Speaker A
Whenever we want to integrate a function over a region, we need to decide what coordinate system we should use. For instance, just a rectangle is really easy to compute in Cartesian or rectangular coordinates. Well, a circle is very easy to compute in polar coordinates but not so easy in rectangular coordinates. However, we can move on to any coordinate system that we wish to define, and if that coordinate system makes it easy to express the region, then it might well be useful using that coordinate system. Consider, for example, this region. There are two horizontal lines, y = 1 and y = 2, and that's sort of the same kind of thing we've seen in Cartesian where you have to specify different y values. But if I look at the two red lines, the line y = x - 1 and x - 2, there's a certain symmetry on the two different sides here, but it's not vertical line symmetry as is often represented in Cartesian coordinates. So what if I do a change of variables? That is, what if I introduce new variables u and v? For v, it's just going to be equal to the y values, and I said I was happy with those horizontal lines; those were perfectly fine. But for the two diagonal lines that are not easily expressed by x equal to a fixed value, if I set u equal to x - y, then the one line is just u equal to 1 and the other line is u equal to 2. These are just constants. Indeed, if I tried to draw the exact same region but now in the UV plane, that is expressing it in terms of u and v as opposed to x and y, then what I get is just a rectangle. It's just the rectangle where the v values are between 1 and 2 and the u values are between 1 and 2. So my plan is that this is going to be way easier to integrate in this coordinate system because my limits of integration are going to be 1 and 2 for u and 1 and 2 for v. So how do I actually integrate in some new generic coordinate system? Let's think back to what we used to do for single variable u-substitution. The idea was the following: suppose I have an integral of some function of x dx. Then what I can do is take x and write it as some function of a new variable, a new variable u. And then what the u-substitution did was it allows us to rewrite this new integral as an integral in terms of u. The function of x is now turned into f of g of u because x is defined as g of u, and the dx transforms into a du. But there's an important point to note there: there is also this new factor, this scaling factor, this g prime of u. And that was u-substitution. I'll note that this formula may seem exactly backwards to how you recall it in single variable calculus. Typically, in single variable calculus, you would actually set u to be a function of x as opposed to x being a function of u. It doesn't matter which way you write it, which you think of as sort of the new variable and the old variable, but I'm going to do it this way because I'm about to have a u and a v, and I want to think of the new v as converting the messy xy into a nice u and v coordinate system. So I like this order of presenting it. Nevertheless, let's go and look at what it's going to be in multivariable calculus. Well, you begin sort of much the same way. You have an integral in terms of x and y, and it's integrated over some particular region. And then you say I need to take the x and write that in terms of u and v, some function g of u and v, and I'm going to take the y and write that in terms of u and v, some function h of u and v. Then if I plug this all in, I want to get an integral with respect to u and v. Well, the function evaluates now in terms of u and v. The limits of integration are no longer the region r; now we call it the region g. It's what happens when you transform under these equations. And then the dx dy transforms into a du dv. But just like in the single variable case, you have to have a multiplicative factor. We're going to call this multiplicative factor no longer g prime; we call it j of u and v. Here j is called j for the Jacobian. At the end of the video, I'm going to talk about the intuition of how to come up with this j of u and v, but for now, I'm just going to state it. J of u and v, the Jacobian, is just the determinant of a particular matrix, a matrix of partial derivatives. And the idea here is that this represents the scaling factor from when you shift your areas, thought of in terms of the xy coordinate system, to being areas in the uv coordinate system. You need to get a scaling factor, and this j gives you that scaling factor. All right, so let's look at the example that we had before where our u was x - y and our v was just equal to y. And then if I take that exact same region, I now represent it in the uv coordinate system, it looks a lot nicer. The first thing I want to do is try to write it in the way that's compatible with my formulas where I had x as a function of u and v and y as a function of u and v. So I'm just going to take these formulas; I'm just going to rewrite them in that way. And if I do that, of course y is equal to v; it's the same thing as v equal to y. I plug that in, and I get x is equal to u + v. All right, so let's compute that Jacobian, this determinant. Well, most of these partials are straightforward, which the partial of x to u, well, it's just one, and so on. So I can compute what these are going to be. It's just the determinant of the matrix 1 1 0 1, and that's just equal to the value of one. So in this case, the Jacobian is just multiplied by one. So here then is my final formula: the double integral over the original region, the sort of parallelogram one, if I take that region and integrate some function of x and y over, it's the same thing as the much simpler region, the region in the uv coordinate system where the u integral just goes between 1 and 2 and the v integral just goes between 1 and 2. The function, whatever it is, I haven't told you, would be represented now in terms of u and v. You multiply by that one for the Jacobian, and then you get your result. Have you integrated with respect to u and v? You might want to do this transformation of variables for multiple reasons. It might be that your region is really messy, and you can transform it into a much nicer region with the substitution, or it might be that your integrand is really messy and you don't know how to integrate that. And then with the substitution, you can now integrate the integrand. So there's multiple reasons why you might wish to do a change of variables. The final thing I want to talk about in this video is to motivate the j of u and v, the Jacobian. Why was that the case? So I want to imagine that I'm in the uv coordinate system, and I'm going to transform a little region, a little area in the uv coordinate system into the xy coordinate system. That is, imagine I just have a little rectangle. This is a rectangle, and it can be described by having two different vectors. One vector is a little change in u and no change in v, and the other one consists of no change in u and a little change in v. When I integrate, of course, I take a region. I think of breaking it up into a bunch of little regions. So in the uv coordinate system, these little regions are simple; they're just little rectangles. But let's see what this little region looks like all the way back in the xy coordinate system. Indeed, this green arrow goes somewhere, the red arrow goes somewhere, and whatever you get defines a parallelogram. As in, the original square defined by these two vectors transforms into a parallelogram defined by wherever those two vectors end up. But where do they end up? Well, if you think of the green vector as the result of just first a small nudge in u and then you transform it into x and y, then the result you get, well, what change in x do you get? It's the resulting change from nudging the u a little bit. In other words, what you get is a change in x with respect to u times that little distance du. And likewise for y, it's the partial of y with respect to u times du. So these are the little nudges in the x and the y direction that result from nudging u a little bit. Then likewise for the red vector, you get a similar kind of...
