 Hey guys, how are you doing? Are you ready for your exams? No Yeah, it was much nicer at the end of August, you know The semester was just beginning and the weather was so nice Now it's winter time in Berkeley Which means like 60 degrees or something? 55 So today is the review lecture. Okay In the last few lectures I gave you an overview from the kind of a theoretical point of view of What this course is about and in particular this last chapter that we have studied and today I would like to focus more on the practical aspects of it of it and Basically on the question. What do you need to know to do well on the final exam? Okay, so I'll go over the most important points of The material which will focus on on the final exam which again will be on December 19th and I'm not going to go over all the information again. Everything is available on online Here and also on B-space I Will I will also have last-minute office hours or the day before on the 18s in case you you know suddenly Don't see the difference between curl and divergence and you panic The night before you can come to see me on on the 18th and we'll sort it out. I'll sort it out for you Okay Also, I want to tell you one more time that the exam will be at her gym in three different rooms okay, and You will the room is determined by your section But is determined by who your GSI is and your GSIs have already announced the room the rooms for for you But you can find all this information again here and here Okay, but you should be aware of where your exam room is. It's important that you are at the right place You don't end up in a you know, it's her gym. So it's big and It's easy to get lost. So be sure you know which room you should you should show up it Okay, any questions about the organizational aspects of the Of the final Okay, good So now let's talk about the material. I already Said that this exam will focus on the material After the second midterm what we call the third season and But it doesn't mean that you should forget what happened in the first two seasons meaning that Certainly problems solving all the problems will require skills which you learned before you should not forget all this stuff Right, especially multiple integrals are a very important part of this of this material So I think it's a good idea to go over that old material Perhaps not devote as much time as the material of the last chapter which we've studied since the last exam but You should probably review it nonetheless Any questions from this part of the audience? No, okay now So I would like to break it into the material So I will only talk about the material since the last exam But again, don't take it as meaning that you should not know anything about what happened before you should But the focus will be on this material. Okay, that's why I'm going over this material now all right, so the first thing I would like to talk about is Integrals in this much in this part of the course We have talked about different kinds of integrals and I would like to summarize again What those integrals are and what kind of information you need to be able to compute them? So the first thing I want to focus is just the integrals What kind of integrals? Well, we integrate over curves and we integrate over surfaces So there are two types of integrals First of all depending on the dimension of the domain Let me talk first about integrals over curves So here we have two types of integrals also Type one integral is an integral of a function. So this is an integral which looks like this you have f ds Where f is a function over curve. So a curve here could be over Could be a curve on the plane That is R2 or in space on the plane or in space But the way we handle these integrals is very similar now How do we compute such an integral to compute such an integral the first step is to parameterize your curve parameterize the curve So which by the way was the very first subject that we learned in this course So this is a good example what I mean when I say, you know revisit Review the material from the previous chapters because certainly to be able to do such an integral you have to know how to parameterize curves And that's something we learned at the very beginning So parameterizing Parameterizing the curve means means introducing an auxiliary parameter, which we usually call t But you may call it something else if you like in writing each of the coordinates as functions of this auxiliary parameter Parameter, so you would have x as a function of this parameter t y as a function of this parameter t and Z as a function of this parameter t that would be in the case when the curve is in space if it's on the plane It you would only have x and y of course Right, and once you do that This integral up and then you have to say that t What is the range for this auxiliary parameter t and the range will be between some numbers a and b So then this integral will be equal to the integral of f where instead of x y and z you try to use you insert those three functions x of t y of t and z of t right And you integrate from a to b But in addition you also have to insert a factor which has to do with the arc length of the curve and That would be the following x prime of t squared plus y prime of t squared plus e prime of t Square and finally did And again if you are on the plane you just don't have the z variable you just erase this But you get the formula from on the plane Now what does this represent? this represents a mass of of Say a thin wire in the shape of of this curve in the shape of this curve If f is a density function Is mass density function Also, remember that if f is equal to 1 if f is equal to 1 in other words You just you just have the square root this integral represents simply the arc length Arc length of the curve so this integral this integral measures the total mass or total length of this of this object this curve Okay, that's what this integral is about It's very easy to distinguish this integral from the next type of integral Because you see that in this integral you integrate functions Whereas the second type of integral you integrate vector fields So in the second type you have an integral of a vector field rather than function f is a vector field So as such it has components I In front of I you have P in front of J you have Q and then you have a third component R if this is vector field on the on in space if it's on the plane again You only have two components Okay, so vector field is a more than a function It's it has three components in space two components on the plane and this integral can be also rewritten as PDX plus Q dy Plus rdc This is the same. This is the same because the dr here stands for DXI Plus dyj Plus dzk and so if you take the dot product of these two you end up with this formula So this are two different ways to write a Line integral of a vector field either this way or that way. It's the same Okay in the case of a Line integral vector field there is an additional complication which we don't have for integrals of this type Integral of this type is enough to just write this You have to specify the curve you have to specify the function and that's it but in the case of a line integral of a vector field you have to You have to specify additional piece of data So which I kind of I want to emphasize it here you need orientation Of course when I say you need to specify what I really mean is that I need to specify it in other words When I give you if I give you a problem compute this line integral I have to tell you with respect to which orientation Okay, if I'm if I don't tell you it means that the problem is not well posed unless you have to find unless You know there's something said which should enable you to find the orientation In other words, this is not well defined if you just have the vector field and the curve You have to also have orientation on this curve. What do I mean by orientation? A curve is going to be to look like this so it goes from some point a to point b and Orientation means the direction which way do we traverse this curve? So there are two possible orientations. Here's one orientation and here is another orientation We can go from a to be we can go from b to a and of course The answer if you choose the first orientation is going to be just a negative of the answer if you use choose the second Orientation, but it's a different answer unless the integral is just zero you will get different answer depending on which orientation you choose So orientation has to be specified and you have to follow that orientation So you have to be careful when you set up such an integral to make sure that you are Computing it in the right direction. You're doing it the integral oriented in the right in the right way Of course here. I have drawn a curve which has two end points There are also curves which do not have end points Close curves. Here's an example of a closed curve in the case of a closed curve It also has two orientation two orientations which we usually call clockwise and counterclockwise Right so counterclockwise is this one There is also a somewhat misleading terminology in the book which is called positive orientation which I prefer not to use because It's positive. It's an it's a matter of convention, right? So let's just call it counterclockwise or clockwise in the book counterclockwise is also called positive But I think it's there's potential for for misunderstanding if you use that Terminology, so let's just stick to clockwise counterclockwise. I think it's fairly it's very self-explanatory Okay, now let's say let's suppose you have this You have chosen chosen such an orientation what will be What will this is how to compute this integral so that you use again Permit realization like this but now it's important to make sure that your orientation that you've chosen is Agrees with the orientation on the auxiliary parameter t this t is actually it just belongs to the one-dimensional space to the line and the line is oriented because Points on the line correspond to numbers and numbers are oriented. We can say which number is bigger Between any two numbers so when we write it like this it means that we are we identify the curve with the interval from a to b and Interval from a to b is always oriented from a to b like this, right? so Let's suppose that under this parameterization you choose the orientation like this So it goes from this point to this point and this point is t equal a and this point is t equal b Okay, let's suppose that in this case this integral is computed as an integral from a to b of P x prime t dt Well, let's just write like this plus q y prime Plus r z z prime But here is a here's a possible trick question. Let's suppose I Will ask you on the exam to compute such a line integral where the curve is oriented In a different way, okay, so if you don't pay attention You would say, okay, that's just you would find your parameterization Where again, let's say t equal a corresponds to this and t equal b corresponds to this let's suppose And then you if you don't pay attention to this stuff to the orientation You might want to write it again like this and that would be wrong You see because you have to make sure that with respect to the orientation which you are told to use the the parameterization is such that You know the lowest one is the initial point the lowest value a is the initial point and the larger largest value B is responds to the endpoint In this picture, that's the case and that's why we end up with this integral from a to b that in other words the integral that I got with respect to t has This orientation which comes from the orientation of the curve But if that orientation is like this You have to go from t equal b to t equal a right because you have to go from this point to this point This point corresponds to t equal b this one corresponds to t equal a that's why so in this case Then you should write it as from b to a and then the same thing Then of course what is integral from b to a integral from b to a is the same as negative integral from a to b So you will actually end up with minus this integral. Okay, is that clear? Ask me if it's not clear So this is a this is a subtle point which you have to you have to remember Now what does this what does this integral represent? interpretation of this integral is that it it represents Work done by force. This is work done by force by force f a long C from the point a To the point b you see from this for if you think about this interpretation it's It becomes more It's becomes more clear why there's a sign because when you talk about the work done by force the force is Moving an object or it may be resisting a movement of the object, right? And and so it's important if you go from a to b or from b to a if the force Let's say the force Let's say the force is comes from the wind and the wind is blowing this way And so the wind is carrying me this way So the what the wind does does perform certain amount of work right and this work is positive but what if The wind is blowing this way, but I'm still I still insist on going there. Okay? Because I'm stubborn, so I go against the wind So in that case ice I prefer I do work not the wind And so the wind is resisting me there for the work of the wind is negative You see so it's important to know not just the trajectory of the object but but it's also important the you know whether The relative position of the force and the and the trajectory in other words the orientation of the trajectory relative to the Direction of the vector field you see so that so it becomes more clear that if I go this way If I go this way against the wind It's a negative Work if I go this way with the wind it will be positive. Okay? You have a question That's right. That's right when I write it like this So the question is about this PQR in this formula. I spelled it out. I wrote F as Of F is a function of XYZ, but now XYZ have become functions of T So I substitute these three functions into F Likewise in this formula, I didn't write it just to save time But what I mean by this is and the same Same similar for for Q and R Okay, so the orientation is important it causes change of orientation Results in the change in the appearance of the sign Okay, so that's that's pretty much all about the setup of integrals for curves So next we move on to integrals over surfaces Integrals over surfaces So what do we need to know here here again? There are two types of integrals that we are interested in and there is a lot of similarity between the first type For surfaces and the first type for curves and the second type for surfaces and second type for curves So the first integrals of the first type over surfaces are integrals of functions So here again F is a function In this case you have a surface M and it is embedded in R3 You could also have a surface embedded in R2 that would be just a region on the plane like this, right? but that's So that's integrating over such regions was the subject of The earlier chapter of the course chapter 15. This is just the usual double integrals So here the novelty is when you look at surfaces In R3 which do not fit in R2 which cannot fit in the plane But only fit in a three-dimensional space like a sphere or part of a sphere Right, so what I'm talking about is something like that would be M So what what does how do I compute this? again to compute I Have to parameterize my surface parameterize Because it is now two-dimensional. I have to use two auxiliary parameters, which we usually call u and v and So we write x is some function of u and v y is some function of u and v And z is some function of u and v where u and v run over points of some domain already on a plane So what I'm doing is that I'm trying to identify a curved Surface like this like this M with a flat surface like this which I called D So each point here would correspond to a point here, and it's a one-to-one correspondence. That's what this parameterization means once you have this parameterization you have a formula for For this integral to write this formula, I will use the following notation. I'll call R uv the vector field which is obtained from this parameterization by Interpreting each of the three functions x y and z as components of a three vector So you will have x of uvi plus y of uv J plus z of uv This is vector R And so to I can differentiate this vector for example R sub u will simply mean x sub u times i plus y sub u times j Plus z sub u times k So again, this stands for partial derivative of x with respect to u There's a partial derivative partial derivatives with respect to u So this is a this is this is another example of something which we use in this part of the course We're building on some material that we learned before because of course we learned Partial derivatives earlier right before the second exam. So again, you have to know how for example how to compute partial derivatives Without knowing how to compute partial derivatives. You would not be able to compute such integrals Once okay, so once we have this we also have R sub R sub v Defined in a similar way. You just put derivatives with respect to v instead of u Then the formula is The following ah, I actually made a I only drew one sign of integration I should have drawn two signs of integration for double integrals. We draw two sides. Well, it's not a big deal. I Will not deduct points for this since I've made the same mistake I'll go easy on you if you put the one integration sign instead of two for double integrals But for triple integrals you should put more than one for double integrals, I will be a I'll I'll be lenient on double integrals in terms of this how many integral signs you put Okay, let me put two signs here though So integral over this M. You'll be able to write as a double integral over D Where here you will put F where again you put X of UV? Y of UV and Z of UV and Then you will put R sub u Cross R sub V. This is a cross product again something we learned before and then you take the magnitude of this vector DA DA is the usual measure of integration on on the UV plane So this term R u cross R v is just like this term which you have for integrals of functions over over curves the first type of integral It's the analog. It's the analog of that That that R u cross R v absolute value is the analog of this Okay, so now the question really becomes how do you parameterize it once you parameterize it once you parameterize it It's very easy because you just need to compute the derivatives and you take the cross product and you end up with With just an ordinary double integral But the question is how do you? How do you compute this? So well, there are there are at least two special cases, which you should know the first special case is when your Your surface is part of a graph It's part of graph of the graph of a function In this case the formula simplifies and in fact, it's a good idea to remember To remember this formula, but you will actually have a cheat sheet So you might as you just write it on the cheat sheet. You don't have to remember Okay, so special cases the first special case is when when the when the surface m is part of The graph of some function g of x y So in this case what we can do is we can just say that x is you Y is v and z is g of u v okay, or we can just Stop pretending that you and we are two separate way are two additional auxiliary variables and just call them what they are I mean, let's just call this auxiliary variables x and y So then the parameterization will be in terms of x and y and we'll have axis x y is y and z is g of x y so in this case actually You can compute This cross product and this has been done. I I did it in class I wrote this formula in class your computation is explained in the book You must have done it many times on homework So you might as well just remember this formula and the formula is like this. It's minus GG DX Minus dg dy So this is i this is j plus k if I remember it correctly Please correct me if I'm wrong. I Think it's correct This formula you just get by applying by applying that general formula in this particular case It's just that when you compute the cross product, so you have to compute this You know the determinant over three by three matrix. It's just that the matrix becomes very simple and it simplifies So it's easy to I'm sorry. That's right. I did not close the parentheses For which I apologize. Okay. No, please please tell me because We we want our worldwide audience to see only the correct only correct formulas. So Okay, and now if you have this you can compute it's a it's a it's it's magnitude very easily and So it's just going to be the sum square root of the sum of squares of this components So it's like this So we actually have not just that my the magnitude, but we actually have a formula for the cross product Which I will use in a minute. Okay That's the first special case. There is a second special case When your surface is part of a sphere the second special case Is M is part of the sphere of the sphere of some radius Let's call this radius R So this this radius R is just a number. So so you got a sphere which is of this radius and Your M could be just the entire sphere Could be the entire sphere Okay, or it could be a hemisphere or it could be Part of the sphere which is within confined within a cone. It's like an ice cream cone Right, and that's just that's just the surface of the ice cream So in each of this in each of these cases a Good way to parameterize such such surface is to use spherical coordinates Spherical coordinates we learned before Sphere a spherical coordinate Spherical coordinates give us an alternative coordinate system for the entire three-dimensional space And we've got three coordinates. We've got row Phi and theta But now we are on a sphere We are on a sphere of radius R. So from the point of view of spherical coordinates This coordinate row is fixed row the coordinate of the spherical chords is equal to R right But we still have two remaining spherical coordinates, which we can use To parameterize our surface Which is good because surface is two-dimensional. So it requires two independent coordinates and phi and theta give us these two coordinates So in other words you and we here are Phi and theta for example if it's the entire sphere you will have theta going from 0 to 2 pi and Phi going from 0 to pi Which is the full range of phi and theta if it's the upper hemisphere again theta will go from 0 to 2 pi and Phi will go from 0 to pi over 2 and Here Theta is from 0 to 2 pi and Phi well Phi it depends how much ice cream you get so It's in your interest to have it as as large as possible. I guess although it becomes more difficult to hold it So let's just settle on the middle Middle ground Pi over 4 okay, so that's how your parameter is an example of parameterization And now in this in this case there is a formula also for There is a formula for put me erase this so that's just I write the formula for our U cross RV, but now we don't have you and V. We have phi and theta and so Our Phi cross our theta in this case And I'll just write the formula for the magnitude is actually what you would expect it to be It's exactly the formula for the Jacobian Except that you have to set row equals R in here. So it's R squared times sine five You have a question right, so the question is about the Cone, how do I measure Phi? Phi is measured from the z-axis right So 0 to pi over 4 would be This kind of range. Yeah, although I have to say in this picture it looks a little bit less than pi over 4 I don't want to short change your ice cream, but This is just an approximate an approximate picture not the best day to eat ice cream in Berkeley though Perhaps earlier in the semester would have been more appropriate Okay, so that's the formula for this and let's go back to This double integral so that's all you need you need this guy you need your function where you stick those Expressions Coming from your parameterization and then you end up with a simple double integral. So you compute it What is interpretation of this? What is the meaning of this the meaning just like in the case of Type one integrals for curves is that it represents something that has to do with the area or mass Something when you average density over your surface. So the typical example is that this is represents mass of Say Aluminum foil in the shape of M where F This function F which appears in that integral F Is the mass density function density function? Now it doesn't have to be mass It could be any quantity which you can measure which can have a density for instance charge This foil could be charged But in a in a complicated way so that it's some parts are more charged than others so you have to integrate you have to average them out and So you can have F may be the charge density function not necessarily mass density function So then you would be calculating total charge By the way, if it is mass you can also talk about about the momentum of the mass Which means that you would be multiplying in descent under this integral f by x x or y or z, right? So you can also find the center of mass for for this membrane for this surface so if somebody if the question is phrased in a falling way find the charge of a aluminum foil in the shape of such-and-such surface if The charge density function is such-and-such function then you immediately know that you have to do this integral Okay So that's the integral of the first kind or type 1 integral over surfaces and Next we go to type 2 integrals Type 2 integrals are integrals of vector fields in this case In this case we are given a vector field f I've given a vector field and We are computing this which is oftentimes called sometimes called surface integral of the vector field But oftentimes it's also called flux flux of f Across so the word flux if you see the word flux it refers to the integral of the second type and I explain what it means it means the flow the rate of flow of for example of a of a fluid through this membrane if f is a velocity vector field so this integral actually was defined originally but as as follows we We said that we have to take the dot product of f with with the unit normal vector field along along Then this will be a function and then we take its integral as a type 1 integral So this definition already makes it makes clear that we need orientation. We need a choice of n so just like the integrals of second type for curves are Not really well defined if you just specify vector field and the surface Integrals of the second type are also Not well defined if you just specify vector field on the surface you have to specify orientation on your surface So it's exactly the same we need orientation and the orientation should be given to you if it's a if you have to solve a problem orientation if you give if you are Given a problem where you have to compute the flux of a vector field Across a given surface the problem has to say what is the orientation or there should be a way to find which orientation It's supposed to be okay Now again, there are two different types of our entity There are two different orientations in case the surface is orientable, but of course will will be dealing with orientable surfaces I explained to you Möbius strip remember Möbius strip is an example of a surface which is not orientable But we will be working with orientable surfaces So an orientable surface will have two different types of two different orientations Which they will differ by a sign So how do you find out which one which orientation to use? So let me first of all rewrite this to make it computable as a double integral So again, I am assuming that my surface is parameterized in this way by this functions in U and V Okay, and In this case this double integral will be written as follows will be double in double integral over this domain D To have there and then here you'll have F dot are you cross RV? I mean in principle you could try to compute this integral by using this n and there are different ways There are different ways to try to find this normal vector Okay, a Unit normal vector So you could actually apply this formula and and then compute this integral as a type one integral You could compute as a type one integral but I Think I suggest I strongly suggest that you use it as you use this formula for it Okay, this because the point is that it's easy. There are many ways to find a normal vector For instance, you know, we talked about normal vectors for graphs of function functions for example, but what you need to do After this you have to first of all normalize it You have to get a normal vector of unit length you have to normalize it and then in addition You will have to compute this DS which itself is a is a rather complicated task, right? You have to compute this DS is which we here compute is the magnitude of our u cross RV The nice thing about this formula is that this formula is obtained from this formula where instead of n you take This are you cross RV which you normalize by dividing by its magnitude. So what happens is that there is some consolation This norm this magnitude gets cancelled by the same magnitude in the denominator So the formula actually simplifies that's why I think that This formula is nicer. So it's easy to get confused if you use a different type of formula I mean you can do it at your own risk, you know if you like if you know what I mean But I'm just I just think that this is the best way to do it in the framework For the kind of problems that we are we are dealing with here okay, so but So here's a formula, but how do you see if this is the right answer and not it's negative? You see so this is a this is a big question how to see To see that you get the right orientation in this formula in other words you could have a you could have a problem Say where you are asked where you are asked to compute To use the orientation to compute with respect to the orientation Upward orientation So let's say you're asked to use upward orientation. How do you know that this formula gives you the? Integral with the upward orientation and not the downward orientation for example, right? So first of all what what is do we mean by upward orientation? It means that you have to choose n of the form Something times I plus something times j where this doesn't matter plus something which is positive Times key This is upward and if I put Less than zero here. It will be downward. So that's the terminology upward vector is a vector whose k component or z component is positive and Downward vector is a vector which has negative z component okay So let's say you're asked to compute Integral overs over surface assuming the Upward orientation and you have this formula and the natural question is does this formula give me the upward orientation or it gives Me a downward orientation. We have to check that because if it does give me the upward orientation I will get the right answer if it gives me downward orientation Then I would get minus sign. I will get negative of the right answers. So it means I will get the wrong answer Right if I get the wrong answer, it means that I don't get the full score on this problem, right? So how do you see that? Well, here's an example Let's look at some special cases where this actually can be computed more explicitly And the first case is that is the case when m is part of this of the graph so for example Let's look at the case when example m is part of the graph of the function g of x y in this case We have already computed are you are x cross RV are y It is written on this formula. This is our x cross our y. So let's let me let me just copy this formula here So this formula in this formula the coefficient in front of i and coefficient front of j depends on the on the on this function that you have of which you are You know, which gives you this graph Right. So this this coefficients depend but this coefficient is always equal to one So this is this coefficient Is one so it's positive And so that means that if you are using this formula For r x cross r y Right, if you're using this formula for r x cross r y then You are getting orientation or you are computing This formula is computing the integral with respect to the upward orientation You see so this is an important point that a priori it could be This gives you a normal vector not normalized not unit but it gives you a normal vector And and by looking at this vector you can find out with respect to which orientation you are now computing And then you have to match that With what you're asked to do in this case R x cross r y so you have to look more closely at this vector and see What does it look like qualitatively? In this particular case the is the simplest information we can infer about this vector is the fact that The coefficient in front of k is one. So it is upward. It's always upward. So now you match You see, okay, so what am I asked to do? I'm asked to do a double integral Surface integral with respect to the upward orientation. Oh, it's my orientation upward. Yes, it is. So this is the right formula f dot R x cross r y Is the right formula? Is the right formula if they're in for for up for Upward orientation But what if I want to trick you and I ask you and I ask you on the test I ask you to Compute the double integral compute the surface integral with respect to the downward orientation If you don't pay attention if you just you have copied this formula On your cheat on your cheat sheet, right? And you've copied this formula on your cheat sheet And so you think you're in good shape You just apply this formula and you get an answer But you get the wrong answer because you did not pay attention to what was asked You were asked To compute with respect to downward orientation, but you computed with respect to the upward orientation Just simply because the formula you are using always gives you upward But I have the right to ask you to compute with respect to downward orientation You know, both integrals make sense. It's my it's um, it's my right and my prerogative in fact to to ask you To define any to ask you to compute with respect to any orientation that I like So in that case you have to be clever and you have to put a minus sign So this is the right formula This is for downward orientation because Putting this minus sign You can interpret it as putting it mine Putting the minus sign in front of r x cross r y if I put it in front of r x cross r y I will get plus here plus here and most importantly. They get minus one here if I get minus one here I got Orientation vector, which is downward so that matches What I'm asked to do right to do the downward orientation. So that's why this is the correct formula Is this clear Ask me ask me Go ahead Right, I'll give another example in a minute for when it's not upward or downward Very good. What if the coefficient was is zero so okay in this case it's it never happens right because in this case It's just always like this So so here's an here is this in the next level of trick question Which is that suppose it's some r u cross r v okay But for some other u and v and suppose this happens that the component of the k vector is zero, right In this case I should not be asking you to compute with respect to upward or downward orientation Because if it has if it's k component is zero it means it's parallel to the x y plane, right? So it's neither upward nor downward And saying that compute with respect to upward orientation would be misleading or ambiguous It will not give you any information because Neither of the two vectors is upward or downward, right? So in this case you actually have to use Another way to describe it Okay, so what are other ways to say we have to describe in words You know the the possible there are two possibilities You have to be able to distinguish between them if the k component is non zero you can distinguish between the sign of k right You can say if positive sign or negative sign But if it is zero for example, you can talk about the sign of the k of the coefficient in front of j Even in this case here's another trick question. I could ask Instead of saying compute this integral even in this case when it's part of a graph Instead of saying it's the upward or downward. I could say compute it With respect to the orientation, which is to the left Okay, so what does it mean? What should you do if you see that? Don't panic it's If your if your if the orientation is described in terms of left right It means we are talking about the y coordinate, right instead of the z coordinate Which would be k and i'm going to the y coordinate, which would be j right So to the left would mean that the coefficient is negative And to the right would mean that the coefficient in front of j is positive Right Now of course in this case, it's not clear. It depends on the function g sometimes Just the fact that the minus sign here doesn't mean anything because the function could be negative So then the total sign would be positive and and so on right But in the particular example which you are computing you should be able to see if it's positive or negative Over that range So what i'm saying is What you are interested in is just the piece of the of the graph So you are working with a particular domain d in u v And you have to see whether over this domain Certain coefficient is positive or negative, right? That's all it is And for example, suppose you find that r u cross r v, let me now switch to the more general setting where you actually have coordinates u and v and it's not necessarily graph of a function, but some other more general surface So you could have r u cross r v is equal to some You know some a of u v i plus b of u v j plus c of u v k This is what you will compute by taking by taking these partials And taking the cross product so you'll get three functions and then you just have to look closely What do they look like? Well, let's say this is u squared plus v squared Okay, so okay, so then it's not one But it's going to be positive Well the only point where it will not would not be positive would be zero zero So let's suppose that your domain does not include point zero zero. So then it's always positive If you look if you see that you know that the coefficient in front of j is positive Which means that the orientation vector is pointing to the right Right because Because this is what this is x Well Of course when we say When we say that it's all relative to how we Organize our coordinate system. What do we mean by left and right? But there is a way which we kind of used to which would be like x goes this way and y goes this way and this is a z, right? So so in this case To the right means that it's a positive number in front of j to the left means it's a it's a negative number in front of j So in this case it's going to be a vector not necessarily like this, but maybe like this So this is to the right and for the left of this way. Is that clear? There is another way to describe orientation by using the word Outward and inward so usually this is often used for spheres or for closed domains Okay, so that's another way Um, and that should be also pretty self-explanatory So at the end of the day you have to compute the formula by using r u cross r v But the bottom line is what you get is either the correct answer Or either the answer which if you put in this if you stick in here you get the correct answer or it's negative So it becomes fine-tuning issue. Is it correct or should I put a minus sign? So then you match the description of the orientation which is given in words Upward downward left right to what you got And usually it should be fairly easy to see and then that's how you see whether you got the right answer So you use that formula without any modification or you put a negative sign. That's how it is Okay Any questions? Yes So what if the sign changes if the sign change of course then the description would not be Correct, I mean well defined, right? So the description should be such that if it says to the right it means that that coefficient should stay positive or negative Okay, all right So so that's about the setup of the integrals and now I want to say a few words about The connection between different integrals These are the formulas which we've learned So this is two formulas So we essentially we have essentially learned three different formulas We have the fundamental theorem of line for line integrals We have stokes formula and then we have divergence theorem Okay, so now I have explained oh also Green's theorem is a special case of this When m actually belongs to the plane is a special case of this. So that's why I didn't single it out Okay So I already explained the kind of the general theory behind this and how you can you can think of all of these formulas as special cases of one And only formula What I would like to focus now is the practical aspects of this how you can you apply this formulas to computing integrals Of the type that we have just discussed so here I would like to emphasize first of all the following aspect of this formulas that in this formulas always there is a trade-off between The sort of a boundary the sort of a geometric boundary of a domain And a sort of derivative of the object that you are integrating There's a trade-off on the left. It's a general domain of integration But The integrand is special in the first formula you're computing Only line integrals or vector fields of this form the gradients The conservative vector fields in this formula integrals only of curl of something which is a curl of another vector field Here you compute a function which is divergence of some vector field So they're kind of derivative Of other objects, but the domains are most general most general curve most general surface most general three-dimensional solid On the other hand on the right hand side You have the most general Object here general function here general vector field here also general vector field But you're integrating over Domains which are kind of derivatives of the domains on the left hand side where in the geometric world derivative means boundary so This is a very important point to remember when you decide which formula is appropriate for this or that exercise So this or that computation So on the right hand side You've got general integrand but special Domain namely boundary Domain This is derivative here is derivative Here's boundary. Okay So the most so so don't try to use the formula where for example Where just on these grounds you can rule it out in other words Or don't use it in a straightforward way use it use it in a more creative way. You see so So let me give you just two examples of how these formulas could be used So example one So let's suppose you have You're asked to compute the line integral of a vector field But the curve the the curve over which you have to compute is very complicated. So this is the first indication that You should replace the domain of integration Okay So in this case Let's say you are asked to compute this So the clever thing to do is to replace this domain by this this Curve by this curve which has the same end points Right and then you can argue that This line integral Is equal to So let's call this c1 And let's call this c2 This line integral is equal to integral over c2 of f dr minus Plus the in the double integral of curl ds Where you integrate over the interior of this domain You see so you can use this formula Let's call this d or m. Let's call it m Okay So the reason is the following that if you take this to the left hand side You will get negative c2 I'm just I'm writing it like this Because I prefer to say this is equal to this plus something but I could Put this integral to the left with minus sign, which means I'm integrating over c1 minus c2 And then you have to see that this is a boundary of m c1 minus c2 would be going with this orientation That would be boundary of m now If I do if I do it like this there is a there is a there is this brings brings up another issue Which is the issue of orientation in this formula So I'm I'm using this formula But in both cases I have integrals which depend on orientation and we have this rule for orientations okay, which Which we discussed already many times So orientation on the left and on the right have to match in this way where if you walk on the boundary And the surface is in your left so The way I did it right now Would mean that the orientation on this surface would have to go that way Okay, so here it will be the surface where orientation is Going through the blackboard The usual orientation which we you normally consider would be this orientation Towards us, right and that's the orientation which we would get with counterclockwise But here I have chosen orientation which goes clockwise. Well, because I also put minus c2 so I reverse this And So this will be the orientation With orientation Let's call it away from us Not towards us, but away from us Because if somebody would be walking if their feet are on the blackboard, but their head points that way So they'd be walking in that in that classroom It'll be like spider-man and they would be And they would be walking also on like this bizarro world which is behind this blackboard Which we don't even know what's going on over there, right, but if that person would be walking Like this the surface would be on their left Right So normally if if I were to walk like this I would have to walk counterclockwise for the surface to be on my left, right So if I wanted to have orientation here towards us, I would have to for this formula to be true I would have to choose opposite orientation But so I'm putting sort of several things into this example because I'm out of time essentially But I'm trying to sort of pack a lot of information into this one example So you should really Look at it more carefully at different aspects of this Okay Variations have to be compatible and there is one more thing which I'm doing here Which is that I'm replacing a complicated path by an an easy path a simple path And the trade-off is that this guy is equal to this one But there is a price to pay and the price is the double integral over this membrane Over this sort of a film think of it is sort of like a film that you get from from soap There's like a frame and there is a kind of soap so you have to integrate over that the curl The best possible scenario where this works is when curl is zero if curl is zero There is no price to pay and in fact this complicated integral over this complicated curve is equal It's just straight equal to the integral here, but sometimes it's not zero Sometimes it's not zero, but it could be very simple So it's not such a big price to pay sometimes You see you see what I mean? So you have to you have to be careful you have to see different different options the different options that you have So this is one example where I'm utilizing formula number two And um, let me give you one more example Which is kind of similar, but for formula number three So maybe I should I should emphasize here That In this formulas Orientations have to be compatible. This is important I'm not going to explain it one more time because we spend a lot of time talking about this in the previous few lectures So example number two Example number two you can kind of a similar trade-off, but where you have to compute a double integral So let's say you have this upper hemisphere And then you also have this disc at the base of this upper hemisphere And together they form the boundary of the upper half of this ball So let's call this b upper half of a ball so then The last formula would tell you that The integral of divergence of some vector field e over this b Is equal to the into the flux Through the surface through the boundary and the boundary now consists of two things Let's call this m one And let's call this m two So the formula then will be that it's in double integral over m one plus double integral Over m two But I have to say with respect to which orientation And the rule in the divergence theorem is that the orientation is outward Which here would mean like this and here would mean like this Here it's outward actually is upward on the top part and on the bottom outward is actually downward But both are both are outward So that's what I mean here and here So how can we utilize this formula for example? What you can do is you can express this As the difference between this and this You see so this way you can express a complicated integral over something very curved the sphere Or upper hemisphere in terms of two flat integrals one of them is over a three-dimensional solid Which actually may be quite simple depending on what I made a mistake like this It could actually be zero this divergence It could be zero so then this actually drops out and then you can express this guy as negative of this guy Right or even if it's non-zero it could be something very simple like one function one or something So then it's easy to compute And this one is certainly easy to compute easier to compute than this one because it's an integral over a flat region So you have this trade-off again Similar to the trade-off discussed in the previous example between an integral over one part of the boundary And the integral over the other part of the boundary with a surplus or price to pay coming from The integral of the divergence. So these are very typical examples of the kind of things that you need To know to be able to apply this formulas Okay So so here's what we're going to do now I'm going to stop and we're going to do a course evaluations Just take a few minutes. Okay. Come on guys. Come on over. So we'll we'll start distributing them and We'll just take a few minutes and after this after we're done with course evaluations I'll I'll be here for for office hours. And so you can ask me questions about anything Well related to the course Okay