 So, let us start looking at the change of variables. So, we saw yesterday in the previous lecture that in R2, there are Cartesian coordinates and there are polar coordinates. So, for a point P with Cartesian coordinate X, Y, the polar coordinates are R and theta, where the relation is X is equal to R cos theta and Y is equal to R sin theta. Theta goes from 0 to 2 pi. In R3, we had three coordinate systems namely Cartesian and then we had cylindrical and then we had spherical. So, we will revise them again today. What are these coordinates? So, this is P, X, Y, Z. What is X? What is Y? What is Z? And similarly, what is X? What is Y and what is Z? So, we will describe them again. But before that, let us start looking at what is called the change of variables, how these are related with change of variables. In one variable, there is a chain rule for differentiation which says that if we have a composite of functions f of g x, then its derivative is f dash at g x into g dash of x. So, that gives you a change of variable formula for integration. So, it says that if you want to integrate both sides, you will get integral of f g x derivative dx is equal to f dash of g x, g dash of x dx with appropriate limits of integration. So, one way of viewing this is if you call this g x as t, then the derivative f dash t is equal to f, then what is g of x is equal to t, then g dash of x dx is equal to dt. That is what is normally a change of variable formula in one variable it is called. So, and this side is nothing but integrating the derivative, so f composite g, if the limits are a to b, so b minus f g at a. So, that is equal to this. So, change of variable formula helps you to compute the integral by making a change in the variable. Instead of looking at the variable x, you look at the variable g x. So, that is a new variable you introduce. A similar thing is possible in say in two dimensions. So, change of variable let us look at. So, to describe this, let us have a variable u and v and we are making a transformation of this to x and y. So, there is a domain here and there is a transformation which changes it to some other domain. So, call it as d, call it as e. So, what do I mean by this transformation? It is a map from r 2 to r 2. So, g is a map from r 2 to r 2. For example, u and v. So, g is a map. So, it is a vector valued map because it is in r 2. So, it will have two components. So, g will have components, let us say g 1 and g 2. That means what? That means g of u v is equal to g 1 of u v g 2 of u v g 2 of u v where g 1 itself is a function from r 2 to r and g 2 also is a function from r 2 to r. See, for every point u comma v, so let us write u comma v that goes to g of u comma v. That is a point in r 2. So, it will have two components. This is the first component that is the second component. So, each component is a function of the point u and v. So, they are functions of two variables. So, now, for example, you can have, if you like in the mind, you can keep an example r theta goes to. So, what is g 1? That is x is equal to r cos theta and y is equal to r sin theta. So, this is g 1. This is g 2. So, polar coordinates representing something in polar and Cartesian, that relation can be thought of as a transformation from polar to Cartesian or Cartesian to polar. Either way, it is because it is a 1 1 1 2 map. So, in general, we will have a transformation g from r 2 to r 2. So, in the domain of r 2, the variables will be denoted by u v and in the range, we are denoted by x y. One would like to know, if I make this transformation, a domain d changes to a domain e. What is the relation between the area of d and area of e? What is the relation between the two areas when you make a transformation? So, to do that, let us imagine a small portion of inside that d. So, this is a point and let us say it is delta u and it is how does a small area element change? We want to look at that and this length is delta v. So, this is a rectangle of length delta u delta v and this is the transformation g. We want to know how does it change. So, it will change it to some kind of area like this. So, what is the relation between the two areas? So, this is x and that is y. So, let us think it off as a movement. At this point, you are travelling in the direction of u by distance delta u. So, when you transform it, this is what you will get and this upper one will be transferred here and similarly, this side will be transferred. It is the image of the small portion. You would like to know what is that area? So, we can think of this small portion on the right hand side in the x y plane to be a parallelogram and to find the area of a parallelogram, if you know what is this vector and what is this vector, then the cross product will give me the area of the domain generated by the two vectors. What is this area? Now, this is not going to be a straight line, but for a small portion we can assume it is a line. How will you move? As you move in du, let us call it p k r s. If you as you move on p k o, you are going to move here along a b and the distance you are going to move in the domain is delta u. So, we want to know what is the rate at which you are moving along in the direction of a to b. So, what is that rate? Rate of change is given by the derivative. So, this a b, you can think it of approximately. So, this can be thought of as a tangent vector. So, what is that tangent vector? So, that vector has got partial derivative of g. This is the partial derivative of g with respect to u direction. So, u. So, this is a vector function. It is a vector and distance is delta u. So, this is going to be the vector. Similarly, the other one, if you look at this one, this is going to be partial derivative of g with respect to v delta v. That is approximately a b and this is approximately a d. We are approximating the curve by the tangent vector. So, what will be the area? This small area on this side. So, delta x delta y will be given by, that is the cross product of delta g, delta u cross product with, I am using the vector algebra that you are supposed to be familiar with, that you are given two vectors a and b, then the cross product absolute value, magnitude of the cross product gives you the area generated by the two vectors, the parallelogram. So, that is the definition of cross product actually. You can also think of. So, v delta v. So, that is this area of this small rectangle or small parallelogram. So, basically what we are saying is, this is going to be changed to, and if we call it delta x cross delta x, delta x cross delta y, that is going to be. What is this equal to? Let us just compute. So, how do you compute that? So, what is this equal to? Cross product. So, let us first write what is delta g delta u. So, that is delta, first component is g 1. g has got two components with respect to u, i plus g 1, g 2 with respect to u, and that is j. These are the components. These are vector differentiation. g has got component g 1 and g 2, partial derivative of g 1, partial derivative of g 2, and similarly, this one, partial derivative of g 2, g 1 with respect to u, i. I do not have to write the bracket. So, delta g 2 with respect to v. So, this is with respect to v of j. Then, what is this cross product? How do you get the cross product of two vectors? So, that is i, j and k. The components, partial derivative of g 1 with respect to u, partial derivative of g 2 with respect to u, third component is 0, because we are in the plane. So, delta g 2, delta of g 1 with respect to delta v, delta of g 2 with respect to delta v and then 0. So, when you expand this, what you will get is the norm of, we want to compute the norm of this. That is a cross product. So, we want to compute the norm of this. So, when you compute the norm and simplify, you will get, this is nothing but, so delta x, delta y. So, this thing comes out to be, these are scalars, delta u, delta v, these are scalars. So, they will come out. So, delta u, delta v and you will have partial derivative of g 1 with respect to u, partial derivative of g 1 with respect to v, partial derivative of g 2 with respect to u, partial derivative of g 2 with respect to v. This is the absolute value of the determinant. So, you should have, this is absolute value of the determinant, absolute value. So, we expand it with respect to ij, find out the norm of that factor. So, that simply comes out the absolute value of the 2 by 2 determinant, because there are 0s here. So, when you expand, you will get only one component. Is it okay? Delta g 1, delta g 2 minus, so that is the determinant coming here. So, this quantity becomes important. So, the small area element in the transformed thing is scaled up according to this factor. Like in one variable, what was the scaling? In one variable, the scaling is g prime. Here, in two variables, both the components contribute and that is the scaling. So, you get delta x delta y. So, this quantity is called the Jacobian of the transformation g at that point, whichever g at that point. You are looking at a point. So, this is a point. At a point, you are looking at. So, basically the idea is small element is transformed like this. So, if you want to look at the full integral, that will be the limit of the small elements. So, that will give you the change of variable formula. So, let me write the change of variable formula to be. So, in general, so this is called the Jacobian, which we defined just now. So, Jacobian of the transformation g with components g 1 and g 2. The first row is partial derivative of g 1 with respect to the variable u, partial derivative of g 1 with respect to the variable v. So, first row is with respect to g 1, two variables, g 2 with respect to two variables. Determinant of that, that is called the Jacobian. If you want the area, that is always a non-negative quantity. So, you have to take the absolute value of that. So, now let us, for a small element, we have seen how the change looks like. So, if we put the small elements, area elements together, we have what is called the change of variable formula. So, it says, let u be an open set in R 2 and g be a map from u to R 2. So, that is, keep in mind that picture. So, g, so this is, so this is a domain and g is a map from a to a. We are assuming this is open. It is an open set. So, that you can have increments in all the directions. That is the basic idea. The g has got components g 1 and g 2. And we are assuming it is a 1, 1 function. This transformation is a 1, 1 function. Otherwise everything can collapse into one point possibly. The areas can get merged. If the g is a constant function, for example, then everything will get merged. g 1 is constant, g 2 is constant. So, we want the transformation should be such that from u to v to x, y, you are able to go and you are able to come back. Similarly, like in Cartesian coordinates, you are able to go to polar coordinates. And from polar coordinates, you can come back to Cartesian coordinates. So, it should be a 1, 1 map. Otherwise, there is no meaning of saying how does the things change. Both g 1 and g 2 are continuous partial derivatives. That is just to say that when you multiply by this kind of Jacobian, everything is continuous and you are able to integrate. So, that is the mathematical condition you have to put. The partial derivatives are continuous of both components g 1 and g 2. So, with these conditions, so this is what the picture looks like. A small element goes there. g is the map. So, what does the statement say? It says, if you want to integrate the function f over a domain d in x, y plane, x, y is the range. There, you are integrating a function. So, it says you make a change of variable. Call x 1 as g 1 of u, v. So, you transform the variable x to u and v by the transformation x is g 1 of u, v. y is g 2 of u, v. If you make that transformation, then this function will be a function of the variable u and v. So, we are saying how does the integral of f over the domain d in x, y plane looks like integrating this changed variable thing in the u, v plane. So, for that, first you have to transform the region d. How does the region d look like in the u, v plane? So, see if that is e. Then, you have to multiply small element d u d v by the Jacobian, absolute value of the Jacobian. That is what we saw. So, what it is saying is that this d x, y, the small element, the area element changes to Jacobian, absolute value of the Jacobian times d u d v. That is the magnification that is coming. In one variable, it comes g dash. In two variables, it is a Jacobian of that transformation that comes. So, one can think of Jacobian as the magnifying factor. The small area d x d y, that is getting magnified as Jacobian of u and v times the change area, times delta u delta v. We can extend it to domains where known overlapping and so on. I think the best thing is to look at some examples. We will come to polar coordinates also.