or different f, so let's call it g. Each point we attach a vector. So we get a field of arrows. Think of a function of several variables in either way. Depending on the context, one is more appropriate than the other. But they're really the same thing. I mean, to try and turn this picture into a vector field picture, I would superimpose this. So let me see if I can. So this one backs up a little bit. So I have sort of little arrows here. And here I have great big arrows. That would be the vector field, of course. We would not put it in the same. But half of the vector field would look something like that. Because these points go very far. So they have big arrows attached. These points move a little bit. So they have little arrows attached. I would attach this arrow to this point. Does it make sense? Anyway, it's not really productive to try and switch back and forth between the two representations. Think of the one that's appropriate for what you're doing. So now we're going to focus on calculus related to vector fields for essentially the rest of the class. And these are particularly relevant for things like physics. But a lot of applications. So, you know, imagine I want to represent, I don't know, fluid flow or forces, some kind of forces. Then a vector field is a particularly appropriate way to represent a type of thing. Now, we did a little bit with vector fields. But now we'll do more. So suppose that I have some function. I'm going to write n, but I'm almost going to draw 2. And I have some vector field here about this one. So this one that I'm drawing. Oops, wrong way. Something like that. So that's f of x, y is negative 1. Is that right? No, it's x negative 4. So when x is 0, the vectors just point down. They're actually longer, they point all the way down to the origin, but they shorten them. When y is 0, the vectors point the way. Or somewhere in between we get this. Is this confusing to anybody? We talked about these a little bit before. I'm good with this. Now, imagine that we have some path here. So this is going to start here and go there. Let's call it gamma. And I want to compute what is the value of this vector field? What is the total behavior of this vector field as I move along the path? So for example, you could imagine that this vector field represents, I don't know, some collection of forces. Maybe it's, well, some collection of forces. And I want to move something along this path from here to here. How much total force did it need? How much total force did it feel as I moved along this path? Well, so since I want total force, this means it's going to be some kind of integral. How much is the total? Well, we have to sort of project these vectors onto that path. So I'll come back to this sort of interpretation in a second. Or we can just define it as some abstract object. So if I look at this curve gamma, so gamma is some, I don't really want to name that. So g is a function, so it starts with r. Right, g is some curve from r to rn, where here t has got some range. And if I look at the tangent vectors to this g, it has tangent vectors. So the tangents to g at any point t are just g prime t. This is what we did in the very beginning. So if I start at this point, t equals, I don't know, 1 half. And the tangent vector here will be g prime of 1 half. And I should have drawn it in red, but I couldn't. And we can calculate if I have, so really this one should be red. I'm going to make it red again. At the point 1 half, how is my vector field? Here's my vector field. And how does the, what is the effect of this vector field when I move in that way? Well, in this picture, they're basically perpendicular. So you don't feel any of that force. Just like force of gravity is straight down, if there's no friction, I can move in this way with no work. But if I want to move this way, it's a lot harder. Whereas some place here, then I need to project this onto that to feel how much it is, right? And so I want to add up one aunt. And I want to add up all of those things as t goes from the beginning to the end, and that will give me something to get to how much work it is to do that. You can define. So if I'm adding up a lot of infinitesimal things, that means that I'm integrating. Take the limit as the increment goes to zero of an infinite sum, then, I mean, of a finite sum becomes infinite, it becomes an integral. So this gives us a line integral, which is f, right this way. So I'm going from a to b. So of f over the line integral of f over the curve gamma. Gamma is parameterized by g of t as t goes from a to b. Will you have cameras? Yeah? Then this is just going to be, I integrate that. This is a vector, but their dot product is a number. This is just an ordinary integral, an ordinary one-dimensional integral once I do all the work, the parameterizing, taking derivatives, taking dot products, blocky, blocky, blocky. And that would be the sum of the dot product between these guys and those guys, which is not really the work because I'm also including how long these are. But let's just say that. If I have some parameterization of a curve, I can compute this integral. That's an example of what it's going to do. So let's do that example. So suppose my function is piece of a parabola, this vector field, and I'm going to calculate the line integral over this chunk of the parabola. And I'm parameterizing it exactly this way. So I'm not worrying about whether I'm traversing at the human speed. So I can just calculate this integral. So the integral from 0 to 1 of f of g of t dotted with g prime of t, gt, calculate what g prime of t is. I have to calculate what f of g of t is, but that's pretty easy. So if I plug t's in here, then f of g of t e minus t squared. And the derivative of g of t is 1 to t. And so that is t minus 2tq to do now. So this is t squared over 2 minus 2 thirds known. 1 half t to the fourth of 0 to 1, which is 1 half minus 1 half. Guess so. OK. So why is that? That's kind of surprising that it's nothing. Well, everybody sees why it's nothing. It has to be that the vector field is orthogonal to the right curve. Right. So this vector field is exactly orthogonal to that. We're out with some of them. Of course, if I change this curve to be something else, then I won't necessarily get 0. For example, if I change this to be, I don't know, just the diagonal line, that should also be 0. If I integrate just along the x axis, then I'll get 0. So now this is not really what we want to do. OK. If we want to interpret this as a physical thing, well, you know, we really should, if I want to parameterization of unit speed, and so when I'm calculating the dot product of f and g prime, I'm not projecting those vectors in there. I'd have to divide by the length. If I really want to calculate, say, work, so that is the effect. Here's my, so I have my curve. I want to calculate what is the composition of this onto that. Then I want this to be a unit vector. Because I want to calculate the contributions without being messed up by this unit again. Certainly, I mean, if I had chosen a little better here, so suppose that, let's say, the integral from a to b of f and g of t dot q prime of t dt is some number, too. And then I look at, then I look at something that goes twice as fast, twice as fast. So really what I want in order to do work is that I want g prime of t. So I want gamma to be parameterized. That's already over there. Maybe. Who knows? Maybe long ago. So I'd like g prime of t to always be a unit vector. Because then the length of the tangent. Well, yeah. So the usual notation for this is we call that s. I'm sorry, but what we really want in the divide-through is we make it be units, I'm sorry. Do you want it to be c over, c prime? Do you want it to be c over? Do you want it to be c over? Yeah, I want these to be units. So I take any parameterization, and I just make it a numerator. Do you have any series? All right. Okay, so let's do one of these. After I'll do one that's a little bit messed up, but then the messed up doesn't get it. So suppose that I have f, three variables, and let's see if I can remember my example long enough to get some of my notes to the board, x, y, z, cosine, sine. But it doesn't matter in the sense that there's a corner. Right? In the x and y, we're fine. Let me try and draw g. So cosine, sine, and this guy, when t is less than pi over two, it just increases from zero. When t is less than pi over two, yeah, I go from pi over two down to zero. So it starts here. It goes down to zero at pi over two. And then that's supposed to be a straight line. So this sits over the circle in the plate. That doesn't look like a circle in the plate. So this guy sits over this circle. He starts at pi over two up here. And as we go on the circle, it comes down. Let me draw it wrong like that. And then it bounces off and goes back up. Except this is sitting over this. Does that make any sense? Nobody has a clue what I'm trying to draw? Here is the graph of t minus pi over two absolute value. It looks like that. And now I'm trying to put this onto this cylinder. It starts here. It goes down. And it has a corner here. And it goes back up the other side. There's a corner here. Right in this sharp corner. So in particular, g prime of pi over two doesn't exist. The pi over two minus t is d increasing on the x-axis. Only absolute value of c minus pi. Why is it going down? Well, because I'm stupid. No, absolute value of minus pi over two is pi over two. And then t increases. Well, it doesn't matter. So if you prefer, you can write it this way. Because it's absolute value. It goes down until we get to pi over two. This is the same as pi over two minus t if t is less than pi over two and t minus pi over two. So this one goes down. Because it's absolute value. It doesn't care. OK. So what was I doing with this? I don't know. So if we wanted to compute, but notice that g prime, well, let's see. So what's g prime of t? Minus sine t cosine t. And then e plus or minus one. P is in pi over two. So it'll be minus one when t is less than pi over two. And it's not defined. So if I wanted to calculate now here, so what's the length of this? Well, this is norm of g prime of t is square root sine squared plus cosine squared plus one square root two. In both cases, as long as we're not at pi over two. So this is almost parameterized by arc length. I just have to divide by root two. And then it'll be parameterized by arc length. So if I want to compute this, actually I don't have to, but I will. Break it up. But in fact, it doesn't matter. So this will be what was that? X, Y, Z. Which is cosine sine this thing. This is going to be cosine t sine t pi over two minus t dotted with velocity minus sine t cosine t pi over two minus t. This group, this piece it is, which will be negative sine cosine plus sine cosine, but this will just be the integral from zero to pi pi over two minus t one over square root of two. It's, yes. The base is pi over two and the height is pi over two. So it's pi square over four. A triangle. And another triangle. So anyway, we can do it. So what was the point of this? The point of this is these are easy to do provided that the grammarization is nice. And it doesn't matter if it's messed up at one point. If it's just split it up, it's okay. Just like regular integrals, if they're messed up at one point, break them up, it doesn't matter that there's no derivative at one point because it's just one point and they're integrating. Okay. But there was something, yes. Then you have to work harder. So if the length is not constant, then you have to adjust g to make it nicer. So you have to work harder. You have to parameterize it so that the thing does have a constant. Which sometimes kind of sucks. Okay. So I did that already. I guess I want to... Wow. So I want to point out some notation that one encounters here. Yeah. And then I'll do the other thing. So if I have... So suppose I have an integral. It's one of these five integrals. Let's do it in, I don't know, R3. So f components, which give me the x, the y, and the z components. In general, I'll have n such things. Well, I can think that g cannot just sort of suppress the of t. Right? So this is just, I'm going to break up my g into 3 component function g1, g2, g3. But I'm going to call them x of t, y of t, z of t. So this, I can rewrite. Let me do it in a couple of steps. But now we can drop the of t's. So I can call this d, y, this d, z. I can just call this f of x, y, z. Maybe actually I should call it 3. f1, x, y, z. f2, x, y, z. f3, x, y, z. None of that. So that gives me f1, x, y, z. f2, x, y, z. dy, x, y, z. Right? So a lot of times you see this kind of notation. Or sometimes you even just see this as, so a lot of times you'll see things that don't seem to make sense. So you might see integrals like pdy plus qpdx plus qdy. Not many people have seen these kinds of things in, say, physics. There's a couple of physics people here. So this is just representing a line integral in a slightly different notation. This is really, you want to parameterize this function and then split it out. And maybe over some curve again. These kind of notations. I still have 10 minutes. One thing to notice, we don't really have to worry about. So these kind of things don't really depend on the parameterization of gamma. Provided that I don't do something stupid like make it run the other way. It goes fast or slow. So what does that mean? Equivalent parameterizations of gamma. So that is, if I have, say, gamma parameterized as g of t, I have some other parameterization. So these both represent the same curve where du dt always the same sign. I guess I needed to. And I can write the curve by two different parameterizations and the derivative of this variable with respect to that variable is positive and it's not zero. Then I'll get the same answer when I do these integrals. Because I'm just going to make a substitution. Right? So that means I have some u of t. I'm going to make the substitution u of t. Is this clear or should I write this out? Do you have an example? Well, let me do the proof. The proof is shorter than the example. So... So that's one. I need to cancel that. So, I mean, a priori we wanted to be unit speed, but in fact it doesn't matter when we can use these kind of things. So let's see what that is. So I have, this would be, so let's just start with this one and go to that one. So I have the integral of f of g of t g prime of t vt t goes from a to b that's this but I'm going to make the substitution u of t. t is some t is some phi of u. So this would be, if I plug in here when I take this derivative I have to, u of u and now my bounds wouldn't go from a to b they go from alpha to beta. This is function h and this is h prime of u. So if I change coordinates then I get the same answer. Provided that I don't turn around and go the wrong way. I need this to be a valid change of variables. That's why I need this to make sure that I'm not going the wrong way. That it's a valid change of variables that the implicit function theorem holds. If this becomes zero then this is nonsense. So I wanted to give the fundamental theorem of calculus put in two minutes and try. So let me just state it and then I guess I'll approve it because I know it's easy. So remember in one variable the proposed that we have well the integral from a to b with some function and I integrate it then this is just the same as plugging in. This is the ordinary fundamental theorem of calculus. That's how I integrate x. I find a function derivative of I evaluation of the integral of 2x from 2 to 3 is I find out that x squared derivative of x squared is 2x and I evaluate it from 2 to 3 and I get 9 minus 4 is 5 because 2x is dx. Something similar goes on here. These line integrals don't depend on gamma if the thing that we're integrating so if f of some function f maybe I should use one of them because g I don't know. So if my function that I'm integrating is the gradient of some other function let's call it p for potential for some curve gamma in just gamma at one at the other end. If not every function is the gradient of some other function but if it is then these are particularly easy. So sometimes this is called independence of path. So let me just say another way to say this in this case this is called a conservative vector field so another way to say this theorem which is the fundamental theorem of calculus says that if f is a conservative yeah. So a conservative vector field needs to have a potential you have a quantity which is preserved. So for example in a physical system with no friction. Stop there.