noc19-ma19-lec33

[Music] [Music] okay the greens theorem we start what is called the integral theorems of multivariable calculus I am NOT going to do a proof here we are not going to prove greens theorem the proof would take half an hour so maybe greens theorem proof I'll put it in put up in the portal but maybe I'll try to do a proof of Stokes theorem or an Gauss divergence theorem if time permits so what's greens theorem greens theorem actually relates a line integral with the surface integral that's what greens theorem does and we have so suppose I consider in the a vector function from r2 to r2 given as then given a so here the line integral is a special form over line integrals so suppose we are looking at a region bounded by a curve so these are domain D which is bounded by a curve C and I will always evaluate the line integral along the curve C if I move in a direction which I'll call the positive orientation in the sense that if I when I will move along this curve the area of this domain should always be on my left hand that's that's the criteria that I will follow so if I am moving along this curve so if I am moving counterclockwise along this curve then the area if I am this guy here then my area is always on the the domains areas on the left hand so Green's theorem states the following fact that for such a kind of domain we will get into more details so this see this curve C is the boundary of the domain D so they write this as del D so I will follow the books approach so it's del D pdx plus qdy this is same as integral over surface integral over the domain D del Q del X minus del P del Y DX dy Q is associated with Y so anyway differentiate it here is with respect to X and P is associated with a 20 differentiating here respect to Y so this is the thing that you have to remember so how do I decide which direction is my positive orientation in these kind of curls you see if I'll I'll take the sub this is my domain the domains out this is outside the domain so this is any vector V from a given point pointing perpendicularly normal if I draw a kind of tangent line here there's a normal which I call the V out vector this is pointing towards our outside now if I am in two-dimensional space so if this is my K vector right which is pointing so this is my two-dimensional space in which I am lying and K vector is pointing in the perpendicular direction so the vector this vector is K cross V out that vector that direction is the direction of positive orientation so if you look at that direction you you immediately understand that if I walk down then my domain has to always remain on the left-hand side so the greens theorem always expects del D to be covered or oriented to be oriented in the positive direction in the positive sense means if I if I move around that would be the way I'll move around so I'm writing green because it's greens theorem Green's theorem actually arose because of some computations in electrical potentials and gravitational potentials Stokes theorem was actually told to Stokes you know private later as the book says I have to check out the history a bit more by Lord Kelvin famous for its absolute temperature absolute zero so the Kelvin scale so and Stokes gave this problem in a math examination in 1854 which and gradually it becomes famous as Stokes theorem now so you see here there is a link snow how do you know that this is a line integral so if I have written F like this so you know here I can take the R curve and then you put R plus d R so this is your D s so D s is measuring the d s here is almost same as D are very small because D R is very small so you can write this as DX I vector plus dy J vector and that is exactly the way you compute D SD s square is due to DX square plus d square root over da square plus dy square you know this already so here if I do my F dot d s and this is nothing but P of Y vector I'm not writing X Y and q of J vector so you know we have sorry making mr. P of DX plus Q of D Y this is so this will give me a scalar so this is what it is so if I write in so if I write integral del D F dot d s that is same as writing integral del D pdx plus qdy so it is difficult to compute this you just compute the surface integral that's that's the message of greens theorem of course there is another requirement P and Q so these are the assumptions under which greens theorem is true P and Q must be continuously differentiable of course we need continuity here for the integration to go through smoothly here all integrals are meant when when you take one one the single integrals there may remain meant in the riemann sense or most cases in the standard Newtonian sense that's that's the really the effective thing that's now let's start giving one or two examples and see how we can understand so we'll take examples from the book and one by one we will try to work them out so here is a problem that says okay you take a curve which is lying along the boundary of this square formed by 0 1 0 1 that is the domain D is formed by 0 1 cross 0 1 so it's 1 0 0 0 0 1 1 1 now I have to traverse in this direction you see I have to traverse in so this is my C which is Del D so this is the direction I should be traversing along the four sides so when I am traversing along this I'm starting from one point and coming back to the same point that is the meaning of del D that you take a complete round of the boundary so start from one point and come back to that point that's the meaning of del D so they are asking you can you integrate on this del D boundary of this particular domain integral Y 4 plus X cube DX + 2 X 6 dy so here is my P here is my Q but I go from one point and come back to another point so you might feel that I've gone from the same point to the same point so it was a kind of very strange so how will I actually I'm in a full circle he's not going from A to B and so I have separately I can work them out I'll work out what are the limits of those two points the limits of the T the parameter and then I'll work it out so here if I want to talk about such kind of line integrals then the best ways to use a greens theorem so in this case I'll now write integral D so it is del Q del X here so what I am going to have here is del Q del X minus del P del Y DX dy so what is my del Q del X is 12 X 5 minus del P del Y that's 4 y cube do del X del Y so here X is running from 0 to 1 Y is going from 0 to 1 so that makes life is much more simpler okay why not write down answer here I'll just write down answer here it is 1 so greens theorem allows you to calculate this line integral and gives you 1 which is not so easy if I give you a complete circular path if you go and look at the line integrals chapter where you have studied line integrals for the first time see all the paths are usually open paths especially if you have a when I how do you compute a total path so you go by one path so you go from A to B by one path so you so when you compute this if you are if you have to compute the line integral here first you compute on this path then you compute on this path then you compute on the third path then fourth part so you start from zero zero compute up to one zero then you start from one zero compute up to 1 1 then you from one warning compute up to 0 1 and 0 1 to 0 0 that is a for like different line integrals you need to calculate if you want to evaluate the line integral that we were choosing but just by applying greens theorem just by applying Green's theorem here you have just solved nothing in one shot and that's the applicability a very interesting application of greens theorem is the area so if let me tell you that before I get into this area business die let me tell you that there can be regions which are much more complicated than the simple region that we have drawn there there can be region like this with holes inside now here what you mean by positive orientation see positive orientation would always mean here when you are encircling the region you have to go so the region is actually here this is the domain so these are holes so this is a C 1 this is C 2 and C 3 that C 1 C 2 C 3 all three together make up the boundary so in such cases how you do these are the cases where this means theorem become useful you do not have to bother too much about it see here along c 1 I move in the counterclockwise direction but along c 2 I move have to move in the clockwise direction and C 3 have to move in the clockwise direction because my positive orientation has has to follow one very simple thumb rule that my area the domain has to always even on the left side as I work on that curve so here on the internal curves you have to go on the clockwise direction on the external curve you have to go on the counterclockwise direction that is something you have to keep in mind and now area so suppose you have a kind of closed curve C covering a domain D and the area a can be computed by this simple line integral and how do we know that hey I can be computed by this simple line integral because suppose now we construct a vector field F where it is minus y vector and XJ vector so let me look at this integral half of integral C minus y DX plus X dy so this is your P and X is your Q so this is my P and this is my Q so this is your f dot d s and then I integrate by greens theorem I will integrate over for the domain D and greens theorem should give me del Q del X minus del P del Y so what is del Q del X it is 1/2 of so here a line integral is been computed just in one short throw double integral so what is del Q del X here it is 1 minus del P del Y is minus 1 so minus or minus 1 so this is so I will get so I will get half of this is not infinity this is the domain D so half of integral domain D 2 of DX dy and that is equal to integral D or the domain DX do and that's exactly the definition of the area that's a magnitude of the area so you see how beautifully this has worked how beautifully this has now there is another interesting way of expressing the I am NOT doing the proof proof I'll send across expressing the greens theorem it is called the vector form of the greens theorem that's what we are going to download vector form so again the same story so you have F vector even as P I vector and Q J vector I am NOT writing the XY z-- so what does it here they say that you can write this line integral as integral call F means call has a sign there is a call sign call F has this sign call of F so let me just remind you this is a sign we use for Karlov call of f dot product with k vector DX dy this is what another form is called the vector form of greens theorem the other form was written in the form of a scalar so we have to see whether this vector form has some meaning or not so let us see how do I prove it so f is P plus P I plus Q J now curl is in terms of three so basically you're here I have to write it in three three components 0 K so call of F is by I j/k del del X del Del Y del Del Z pq0 so it's a vector into this part which is zero minus del Q del Z minus J vector which is zero minus del P del Z and this del Q del Z is 0 and del P del Z is 0 K vector into del Q del X minus del P del Y and you see this is completely 0 because Delta is no Z in Q it is in terms of X Y so basically this is 0 this thing is 0 and this thing is 0 because P is a function of X Y only not XYZ so I am writing everything stepwise so I wrote this so it is what we have here is nothing KK dot del Q del X minus del P del Y and what do what do I have here so I have here a vector if I take a dot with K vector I'll simply have this as del Q del X minus del P del Y and this simply gives me back the form in which I know about Green's theorem so this is another way to integrate the call of F dot K this this call of F will become important when we will be talking about Stokes theorem right so there is a kind of result where we will not talk much about it it is called the Gauss divergence theorem in the plane because goes die Gauss's integral Gauss divergence theorem relates the integral of the divergence of a function the like the line integral basically the the surface integral the surface integral of divergence of a function over a given surface is related to the volume integral over the volume or the triple integral volume integral is a triple integral with respect to the volume enclosed by the surface so that's how Gauss's what scouts divergence theorem in that that is very important in electrostatics is also important in mechanics so ie of course in fluid mechanics naturally so fluid and classical both so there is something called Gauss divergence theorem on the plane so I'll just make a statement of this and in the talk here Gauss divergence theorem on the plane so it says that this line integral here is is a not all surface but a small arc out length is a surface integral in terms of the divergence of the proof here is slightly not simple I would not like to give the proof here at all so there it needs some more ideas from vector calculus which will confuse you so will not go to all these things so there's several for example as I just as an example I will end before in the talk for example there is a problem here evaluate integrals C Y DX minus X dy do not immediately jump to the conclusion that this crime cut the area because look at the area look at the area stuff here area is X dy minus y DX here it is y DX minus X dy kind of negative of area right so here you have the domain D as minus 1 cross minus 1 and you see is the boundary of the domain oriented in the positive sense how will you evaluate it there is one simple thing to observe first observe that I can write this let me write down that that this integral can be written as minus integral of X dy minus y DX over this C so what is area X dy minus y DX so X dy minus y DX is area in fact twice the area so this is minus twice the area so what is the area here of this domain one side is two and other side is 2 it is 4 so minus 2 into 4 is minus 8 but if I go in the clockwise direction opposite direction because this would be if I go in the clockwise direction the value of the integral would have a negative sign here so if you become 8 so the problem says you have to go in the clockwise direction but let me just keep it in the anti-clockwise direction make me cool and so the area of here this is this is nothing but minus or twice there and this is minus 8 that's answer so that is how you apply what you have learned here so with this I end my talk here and let me tell you 7th week is finished there are 4 lectures in 7 weeks because the two lectures on surface integral has been combined into a single lecture which you know is pretty involved Green's theorem once you know the formula you have the statement of the theorem you know how to apply it and that's what we concentrate on that stats water and proof of greens theorem will come later as a part part in the portal it's it's not a part of the whole thing let me see what I can do if I can give a problem if the proof of Stokes theorem can be done in the class but I ignored thing it can be done in the class maybe I'll just for the just for you know kind of a mathematical work to be done so that you do some math and think about how proofs are done here maybe I'll try to do the Stokes theorem here in the class but Gauss divergence theorem I do not know how much I will have time to do but will do do something about it so let me tell you hear the exact divisions may not be maintained so we will cover Stokes theorem it's applications Gauss divergence application and we'll speak about fields in the eighth week because Fields has lot of link and how differentiation and integration is linked here in this vector situation and how these ideas are called gradient divergence can be applied to several physical situation those things will be discussed so overall we have to I have to provide your content two hours and 30 minutes so I'll try my best to cover the thing in three hours which could be in a talk of four which could be in a talk of one two maybe four and five so it could be any talk of four or five so but the seventh we call signs with four talks because two talks have been combined in one the surface integral idea is that why I combine them was because if I just separate them out the integer it loses the link so I cannot talk about line integrals and suddenly I don't speak about orientation because that is how the idea of sorry I don't speak our surface integrals and I don't speak about oriented surfaces because otherwise the parallel that you have learnt in case of line integrals would become lost so that is the way this or if I divide this thing into two parts a little bit thing for one lecture then another lecture a little bit thing then the link gets disturbed so it's better to go in one flow so two talks have been jammed into one so there'll be four talks one having almost an hour so thank you very much thank you for being patient and we are ending the course and hope the journey was fine and hope many of you find something useful from this course to take anyway the YouTube is with you and so you want to discuss with us thank you [Music] [Music] [Music]