 So, let us look at the example. So, it says look at the x is bigger than 0, y is bigger than 0, x plus y less than or equal to 1. So, what does it look like? So, domain is, domain D is x, y such that x bigger than 0, y bigger than 0 and x plus y less than or equal to 1. So, let us try to picture x bigger than 0, y bigger than 0. So, where is the domain? It is in the first quadrant, x is bigger than 0, y is bigger than 0. So, on the first quadrant, so y x, x plus y is less than or equal to 1. What does that mean? x is bigger than 0. So, you can say y is less than or equal to 1 minus x or x is less than or equal to 1 minus y depending on, what is that x plus y equal to 1? That is a line. So, that is a line cutting x, x is at 1, cutting y, x is at 1. So, that is this line. So, that is this line x plus y equal to 1. So, this is 1, this is 1, this is 0. So, what is our domain? Our domain is precisely this triangle. So, if you want to write this D, you can write D as, so this D can be written as x, y. x lies between 0 and 1 and for every x, what is the height vertical line going, moving parallel to y axis? It goes from 0 up to, so y goes from 0 to 1 minus x. We have to interpret that line as a function of x, y equal to something, as a function of x, y equal to some psi x. So, that is, I can also write probably this as x, y. y goes from 0 to 1 and if I want to look at every point y, it starts here and goes up to that line and that line remains the same, the limit remains the same, wherever be the point. So, that is x between 0 and goes up to 1 minus y. It should be a function of y, so 1 minus 1. So, it is of type 1 and type to both. So, what else? f is a function given by, so you can integrate that function, if you like. We will not go into integration. So, f x, y, integrate between 0 and 1. Here, what is fixed here? I am changing that example or what? So, you can integrate. So, let us integrate as a function of, is it okay? Because the function was, there are two parts of the function, x square plus y square. So, integral of x square plus integral of y square. So, that is two parts of the integral, integrate with respect to dx, with respect to y and then integrate that with respect to x. So, this is a simple example of, you can try again yourself later on. Let us look at something. Then, by Fugini's theorem, this is given by, that is okay. For a rectangular region, what do you think? For a rectangular region, what should be the volume? If the region at the base is a rectangle with sides a, b and c, d, then Fugini's theorem gives me this integral. If the function is constant function, what it should be? A rectangle is both of type 1 and type 2. You can calculate either way. If the function is constant, that comes out. Integral 1, whether you write type 1, a, b cross c, d. So, you integrate over c, d. So, d minus c into b minus a. So, that is the volume of the parallel of the any way. So, that is, let us not go into that. Supposing your function splits into two parts, fxy is a product of two functions, phi x and psi y. So, when you integrate fxy with respect to x, fix or y fix, other function is not going to play any part in the iterated integral. Is it clear? When you integrate fxy with respect to x, then phi y is a constant. So, that will come out as a constant. So, your double integral splits into product of two integrals, integral of phi and integral of psi. So, that is what it says, because the function itself is splitting into two parts, independent parts depending on x and depending on y. So, these are the things. Let me look at some more interesting examples. These are all simple examples. x square plus 2y, dxy over this. So, what will be the integral? Integral of x square plus, or you can calculate directly, x square plus 2y. Integrative with respect to x first or y first, whichever you like. So, when you are integrating this with respect to x, it will be x cube by 3. 2y, y is a constant when you are integrating. So, it will be 2yx appropriate limits. So, only keep that in mind. So, when you are integrating with respect to x, it comes out this and then integrate that with respect to x between limits 0 and 1. So, a simple example. And if your domain is both of type 1 and type 2, Fumini theorem says you can calculate it either of the ways, iterated integral, either by fixing x first or y first, then the two should be same. So, your iterated integrations, if the domain is of both type 1 and type 2, then the double integral should be same as integrating with respect to domain type 1 and also equal to integrating as domain type 2. Both integrals should come out equal, because both are equal to the double integral. So, keep that in mind. To check sometimes, I think let me look at. So, these are all simple examples. Type 1, type 2, disk and all that kind of thing. So, I think type 2, I should discuss something more which requires. So, you have to decide upon, given the domain, whether you want to write it as type 1 or type 2. It may or may not be possible to write every domain as type 1 or type 2. You have to see which is possible and accordingly use Fumini theorem for that. So, we have seen examples of those domains. Let us look at this, where D is the region bounded by the parabolas 2x square and 1 plus x square. So, now, we have curve sketching of one variable will come in handy. What are these parabolas and what is meaning of saying is region bounded by the parabolas. So, let us look at 1 is y equal to 2x square that is 1 and the other one is y equal to 1 plus x square. So, what are these parabolas? y equal to 2x square that is our standard parabola y equal to x square anyway, 2 times. So, let us draw that. So, it will look like something like this. What is y equal to 1 plus x square? I have copied 1 plus x square. It is again y equal to x square essentially, but y is shifted by 1. So, when x is equal to 0, y is equal to 1. So, if I draw this parabola now, I am saying it should look like this. It should intersect the other parabola y. See, if you look at y equal to 2x square for the same x, y is varying twice. Here y is only x square. So, that is increasing at double the rate. So, what is the region bounded by it? So, what is it? What we are looking at? We are looking at this point. So, this is the region. So, this is a domain D. Is it of type 1 or type 2 or neither? If it is of type 1, then I should have x lies between something. So, it looks like x lies between this and this. So, I have to find these points of intersection. So, what are these points? How do I find the points of intersection? 2x square equal to 1 plus x square. That means x square equal to 1. So, x is equal to plus minus 1. These are the coordinates of the points of intersection, the x coordinate. So, x lies between minus 1 plus 1. And for every x, where does the y vary? So, that is the question we have to analyze. So, for every x, here is the y. So, it goes from the lower parabola to the upper parabola. So, y goes from, for every x, y goes from the lower parabola, that is, what is the lower parabola, y equal to 2x square, 1 plus x square. So, type 1. I can integrate and be happy about it, whatever the function may be. Let us try to write it as a type 2. Can I write this as type 2? All x, y. So, I have to find out what is this value of y. So, what is this point? What is this point? y equal to 2. So, y goes from 0 to 2. Now, if I want to find out what is the limits of x, I have to go parallel to the x axis, because x I want to verify. So, for that, for any point in between this, if I go, I go out and then I come in again. So, up to some part, it starts here and ends here. If I go here, then it starts same and ends same. So, it does not look like a nice idea to write this as a domain of type 2. You cannot, but if you like, you can try to cut it up into many pieces and do it. So, when you want to integrate over such a domain, you should decide to write domain as type 1 and use that to calculate the integral. So, that choice we will have to make. So, you define. So, all this is described here. So, when you want to say whatever functions fx plus y. So, integrate 2x square to 1 plus x square. Integrate with respect to x is outside. So, type 1 we have written minus 1 to plus 1. So, that is, if we try to write it as type 2, we will have a problem and so on. So, more examples of double integrals. So, I think other examples you should try to do it yourself. Now, I would like to spend some time on. So, this is, when your function, when your domain D contained in R 2 and D is interpreted in Cartesian coordinates R 2. So, there are two in the plane. There are different ways of locating a point. You can locate a point differently. Why are coordinates defined at all? What is the need of having coordinate geometry? Because it helps you to locate a point in the plane. For example, in the line, what is the need of putting a 0 somewhere and 1, 2, 3, 4, so on? Because calibration of the line helps you to locate a point. The point is minus 5. That means what? That means somewhere you have marked as 0, some point as a reference point as 0. If you move on the right side, you write plus. If you move on the left side, you write it as minus just for your convenience. Then you mark equal units. On the right side, if you go 5 units, then you are reaching a point which has got coordinate plus 5 as only left side. So, on the line, marking those things are putting coordinates for every point on the line. It is a very interesting story. If you imagine a planet which is all linear, everybody stays on the line. Two persons meet on that line somewhere. One person invites the other one to come to my home in the evening. He asks, where do we live? What shall we tell? How does the person tell the other person? Where does the other person live? Imagine a road which is very long. There is no signpost anywhere. So, you do not know which side the habitation is, which is on this side or on this side. Which side you should start moving? Like a hero is lost on a road. He does not know which side to go. So, that is where these mile posts, those help. If you walk somewhere and you find it says, 20 kilometers on this side will be a city called so and so. So, on the line, on a line land, if you want to give every point an address, you have to choose one reference point. Call it 0. Every city has a gantager reference point. So, that is the reference. What you do? From the reference point, it also gives you a direction on the line. One side you can call it positive, other side you can call as negative. That is mathematical way of saying. One side you can call as north, other side you can call as south. Whatever you like, one side you can call it as ram, other side you can call it sham. It does not matter what name you give. Because we are doing mathematics, we say on this side, what is left and right on the line? Depends on where you are standing and where you are looking at. If you are looking this way, this is left, this is right. If you look other way around, that is left, that is right. So, you have to say this side is called plus this side is called negative, whatever it is. Now, every point gets a coordinate. How many units mark up? That is coordinate geometry in line. You need one reference point and one direction. That is one reason why line is called one dimensional, only one direction. Now, imagine a plane. Imagine a plane on which no signpost, nothing is visible anywhere and the same problem. How does two percent have each other, where to come? To come to my house, where is your house? So, to give an address to every place, every point on the plane, you need first a reference point. Then, one direction is not enough because one direction will allow only to move around one line. To get out of this, you need another direction. So, you need two directions. One direction is given by one line. The second direction is given by another line passing through the reference point, but not parallel to this because if it is parallel, it is not giving you any direction. Neither coincident with it, then again no direction is given. So, you need two intersecting lines which are not coincidental or parallel. Then, move along one direction, some units move along other directions, some other units and you reach a point. You do not need a rectangular coordinate system. For coordinate geometry, you do not need. You only need two directions, that means two lines and a reference point. That is why the plane is called two dimensional. Only rectangular because it helps you to compute things easily. The distance between two points is given by Pythagoras theorem very easily. If the axes are not perpendicular to each other, you have to apply trigonometry. That is the birth of trigonometry. Y trigonometry is required. Anyway, what we are saying is what we have interpreted D as? We have interpreted D in the plane by coordinate system. That means x coordinate and y coordinate. What is this? We have got two lines, reference lines. One is called x axis, other is called y axis. They are perpendicular and this is the origin. A point, how do I locate a point? You draw a perpendicular, call it as y. See how much it is from other one x. So, this has got x comma y. These are the coordinates. So, this locates the point. Given a point in the plane, given two axes and a point of reference, you can write down the coordinates. So, every point gets a unique coordinate. Conversely, it is one to one map. Every point gets coordinates and every coordinate gives you a point. So, supposing you have 5, 6. I say a point is 5, 6 coordinates. That means you go 5 units here and then go 6 units. You will get a point. Every point gets coordinates. Every coordinate ordered pair gives you a point. So, points in the plane can be located in one to one way by coordination coordinates. But this is not the only way. Here is another way I can locate. This is only for my imagination. So, let us put a reference point. Imagine they are not there. So, let me remove them so that you do not feel. So, reference point is required. Now, after this reference point, this is the point I want to locate. This is the reference point. How do I locate it? Without rectangular coordinate system or coordinate system at all? First of all, let us see how much is this distance d? d is at some distance. How many points are there at this distance d from origin? They are infinite. So, these are all points. That is a circle. Now, this point I am looking at. Why I am looking at that point? Because I want to fix this point. So, I look at a reference line and look at how much is the angle I make. So, this point is known how much is the distance d? How much is the angle theta? If the angle is theta is here, then this is the point I am locating. So, every point at a distance d is uniquely known by the angle it makes with the reference line. If there is some different point, then distance is different. So, all the points in the plane are located by treating them as points on a circle of some radius and how much angle it makes. So, d and theta. Again, you need two reference things. How much is the distance? How much is the angle? These are called polar coordinates. The angle comes into picture. So, we will see next time how do you evaluate an integral when the domain is best described by polar coordinates. Because supposing you want to integrate over this sector, then there is neither one type 1 nor type 2. You have to make efforts. So, we want to change the variables from Cartesian to polar and see how does the double integral change evaluation of double integral change. So, we will look at what is called change of variable formula for double integrals from Cartesian to polar. So, let us stop here.