 I mean, it's right now. It's on camera. It's on camera. It's on camera now. He's like sleeping. Actually, I'm sleeping. I'm sleeping. I'm sleeping. I'm sleeping while you're sleeping. I'll make sure that we have the phone. I'll make sure that we have the phone. No, no, no. Okay, so... You walk and sit down. No chance. So, a couple of... What am I doing here? So, the final... It is... Oh, shit, I got things in there. Tuesday. It's at 10. It's at 9th time. It is not here. Oh. And of course, I forgot exactly which room it is. I believe it is in Javits. So, I forgot we have Javits. It's not the big Javits. It's like Javits 102 or something. No, I don't. It's one of the smaller Javits. So, it's a little bigger than this room, so there's room for people to spread out. I will post that on the webpage. I will also send everybody an email saying, XAM is in. Javits... I'm going to make up a number. 748. Because there is some Javits 748. Yes. You have two and a half hours. Excellent. Yeah. I will try... I mean, well, okay. The previous exams, I thought were an hour long. Many of you took more than an hour. So, I will make this exam about an hour and three quarters long, which means you'll have two and a half hours. Okay. Well, okay. So, another theme is... Abraham has offered to do a... a new kind of thing, an extra session, either on Friday or Monday, as you guys wish. So, he can do it Friday. So, Friday at 2.30. Or Monday and to be determined, but afternoon. So, let me first ask to any of you have... No. There are no finals on Monday. Monday is a reading day. So, nobody has a conflict with Monday. So, let's just see. How many people would prefer Friday? Thank you. Oh, of course. I mean, I count for the moment. One, two, three, four, five, six. You count up. Or are you raising for him? Oh, I guess I would... Three, four, five, six, seven, eight, nine, ten. And how many people would prefer Monday? One, two, three, four, five, six, seven, eight, nine, ten. I'll go Friday. Monday. Do you like a climate? Oh, jeez. We seem to be tied here. Friday, Friday, Friday. Oh, my God. Oh, I got a quarter. You got a quarter. Okay. It's good. It's Monday. It's tails. Friday! She never saw tails this Friday. So, let's do this. Does anyone have a class Friday at 2.30? Yes. I don't have a class, but I have it. You have a class? Wait, you have what? I can't be there. Oh, it doesn't move. I mean, so how do you have a class Friday at 2.30? I can't be there. At least 200. Okay. So, let me see if he is willing to do both. What? Yes. Let me see. I'll ask him. If this one, if it is, would be in his math, P131. If he's willing to, I'm going to be a JNK. No, I will be in Toronto. So, I can't videotape it. I like it. Anyway, I will be here. I'm going to be in the country. So, but if he wants to videotape it, it's going to be difficult, though, to get the videotape available to you for the fun. So, there's sort of no point in the long term. Yeah. You talked to him, Ronald? Yes. I land Monday afternoon. Monday around three. Do the job on Monday afternoon. I can do it Monday night. I can show up. I can do it Monday night. I don't know that I can do it. I want to show up Monday night. I mean, I land at JFK. Are you going to India again? No, I'm going to Canada. Don't need a visa for Canada. Yeah. Maybe if Artem is available for both, you can bring in Artem again? Artem is also going to Canada. Huh. Well, providing his visa comes through. Wait, why does he need a visa? Why? Because, I mean, it's a conference. Oh, I meant, like, why does he need a visa? What? Because he is not a U.S. citizen. Oh. Did you say that? Oh, yeah. Or a Canadian citizen for that. You should cry a little bit. Okay. So, let me confirm with Arthur. So, let's do this for sure. And let's see if he would also be willing to do this. So, best would be if we could do both. But, I mean, this room is available. I know it's available. You don't have a room here. Rooms are harder to get on the reading day. But, I will see if both can be done. Okay. I'm sorry. In terms of Monday. In terms of Monday afternoon. Are there people with a conflict Monday afternoon? Are there people with any sort of time restrictions Monday afternoon? I mean, later would be better. Later is better? No. Monday is 6 a.m. Later is better. Exactly. So, we could try and do it sort of at class time. Yeah. Right? Because most of you, if you have a non-Stony Brook commitment, you probably do not have a non-Stony Brook commitment. For example, around 4 o'clock on Mondays. Because you kind of do something else at 4 o'clock on Mondays. So, let's try for 4 p.m., if you know where. Provided that he is available at all. Okay. So, let's try for that. I don't know. I will post whatever the results of this are on the webpage. And I will go on Blackboard and send an email to everybody. Yeah. Yeah, everything. There we go. Yeah, let me discuss that. I'd like to try and get through at least a little bit about Soap's theorem and Gallister's theorem before we do that. And then I'll come back to that. Surface. Surface. Surface. Surface. Surface. Surface. Surface. Surface. Surface. Surface. Surface. Surface. Surface. Surface. Surface. Surface. Surface. Surface. Surface. Surface. Surface. Surface. Surface. Surface. Surface. Surface. Surface. Surface. Surface. Surface. Surface. Surface. Surface. Surface. Surface. Surface. Surface. Surface. Surface. Surface. Surface. Surface. Surface. Surface. Surface. And gamma is the boundary D. So I have something that looks like, I don't know, this. This is D. Here is gamma. This is also part of gamma. Then the line integral over gamma, oh, and I guess I need to say something about my vector field. My vector field has goodness. So my vector field is continuously differentiable. And since we're in the plane, it consists of two pieces, p and q. Then, if I want to integrate the line integral over gamma, the p dx over q dy, then this is the same thing. And let me actually write it as a double integral to emphasize that it is a double integral. This is the integral over the inside of the curl of f. Let me write dx dy. Of course, it might be dy dx. But whatever. Okay? So that means that we can return appropriately chosen double integrals into line integrals over the boundary or line integrals over closed curves into double integrals. And there he is. Okay. So you were asking me something that I didn't quite understand. Yeah. So let me not do what kind of do. Suppose we have, I'll just sort of set this up, but not really. Suppose that we have this ring, this annulus, where this is the circle of radius I don't know, 1. And this is the circle of radius 2. We have some vector field. I mean, should I do it fully? Or, okay, yeah, do it fully. Okay. Suppose this is my curve here and here. And so gamma is both of these together. Right? Let's call this gamma 1 and this gamma 2 because it has radius 2 and this gamma 1. So, and what function do I want to integrate? I want some vector field. It doesn't really matter. How about, let's do the vector field. I'm just going to set it up anyway. So let's do minus y of x. Okay? So this is the vector field. That's my region. And so the line integral over gamma of the curl. So I guess let's do the curl of f. So that means that the planar curl of f is dx of x. Oh, that's a stupid one. Do negative y squared. Yeah. x squared. Okay, sure. d dx of x squared. Since I'm just setting it up, it doesn't matter. d dx of x squared minus d dy of y squared negative. That's the curl which is 2x minus plus 2y. So Green's theorem says in this case that the line integral over, let me write it as gamma 1 moving down to 2 of this 2x plus 2. Yes. No, wrong. Duh. Of minus y squared dx plus x squared dy is the same thing as the double integral over this region. Let me not parameterize it yet. Of 2x plus 2y dy dx. I don't care. That's what Green's theorem says. So what does that really mean? So in this case, I can do this integral in two ways. So one way is I can parameterize these two curves. So I can write gamma 1 as the curve. So I'm calling this way. So that would be cosine minus t psi minus t. I know this simplifies. As t goes from 0 to 2 pi, cosine of minus t is the same as cosine of t. Sine of minus t is same as minus sine of t. Again, t goes from 0 to 2 pi. So that's my inner curve since I brought color in this. That's my blue curve like that. And my other curve, gamma 2, can just be 2 cosine of t, 2 sine of t. Again, t goes from 0 to 2 pi. Yeah. In that case t is negative because you're tracing the wrong way. Right. The clockwise. I really think of t as theta. Right. And so I could write it going from negative to negative. Yeah. Okay. Cool. And you can parameterize it other ways. It doesn't matter. I mean, it doesn't. Let's do that. Okay. So let's just write out what this guy is. This is not the best way to do this, but let's write out what it is. This is, so why, let's over count. So this is a line integral over gamma. The blue curve of minus y squared dx plus x squared dy plus a line integral over the black curve of minus x squared dx plus, oops, y squared dx plus x squared dy. Right. So I can just do them separately. For this blue curve, y is cosine of t, x is my, I'm sorry, x is, so on the blue curve, x is cosine of t, y is sine of t negative dx is the derivative of, is, is minus sine t dt dy, so let's emphasize this together, dy is minus cosine t dt, not td. That's what we have on the blue curve. On the black curve, x is 2 times cosine t, sine t, and dx dy is 2 times minus sine t dt. That's what we have on the blue curve. On the black curve, x is 2 times cosine t, sine t, and dx dy is 2 times minus sine t dt. That's what we have on the blue curve. So this is, now we can just plug in, this is an ordinary integral now, a regular one-dimensional integral of minus cosine t, let's put this square there, times dx, which is minus sine t, so dt. Right. This is x, this is minus y squared dx. Did I write dy? Sorry. It's another cosine. Right. Minus y squared is negative sine t. Minus y squared is minus cheese. So minus y squared, y is sine, so that's minus sine squared t. This is minus y squared. dx is minus sine t dt. That's the first bit. We're going from 0 to 2 pi, plus then the next bit. We have x squared, which is cosine squared t, dy, which is minus sine t dt. Probably it is. Yes. So that's what we have, and that integral kind of sucks. Right. This is sine qt plus cosine minus cosine qt dt, 0 to 2 pi. Kind of sucks. It's doable, but it's not my friend. That's only part of it. Right. The other part is almost the same part. So let's get the other part. 0 to 2 pi. You got this. Huh? You got it? I should just go on? No, no, I mean like you got it. No, I think bold. Yeah. I think we understand why we don't want to do this. Sometimes it's the easiest way. But not in this case. But it is. So minus y squared dx, same nonsense here. No, it's a different curve. So that's sine squared t is minus sine squared t is minus y squared dx is minus sine t dt. In fact, I think it's the same, right? And I lost the factor of 2, right? Because there's a 2 here. And there's a 2 here. So I really lost 2 to 2. Please introduce 3 tos. Sine squared. Sine squared is 3 tos. Yes, 3 tos. Ah, yeah. Right. 8. It's 3 tos that I lost. And then here, I've lost track of what the thing is. X squared dy. So that's cosine q to t and dy is 1 of these is minus. No, that's not 1. This is the cosine q dt. Is this one minus? No. dx was 2. I already changed the minus. Okay. So I get the same thing so this gives me a 9 here. One minus one plus. Yeah, one minus one plus. Oh, that sucks. Okay. Oh, my God. Okay. There it is. You don't want to do it any too. Okay? So there's that. The other way, which is easier, is we do Green's theorem. We turn it into a double integral. But we really want to do the double integral in polar coordinates. Yeah. If you want to, if you do it in rectangular coordinates, your life is horrible because you have to do it in a double integral region. You don't want to do that. But in polar coordinates, it is not bad because if you just integrate r goes from 1 to 2 and theta goes from 0 to 5 and we integrate 2x plus 2y. So that's 2. x is our cosine theta, y is our sine theta, r dr d theta. That's easy. Right? This is just 2 times the integral from 1 to 2 r squared dr cosine somewhere. Cosine theta plus sine theta. I wrote it backwards. d theta dr. I can do it dr v theta. It doesn't matter. So that's an easy integral. Yeah. You wait. Yeah. One is going to talk once and then two is going to talk once. Yes. It's just. So for Green's theorem, I need the things to be oriented so that when I walk on the curve, my left hand can touch the inside. So when I walk on this curve going this way, again, that same hand is touching the inside. So this is oriented correctly for me. If this had been oriented this way, it would have had to work a little harder to use Green's theorem. Right? But it's not. So yay. If the inside were oriented counterclockwise, how would that change the answer? Like would it just negate it or? So here, in this orientation, this would be minus. That's easy. To think about it in Green's theorem, I think we kind of have to do this guy minus the little guy. Right? We do the double integral of this guy minus the little guy, which will be slightly very close. Yeah. Yeah. Well, this would still work if I had a weird singularity. So because of the way this thing is oriented, I don't care what happens anymore. Right. So even if this guy blew up out here, I don't care. Oh, so that still works. So this is measuring, I mean in terms of physical stability, it's just measuring the circulation of the zone. It depends on what your intuition is. If you're thinking in terms of physics and forces and stuff, this is the circulation of this, physics of this, this is measuring this vector field, how much twistiness or how far away from being gradient this vector field is perhaps. Over this area. Right. So how much twistiness there is of this vector field over this area if you're thinking in terms of physics. If you're thinking in terms of, I see you just second. If you're thinking just another interpretation, if we think of it as a line integral, this is how much work to drag something around this path and then also to drag it the other way. Maybe you have, you know, you have two things. You're going to drag this around here and this the other way and how much work was done. It's the same. Seeing that they're the same, I don't have physical intuition. I'm not a physicist. It doesn't seem to me that they're the same. But they are. Yeah. How would we, if it was pouring to the other end of the center, how would we set up the double integral? Well, so we'd have to think of it as really two things that we subtract. Right? So this would be, instead of integrating from just one to two, I would integrate from zero to two and subtract off the other integral. Yeah, wait. Wouldn't you add one from zero to one? Yeah, I would add. I'm sorry. Because I have to reverse it. Yeah. So you would just go from zero to two. I would go from zero to two, but I would have to add one. Like going from zero to one twice, not one to two. Yeah, I count the middle twice. Because it's not this problem. This problem is the zero to two problem. And this is the zero to one problem. Yeah, so I think I have. Okay. We're good? Okay. Yeah. Oh no, that means yeah. Okay, I thought that was the thing. It's a thumb. It's a thumb to our fingers. No, they're thumbs. So let's move away from Green's theorem, but not quite. So this is called Stokes' Theorem in Plain, which if we can make sense of it, it's exactly the same theorem in higher dimensions. Whatever that means. So I need to say what that means. But then there's another version. I'm going to erase this example. Delete that. Okay, erase that. So there's another friend called Gauss's Theorem, which I'll suppose by the name Divergence Theorem. So there's another, and I still want to just talk in the plane for a minute, which instead of measuring the curl or the circulation of my vector field, you can oppose what I want to measure. And let's just say I have a similar region. There it is. I have some vector field, and I have some stuff flowing around in the plane. Actually, it's used for the same vector. Doesn't it? I have some vector field. I don't want it to be always this way. Some vector field. I have some vector field like this. Stuff flows through this region. I could think of this. So what I might be interested in is what's called the flux over my region. I guess I'll still call it. Which is the amount of stuff. In fact, instead of drawing it that way, I'm sorry. Let's draw it this way. This one's easier to draw. So I have some, think of it as electrical current or whatever. Think of it as stuff flowing. And I want to know how much stuff that flows according to this vector field is created or lost as we flow through this region. So if we think of this, we're thinking of this in terms of meteorology, and these are wind currents, then we want to know whether we have, looks to me like we're going to a more low pressure zone. Stuff is spreading out. So I want to compute how much is gained or lost as we traverse deep. Right? Yeah. How much in area? Yeah, okay. Think of it as area. So I have a sort of a unit area here, and as it flows through here, there's some gain or loss due to the stretching of the vector field. But I'm really thinking of sort of things are added, things go into D here, the way my vector field is. Things come in here, so if I take this curve, it flows in and then it flows out, and I get more stuff out than I got in. Do you understand what I'm saying? Divergence is negative. Yeah. That's confusing, because I wouldn't expect it to be just less than one or positive. So how do you think, how do you think... It's the exponential. Okay. So if you exponentiate divergence, then you get... Okay. Because we're integrating is like exponentiation in some sense. Now, I've lost it. Okay. So I want to know sort of the net amount lost or gain as the stuff flows through D. So that would be... I mean, we already know how to do that in terms of a line integral. That would be, I just compute the dot product of the vector field with the normal vectors as I go around this curve. So that should be... Well, it's not zero if the vector field is zero yet. If the vector field is divergence free, then it's zero, but if the vector field is not divergence free, then it's not zero. So that should be the integral. Let's call this curve gamma over gamma of f dotted with the normal vector. Let's try it that way. It's the dot product of f with the normal vector. That measures how much is gaining here. It is the projection of the thing in the red direction. And here, since it goes the other way, this projection is negative. So the stuff coming in cancels out with the stuff going out. And we can just project... We can just calculate that. And so Gauss's theorem in the plane says that... What's the differential? D... X vector? D... So in other words, you parameterize everything in sight. You calculate the dot product. D... But yeah, that should be... So when I do f dotted with n, I'm going to wind up with something that's going to look like a p dx plus q y that will be different p's and q's in the other one. Let me get there in just a second. Well, in fact, let me get there now rather than saying the other two. So far, I haven't said anything about Gauss's theorem. I've just said about flux. I'm working out to Gauss's theorem. So now we can be clever and turn this vector field into a different vector shift. Since I'm dotting with the normal, I could instead... Instead of looking at this vector field, I could instead look at this vector field which is just taking all of these vectors and taking them perpendicularly. If I'm looking at an orthogonal vector field where I've just swapped the components and changed the sign so that f and h are perpendicular to one another with the right orientation so everything is great. And so now I can actually write this in this form. So this is the same thing as h... Yeah, h dot dx. I can write it this way which is the same if we parameterize it as minus q d... plus p y. Yeah. Yeah, this is all in R2 right now. In general, we could have it... My goal for today in the next 25 minutes is to crank this up to three dimensions as well. Works in 27 dimensions too if you want. I want to increase this. But this is really just the same now I can just use Green's theorem. So by Green's theorem this says that... Oh, this is still... I need to go over gamma. By Green's theorem... By Green's theorem that just says that this thing is... Okay, why did I just... Yeah, so by Green's theorem this is the double integral over by region d of... Well, I want to take the partial of this with respect to x and I want to subtract off the partial of this minus minus. This is minus minus. And let me just write it. So all I did is I said okay, I don't know who P and Q are I just want to do Green's theorem and Green's theorem said I can turn a line integral with the proper orientation into partial of this guy with respect to... Partial of this guy with respect to x minus partial of this guy with respect to y and that's what it is the same thing as dP dx plus dQ dy which is the same as the quantity that we already talked about called the divergence of f which is just I mean just for mnemonic purposes it's the same as integrating grads dotted with f So we can transform this integral relating to normals and my integral relating to normals into essentially by Green's theorem into a different double integral namely integrating the divergence rather than integrating the curl over the area and that will measure the flux So we can measure the twistiness of the vector field by integrating the curl or we can measure the spread outiness of the vector field by integrating the divergence So Gauss's theorem in the plane or the divergence theorem in the plane says that this equals that f d boundaries be nice nice means the same stuff it always means the flux is the same conditions the same conditions the flux of f over d is the flux of f over d is f dot n d, I guess it's ds I guess it really has to be Well it doesn't matter since you know is the same as double integral divergence of f So again really these are both Green's theorem wearing slightly different clothes and they're all Stokes' theorem Does it work the same of the Green's invention? So there, now I have to talk about Stokes' theorem Here it is provided you know what this means and Green's theorem I mean Gauss's theorem here it is provided you know what this means So the problem with going to more dimensions is what is this line integral mean? Is it a line integral? Is it a surface integral? Is it a rabbit wearing a fake beard? I don't know something I'm going to talk about sort of next I hope I don't get there So let me try and set up a little bit to get there I'm not sure what flux actually means Let me spread out How about that? I'm going to line it somewhere here and not over a line So I'm going to go on a line Yeah it's not a stick I'm going to pour some paint on the river I'm going to have an oil slick I have an oil slick that falls out here and then I let and the river flows and what's the oil slick over here spread out or maybe there will be some compression there might be all sorts of crazy stuff going So it's really the expansion of area It's a measurement of how much stuff spreads out I could put a point mass I could put a drop of oil in the water and then let it flow and see where it goes That's what? Yeah it's the amount passing through the surface So some is going to flow in So imagine that I have a line here and I have a line worth of stuff here and then let it flow and it's going to spread out in some way and I want to know how much is added as it crosses this region And if this vector field is conservative the answer is none So because it's divergence 3 So there isn't any So Ok So stokes there and gausses there are the same in more dimensions provided you know what more dimensions means So in order to say what more dimensions means we have to talk about surface integrals at least a little bit So we have to remember how surfaces work So let's just do everything in 3 dimensions You can generalize to higher dimensions but let's not So if you remember I have some region in let's call it un and I can think of a surface as I don't want to think of it in the base So let's start So for a curve I take an interval where t lives and I throw it into space or the plane and I get some per gamma of t I have a function gamma which takes in say 0, 1 which gives me out a piece of a curve For a surface I take some square in u, v, l and I apply some function g which is a function from r2 to r3 and out will come some wiggly thing which is a surface this surface could bend around and underneath it doesn't matter The graphs of surfaces that we dealt they don't bend around but this could be it can't be quite a sphere unless we can track but this is what we get for a surface and furthermore we can think about at each point if I fix v and allow u to increase I'll show you how to color I should use it I fix v and I allow u to increase that gives me some infinitesimal direction here and it's image is going to be some vector here which is given by the partial this is dg du is that blue vector and if I take an infinitesimal fix u and allow v to vary then I'll get some other vector here dg, dv so that gives me a pair of vectors which are both tangent to the surface but we did this kind of thing at the beginning of the class and then either you forgot it or you didn't but we can describe what's going on we have a tangent plane here which is spanned by these two vectors dg, dv but now we know a little we can also think what do I get if I take the cross product of these two values so if I look at dg, dv and I'm thinking of these being at a fixed point dv I want to keep writing u's and v's everywhere but this is now this is a vector it's not a function and I cross it with dg, dv this gives me a normal vector the normal vector to my surface let's write it as s gives me my normal to my surface at uv so I can find so at each point actually around my surface I have a well-defined normal based on how this guy is mapped there and if I choose a different parameterization so if I choose a different way of representing this by u and v these vectors will move around but they will move around in the same plane because I have a given tangent plane so at worst my normal will point the other way so this normal vector well if I unitize it and divide it by its length then this becomes unit normal and now I get a vector which either points in or out but doesn't depend on my parameterization other than whether it points in or out right so this vector n doesn't care how I use my parameterization g, it only depends on the surface except that it might point the wrong way so if you think about some three-dimensional object like a sphere so yours are easy to visualize then at every point on the sphere I have an outward pointing unit normal and then there's another one that points in but other than that it's unique and this whether it points out or in well there's an analogy to curves whether I choose to traverse them this way or this way if I have a closed curve let me make it closed if I have a closed curve I can either go around with unit speed this way or I can go around with unit speed that way and we have the same analogous thing here if I have a closed surface it doesn't matter if it has holes in it I have either a normal which points out or I have a normal which points in but other than all that depends on is how I chose my tangent plane which vector I call x whether I'm choosing my coordinate system this way or this way okay so by convention for the same same reason we choose counterclockwise here as our standard orientation we choose outward pointing normal so we say that a surface a closed surface is positively oriented this means that the normal that we choose in this parameterization is it pointing out to be working? well it means it's way from the inside well if it's closed there isn't an inside closed it's not built into the surface if it's not closed I can't say for this surface which way is better this way or this way depends on how I chose my coordinate system and there's no canonical coordinate system to choose which one but if I'm thinking of my skin then this is better I don't want to show you the other one because it'll hurt because I am a closed surface even if we tissue my mouth I don't want to okay your hand you'll have a mess I don't want to stick my hand down okay so we have this notion of positively oriented not all surfaces can be oriented right if we think of a good example of a surface which is not orientable is a mobius right so this isn't closed right this guy where I put a half twist a mobius strip like that well it's not closed but it's also not orientable because if I choose an outward pointing normal and I bring it around this mobius strip it's an inward pointing normal so I can't orient this guy if I wanted to make this closed I would have to be in four dimensions I would get a prime model but same thing you're saying it's not outward pointing because it goes to the other side when it gets to the other side it's pointing the other way but it's not pointing the other way that was actually a flash strip no it's not a tube it's a flat thing and so I have two orientations it's not I can't make one I can't make not orientable surface without self-interfections in space because you live in an oriented space so you don't have a choice there so a standard example would be a prime model which I will try and draw here so this goes in but it doesn't really go in it turns around so this looks like a tube except it sort of turns itself out as you go around I guess I should turn it over the other way so you can try to fill it up but it won't fill up but this passing through is not really there it's a little hard to visualize this but I can certainly write a mapping from a collection of clothes just guys to get okay so we're not worrying about non-orientational surfaces and so then I ripped up my notes good that's nice so some things about the utility of this normal object just like we can calculate things by integrating the tangent vectors so it's the integral gives me the length sorry the length of the tangent vectors add them up that gives me the length of excuse me this gives me the length of the curve we can do a similar thing we can compute the surface area by computing the length of these normal vectors but I have to do the double integral over the surface of the length I just take the cross product which this is just the normal vector and I integrate it over the surface I integrate its length over the surface and this gives me the area why does this give me the area well one thing that I didn't emphasize is if we have two vectors u and v and we take their cross product then the area of the parallelogram they span is the length of u cross v and so because of that fact if I integrate all of these little parallelograms over the surface it gives me the surface area I'm almost ready to stay no way am I going to prove it the higher dimensional analogs of these guys so just like we think of ds as this quantity this is my length element this is parameterizing by by arc length we can parameterize by unit area so often I don't think there's a standard notation but often this is d sigma just dg du cross dg dv let's say so this is parameterizing by unit area rather than parameterizing by arc length and so another way to say this is this is just integrating over my surface d sigma getting there this is the surface area surface area so the surface area is just what I get when I integrate the area form over the surface no but this will not be over the whole surface so if I want to compute the top half of the sphere I can find the parameterization that maps a square let's say the top third sphere I can find the parameterization that maps it onto that and I can do the double integral over that square that maps onto the stuff above the what is that called what is this this traffic of cancer is down here anyway so I just integrate everything north of the traffic of Capricorn there I go um so I can integrate this over that or it can be closed and then I can compute the area of the whole thing alright so what does what does Stokes theorem say so Stokes theorem in the plane says that if I want to integrate the curl I mean if I want to I'm going to blow this so what okay so let me let's now think of flux in terms of a surface so suppose that I have now some surface let me draw it instead of the square boundary let me draw it with a brownish boundary suppose I have some surface here bounded by some curve and I have stuff over this vector field which flows through it let's do it open for me I can also do it closed but let's do it open for me and I want to measure how much stuff passes through this net so that would mean what do I want to do I want to measure this vector field dotted with the unit normal over this whole surface so the flux over this surface let's say of f over my surface s suppose it would be a curly s is going to be what I get when I take f dotted with my unit normal and I integrate it I guess it's a triple integral now now it's double it's a surface sorry I integrate each this over over my surface well it's a surface integral so let me do that I don't want to do that sorry I'll do that in a second it's a surface integral right so it's a double integral but it's a double integral over this surface so in practice what this would mean is that I have some parameterization of the surface so suppose that g takes u v into r3 so that g of some box is my surface right the image of g is my surface then that would mean that I would compute the double integral of f which depends on three variables so I would plug in g of u v at every point this would now give me a vector there's a vector field that I evaluated at each point and I need the normal vector so the normal vector is going to be dg du crossed with d and I take a dot product right this is going to be well this needs to be at g of some point so if I can parameter if I have the parameterization of my surface this is my normal well it's my unit normalize to divide by its length wait where did you get it are we going to use dsign or are we just going to add another one okay if I'm going to dsign yeah it's the same thing so in practice I would do some calculation like this and I would integrate over u v okay what Söder says is that this sucky integral can be turned into a line integral a line integral over the boundary and that so if we're calculating flux we need to align this normal with the surface what guarantee do we have that you choose a positive orientation of your surface but do we only know that that exists at this close well if your surface isn't closed then this is positively oriented boundary so then they'll match if my normal goes the right way then my boundary will be oriented the right way and if my normal is going the other way then my boundary is oriented the other way so it doesn't match so the cross product is going to be oriented with respect to the right hand move that means if you choose your orientation so that the image of u goes this way and the image of v goes that way then the cross product will go this way orientation so that the image of u goes this way and the image of v goes that way then it goes in but if your surface is open then your surface is oriented and if I do this one then the orientation that I inherit from the boundary matches because if I do this one then I put an orientation on this boundary curve based on this guy so it goes this way goes this way so it's oriented clockwise if I put the other one that goes the wrong way then the orientation I get will go the other way so that's okay yeah yeah it's kind of funny but that's how it works what about the next step okay and so alright so one more we're almost there so we have thought in gauss's theorem which is the divergence theorem is related but not quite the same so I don't I mean do you have a question so gauss's theorem still left on the board says the same thing except a different thing so so let's think about what stoves theorem is telling us morally we think of this curl as a kind of divergence so in hand waving this curl is a kind of divergence a kind of derivative a kind of derivative and it says that we can integrate the derivative kind of thing over the edge or we can integrate the function over the inside there's a relationship between the two gauss's theorem says the same thing but a different kind of inside if I have a closed surface I guess let me say one over no if I have a closed surface and I want to compute the divergence of this vector field so I can do the surface integral this is the surface integral of the divergence of as just general then this is the same thing as the triple integral sorry this is the so this is some closed object think of it looking like a beam let's call it B for beam so I have some B for beam I can do a triple integral over the inside of my B for beam of F so I can integrate my vector field what if it's the other way it's the other way sorry yes so this is one does it just like a flip back and forth integrate over R that'd be really cool like odd ones so I can do this surface integral over the edge or I can integrate the divergence over the inside so I mean we could keep pushing things down you could cut it in half you still have the theorem on the boundary and put it together so it's this duality between derivatives and boundaries so this is the boundary that's the surface if I think of V as being a solid blob then its boundary is this surface that goes outside so I can do the surface integral over the outside or I can do the integral of the divergence over the inside this is Gauss's theorem and okay Gauss's theorem just like Green's theorem has the same thing that if it has holes in it you have to have the right kind of orientation and block and block and block I guess one other thing that you can check is that if I take the curl of the gradient that's zero which in terms of Stokes' theorem we have independence of surface for gradient flows just like we had independence of path for gradient flows in Stokes' theorem in higher dimension we have independence of surface if this is the gradient of some function if I have a preserved quantity then I can make this be a nicer surface if it's easier to integrate over a flat is then if this is a gradient flow I just make it be a flat disk as long as I keep the boundary the same I can just make it flat if it's not gradient then I can't do that okay so I've kind of run out of time let me I don't think it's fair actually given that I just went for one class I'm not going to ask you any Gauss's theorem or Stokes' theorem is fair game okay right I won't put on the final a Stokes' theorem question or Gauss's theorem question but I will you can expect there's a good chance there will be some kind of line integral where you might want to use Green's theorem or you might want to work really hard I don't like things going the opposite direction that's a part of Green's theorem right just remember that counterclockwise is good I don't really want to do it I will throw up the list of topics we'll be back