 Hi, I'm Zor. Welcome to a new Zor education. I would like to start a new section of this course. The section which is called double integrals. Well, I think this would be my last section in this advanced course of mathematics for teenagers. Usually the concept of double integrals is not covered in schools, but I think it's relatively simple concept. It's just one step further from regular definite integrals. So I decided that it probably makes sense, and it would be kind of a logical conclusion of this whole integral and calculus in general theme for the high school students. Alright, so if you remember when we were talking about definite integrals, we basically introduced the concept of area under the curve. So if our function is defined on a segment AB, then this area under the curve is basically an integral from A to B. Now, the way how I define this is basically as a sum of rectangles built under the function, under the function's graph, and the sum of these rectangles, sum of the areas of these rectangles. That actually approximates the area under the curve. Now, obviously we should have proven a theorem that regardless of how we divide our segment AB into smaller pieces, smaller and smaller, in the limit when the biggest of these pieces is shrinking down to zero, then the sum will have one concrete limit which is called an integral. So integral is basically a limit of so-called integral sums. Now, I would like to expand this picture to a three-dimensional case. Now, this is in a two-dimensional case, we have function of one argument and area under its graph. Now, in three-dimensional case, we have function of two arguments, function of X and Y, and it's defined for X between A and B and for Y between C and D, so within this rectangle. And this is basically the graph of my function, obviously not very artistically pleasant, but anyway, I'm sure you understand. So, this is the function and I would like to know the volume in this case. We are talking about three-dimensional case, right? So, we are talking about the volume under this surface. Now, not the curve, now we are talking about surface in three-dimension, right? So, on the top, it's limited by this surface which represents the three-dimensional graph of my function. F of X, Y. Now, on the bottom, it's X, Y plane. And on the left and the right, we have planes again bounding our area. This is plane X is equal to A, this is plane X is equal to B, this is plane equal to C and D. So, that's how my space is actually defined. So, I would like to know the volume of this figure. Now, obviously in the future it might be expanded to not a rectangular area where it's defined, but some other more complex thing, but let's just talk about this right now. That's the simplest case and that's how I'm going to introduce the double integral. All right, so, I will do exactly the same as I did for two-dimensional case and the function of one argument. I will divide my rectangle where my function of two arguments X and Y is defined into smaller rectangles like this. And on the top of each small rectangle as a base, I will build a rectangular parallelepiped, which means it's a flat on the top, it's flat on the bottom and it's flat on the top, but it touches the surface. So, it grows as much as to touch the surface. So, what will I get? Well, obviously the volume of each particular parallelepiped, I know, I mean that's the multiplication of three dimensions. And then I will add them up together and then there are obviously the same kind of a theorem similar to the two-dimensional case that whenever I'm shrinking the size of these small rectangles, so the biggest of them goes down to zero, I will have a certain limit and that limit would be the volume of this particular figure, object, whatever. So, now let's just try to do this. So, we will divide the segment from A to B into n different intervals, let me use the capital N, rather n. And from C to Yn would be my division of along the Y coordinates. So, these are x's and these are y's. And on their crossing I will have this subdivision of my area where the function is defined into small rectangles. So, let's talk about one particular rectangle. Now, I will use double indices, so rectangle, I know how to call it r, i, y, i, j. Now, this rectangle is on the x side, it would be from xi minus one to xi. And for on the y side, on the y it would be from y, j minus one to y, j. So, this is a rectangle, it's somewhere here where its right boundary is xi and bottom boundary is y, j. So, the area of this rectangle obviously is equal to xi minus xi minus one times i, j minus y, j minus one. So, that's the area of this rectangular parallel pipette which I'm going to build. Now, what is its height? Well, I was saying that I will grow this up until the point where it touches the surface. I think it would be a little bit better and it would be exactly the same result. If I will just take one of the corners of these small rectangles, the one which has the coordinates xi and y, j. And whatever the value of the function is in this particular corner, I will set as the height of this rectangle. So, my volume i, j would be equal to function of xi, y, j times xi minus x, i minus one minus one, y, j minus y, j minus one. So, that's my volume of one particular rectangular parallel pipette. Now, what do I have to do to approximate the whole volume? Well, I have to summarize this by i and by j. So, i is changing from one to m and j is changing from one to n. Now, let me just write it slightly differently. So, this is my volume. I will call it mn. So, it's not the real volume. It's a volume based on approximation of division by m on x and by n along the y-axis. Now, then if I will increase the m and n to infinity and simultaneously make these smaller and smaller, the theorem is, which I'm not going to prove, but again, I spent a lot of time in a two-dimensional case. It's basically going along the same line. So, the theorem is that no matter how you divide your rectangle into smaller rectangles by these points of division, as long as the maximum among xi minus x minus i one goes to zero and the maximum among these goes to zero, which incidentally means that m and n both should go to infinity. So, as long as this is done, under certain assumption about the function f, it should be smooth in some way. So, continual, differentiable, etc. I mean, there are certain functions which are only considered for these particular purposes. And in theory, for instance, the continuity is sufficient. But there are some weaker actual requirement, but we are not talking about this. Most of the functions you will be dealing with are relatively smooth. So, for all these smooth functions, the limit of that thing as this goes to zero and this goes to zero and m and n goes to infinity exists. And this limit will be the volume of this particular figure we are talking about. Now, let me just change slightly the notation here. I will use the following. It's sigma, sigma. Let's say first we sum by i and then we sum by j. Now, f of xi yj and I will put here delta xi and this is delta yj. I will use this notation. Why? Because it's a little bit more familiar. Because if you will consider yj as a constant, which is not really changing yet. And just consider this piece. What does it remind you? As delta xi goes to zero and m goes to infinity, this part must have the limit. Now, what is this limit? Let's just think about it. This limit is the following. So, you fixed yj, which is somewhere here, which means there is a yj and yj minus one. So, you fix this particular slice, if you wish, of this figure. And then you are making these x smaller and smaller and smaller. So, that's what it actually means. So, eventually you will get the volume of this slice only. If ij and ij minus one are fixed, then you have a slice of this figure. And to basically make a volume of this slice, you basically do whatever I just said. Now, from regular integrals, from definite integrals, you remember that this thing, when m goes to infinity and all delta x's are going to zero, that's the definition of the integral. From the beginning, which is a to the end of function f of x comma yj dx. That's what it is, right? So, if I will replace this with this, it will be a function of yj, right? So, I have now function sigma from j one to n dy, delta, first delta. Now, if this is considered to be a function of yj, so what is this? As n goes to infinity and delta y goes to zero, that is basically, again, an integral. In this case, it's by y, from c to g, g of y dy. Or, in our expanded notation, it would be integral from c to g is external integral, from a to b is internal, f of xy dx dy. And that's the definition of the integral, of double integral in this case, right? So, I didn't really do much above and beyond from the understanding standpoint, to whatever we were talking about when we introduced the definite integral for the function of one variable. So, this is kind of an equivalent for the function of two arguments, f and xy. Now, here is a very interesting consideration. What I am saying is, I really don't want to have these brackets. And here is why. Think about the logic how I came up with this one. I had a double sum, when first I fixed my y coordinate, and then I sum by x. At the same time, I could come up with the same volume by fixing x, coordinate from here to here, from xi minus 1 to xi, and make a slice of this figure, and calculate the volume of this slice. And then by actually making y smaller and smaller and smaller. And that actually means that I have reversed the summation. The summation in this case would be from i from 1 to m, then for j from 1 to n, f of xi j delta y j delta xi. So, I just change these two deltas, which doesn't really change anything at all. It's still double sum. Double sum can be calculated like columns and then rows, or rows and then columns. It's still the same sum of all the elements in this whole table, two-dimensional table of x's and y's. And now this looks like sigma from 1 to m integral f of xi comma y, that's yj, right? dy, that's delta. And now, again, this is some kind of a function of xi. And the whole thing actually, as a limit, when delta xi goes to infinity, it goes to dx and integral from a to b. By the way, this integral from c to d, by y. Which is exactly the same. So, that's why I can really get rid of these, because they are exactly the same. I change the order of integration. I don't need i anymore. I change the order of integration and I've got exactly the same result, which is the volume under this particular surface. So, my point right now is that I have introduced a new concept, which is called double integral. It's really analogous to definite integral for functions of one argument. But in this case, that's the function of two arguments. And obviously, the graph of this function is surface now. There is one complication if you will compare it with a one-dimensional case, with the function of one argument. You see, function of one argument can be defined only on a simple segment from a to b. Function of two arguments obviously can be defined on a rectangular domain, x from a to b and y from c to g. However, you can actually consider some other function, which is defined on a much more complex area here, which is not really, maybe it's a circle. Function is defined on a circle. Is it possible? Consider a sphere, for instance, a sphere which has a center at the origin of co-ordinate. It's actually defined on all those x's and y's, which are within circle, right? Because that's how sphere actually is defined. Outside of the circle, there are no values of my function. Well, let's talk about half a sphere, right? Because the whole sphere has top and bottom, but half a sphere is really some kind of a surface which can be represented as a formula. So, that would be a complication which I will address in the next lecture. But it might be really a slight complication. However, in case of rectangular area, which is actually domain of our function f of x, f of x, y, situation is really like this one, which means you can really exchange, because these are constants from A to B and from C to D. They define this rectangle where it's defined, where the function is defined. So, in this case, it's really kind of symmetrical. You can change the order of integration in case of, you know, more complicated domain you cannot. But we will talk about this separately. Okay, so far, this is basically the introduction, this is the overview of what is a double integral in the simplest case when domain is represented by a rectangle from A to B on x and from C to D on y. That's it for today. Thank you very much and good luck.