 Hi, I'm Zor. Welcome to a new Zor education. Today we will consider a little bit more complex case of double integrals. Now the previous lecture was about double integrals on rectangular base. In this lecture, I will present the circular base on which we will be integrating this particular surface. Now this surface represents some kind of a function of two arguments f of x, y and it's defined on this circle of radius r with the centers at origin of coordinates. And I would like to know the volume under this surface, which is bounded by the surface on the top. The x, y plane on the bottom and this circular surface on the sides. Now I obviously assume that the function is smooth enough and I will use exactly the same approach as with double integrals on rectangular base, which basically is the same as when I was talking about single integrals, when I was talking about area under the curve. Which means I will divide the base into small pieces. In this case small pieces would be obviously something like this. These lines would represent division by x and these lines would be representing division by y. Something like this. Now on each little rectangle, I will build a rectangular parallel pipette, which will go all the way up to the surface. Now obviously we have to consider the goods case when the surface is smooth enough and the sum of these volumes of parallel pipettes as I am shrinking all these intervals in which I divide my circle at the base. So the sum of these volumes will have a limit, which we will call basically the limit of this volume under the surface with the base circle. Now this volume does not depend again in smooth cases of the surface. It does not depend on how exactly we are dividing our axis and y's as long as the largest dimension of these rectangles goes down to zero. And obviously the number of these rectangles is going to infinity. I do realize that at the edges we will have a circular base, right? We are dividing it into rectangles, so obviously there will not be as smooth inscribing, if you wish, of these rectangles into a circle. But again, the theorem states that as we are diminishing the dimensions of these circles, there will be tighter and tighter inscribed into the circle and the perpendicular, which I am constructing, which form my parallel pipettes, will be again tighter and tighter inscribed into the surface. And at the limit, we will have the same limit no matter how we divide this thing. So our purpose right now to summarize now each particular volume, let's call it delta V i j. What is this particular volume? Well, it's obviously the it's a parallel pipette, right? So it will be the area of the base. Base is delta x i times delta y j, right? If i and j i are numbers of these divisions times the height and the height would be f of x i y j. So somewhere here, there is a point which is x i and i and y j. Now there is a square which is, which this particular point determines and that's how my individual parallel pipettes volume would be expressed. Now, what we have to do to calculate the sum, sum would be basically sum of delta V i j and somehow we have to summarize by a and by j, by i and by j. But this is exactly what's the difference between the circular base is and the rectangular base. In case of rectangular base what we could do is, now imagine this is a rectangle. We can actually do the summarization by x from zero to whatever the maximum m and for each of those have summarization by y again from zero to to n or we can do the other way around. We can do by y and then by x doesn't really matter. In this case the situation is a little bit more complex. We can obviously start our summation by one of the dimensions. Let's say by x. Let's call it primary, right? Now, so x is actually going okay, this is my x. So going from the beginning which is x zero to the maximum which is x m's where x zero corresponds to minus r, where r is the radius and x m's the last one corresponds to plus r. So we can do this, but however for each x wherever we chose x, let's say we chose this particular x. y is only within these boundaries. It's not from maximum to minimum or for minimum to maximum from zero to n. No, it's from some kind of number which is inside the circle to some kind of a number which is also inside the circle. So in case of rectangle, I will just go and replace this with integrals if you remember from a to b and from c to d, where a, b, c, d are my dimensions of the rectangle. This is by x and this is by y. And then I put f of x, y, dx, dy. Now in this case, we cannot, we cannot do this. In this case, we can say that our sum would be actually, well, let me call it, instead of approximation, I will put it converges to. It converges to different kind of integral. Integral by x would really be from minus r to r, right? By x, from minus r to r. But let me do it differently. I will put dx here and then I will open parenthesis and I will say exactly what I would like to integrate. Now I would like to integrate now by y, right? But y should not be counted from minus r to plus r. Let me just draw a circle if this is minus r and r by x. Now whenever my x is fixed, how y is changing? What is this particular? This is if this is x, this is r, then this is square root integral square root of r square minus x square with a minus sign and square root r square minus x square with a plus sign. And now I can have the function which is the height of this perpendicular, height of this parallel pipette and dy. So that's what my summation and the resulting integration should actually be. So the limits of my integration by y depend on value of the x. That's why I put it in in parenthesis. Now how can I take this integral? Well, very simply. I assume that x is a constant and take this integral, which is basically an integral of one variable. So I will take this antiderivative, indefinite integral and substitute by formula of Newton-Lagnenitz, the top and bottom, and I will get the result. And then the result will depend on x. So now I have the integration by x of the function of x and I can do basically this. So in theory, what people really do is they do not put these parenthesis. They put something like this integral from minus r to r, integral from minus r square minus x square to r square minus x square f of x y. And now what's very important is the order of differentials. First I put dy here and then dx. It means that this belongs to the first integration. And then when I have performed this integration, the second one would be only by x. So it's two single integrations. Basically, it's the same as in case of rectangular base. The only difference is that in case of rectangular, I have two constants here. But it still depends, but still inside integral will still depend on one of the parameters. If I integrate by y, my x will be a completely independent variable and then I integrate by x. Now in case my limits also depend on the same x, I will still have a function of x as a result of integration. So it doesn't really matter whether these are variables or constants as a result of the first integration, I will still get the function of x. And then I will integrate it the second time by x. So that's basically the whole story. Now why I specifically decided to choose a circular base? Well, for one very simple reason. This particular technique would also work in many other cases. What's very important is to understand that whenever I have my limits also dependent on the x in this particular case, on the primary variable, it does not have to be a circle. It can be anything. It can be ellipse. It can be anything else. So basically just as an example, it can be, for instance, it can be, let's say, a pyramid, in which case my x would be, if this is my x, this is my y, my x would be from zero to some maximum, whatever it is. But for every x integration by y would be along this line, which obviously depends on how far along the x-axis I have moved this point. So it's different. So integration by y will depend, it will be always from zero to something, and that something which would be dependent on what exactly my x is on this distance. So this depends on this. Integration by y depends on where exactly my x stops. It would be something similar to this. So this would be probably zero in this case, and this would be something which depends on x. And I'll probably do this maybe in the next lecture, just as an example. Now, in this lecture, I would like actually to do one example, relatively simple. So you would understand how it works. So let's assume that I have a cylinder, a real cylinder. So my function is a constant equal to h. So it's always h. It's defined on this circle from, with a radius r. Now, what should I have as a volume? Well, I should have the area of the base times high. That's what I have to have. Well, let's just do the integration and see if we have it, right? So integral from minus r to r, this is by x, integral from minus r square minus x square to r square minus x square of h, right? Function is equal to h constant, first dy and then dx. All right. Okay. So this integral, inner integral, well, this is a constant, right? So the result would be h y in the limits of minus square root of r square minus x square to r square minus x square equals to, we have to substitute into y. So it will be 2h square root of r square minus x square, right? So that's my result. From this, I subtract this, h is a multiplier. So I will have 2 square roots. Okay. So fine. Let me substitute inside of this thing. My second integration would be, now 2h I can obviously put outside of the integration and I will have this. Great. Well, there are many ways of doing some integrals like this. Probably the easiest way is to substitute x is equal to r sin t. Obviously, x cannot exceed r, so this is a legitimate substitution where t is from minus pi over 2 to pi over 2. That's how x would change from minus r to r, right? Sin of minus pi over 2 is minus 1 times r, it would be this. And, okay. So what is this integral in this case? 2 integral, that would be from minus pi over 2 to pi over 2, right? If t is equal to minus pi over 2, then x is equal to r and sin of minus pi over 2 minus 1, which is minus r. And this would be plus r, right? So that's fine. Now, what is square root of r square minus x square? Well, that's r square times 1 minus sin square, which is cosine square, which is r cosine t, right? Okay. So it's r cosine t. What is dx? dx is differential, which means it's a derivative. So r, derivative of sin is a cosine, so it's r square cosine square dt, right? Okay, fine. I got that. So it's this integral. How can I simplify it? I don't like this cosine square. Well, I will do this. Remember cosine of 2t is equal to, let me take another cosine 2t is equal to cosine square minus sin square, or cosine square minus 1 minus cosine, so it's 2 cosine square t minus 1, right? From which cosine square equals to 1 plus cosine 2t divided by 2, right? So that's what I'm going to do. I will substitute instead of cosine square, I will substitute this getting the following. 2 integral from minus pi over 2 to pi over 2. Well, r square can go outside. So instead of cosine square, I will put this. Now, this is divided by 2 and I have 2 here, so let me just get rid of the 2 and division by 2. And I will have only 1 plus cosine of 2t dt. That's what I have. Now, this is easier because this is just the sum of 2 integrals, right? So it's integral of 1, which is r square integral from minus pi over 2 to pi over 2 to 1 dt. That's one thing. Plus r square integral of minus pi over 2 to pi over 2 of cosine of 2t dt. Okay, so we will do it separately. Now, this one integral of this is r square times t, which is anti-derivative in the limits from minus pi over 2 to pi over 2. So it's pi over 2 minus minus pi over 2. So it's pi, right? So it's pi r square. That's what this thing is. By the way, somewhere along the line, I forgot this multiplier h, which was in the very beginning. Sorry about this, but it's always multiplied by h, so everything should be multiplied by h. This, this, this, everything. I wiped it out, but that was from the very beginning, obviously, right? Because the very first integral, if you remember, I had to multiply by h. All right, by height of perpendicular. All right, now, so I will have, this is my first integral, this one. Now, my second integral, my second integral, let's put it here, minus pi over 2 to pi over 2. What's the indefinite integral anti-derivative of cosine of 2t? Well, that's sine 1 half sine of 2t, right? From minus pi over 2 to pi over 2. Now, if I will substitute pi over 2 into the t, it would be pi, right? Sine of pi is 0, minus pi over 2 also 0. So it would be 0 minus 0, which is 0. So this thing is 0. And my result is this, which corresponds to my original geometric formula. Okay, basically, it works. Maybe in the next lecture I will do another example, when I promise I will not forget the multiplier. In any case, my purpose was to basically show that integration works and you will get the correct formula. Now, can I say that this is the proof of this formula? Well, no. I mean, this formula is obtained through also some kind of limiting procedure. In case of a cylinder, I remember we were just slicing it by horizontal planes and we also kind of went to some kind of limit, which is basically the same thing as integration. But when I was talking about geometry, I did not use the term integration. I was using, well, this is a sum and let's just go to the limit and then we had to prove somehow that the limit exists. And all this is actually a masked integration, so to speak. So basically, it all goes to the definition. And the most important theorem is that no matter how we break into small pieces, our basis to get the volume as a limit of the sum of volumes of parallel pippets, or in case of flat figures whenever we are talking about area under a curve, in all these cases we were dealing with integration, which is summation stretched to a limit, summation of infinitely small pieces, infinitely large number of times. And the existence of this limit and its independence on how we break our base. These are extremely important and not very easy theorems of higher mathematics. For now, I think you should really be just thinking that these theorems exist. They cannot, they can be proved. And there are certain restrictions on the surface as much as on the curve, whenever we were talking about areas, restrictions of smoothness. If these are smooth enough, for instance, for integration for area and for volume, the continuousness is a sufficient condition. Not necessary, but sufficient condition. So you don't really need differentiability of the surface or a curve, etc. to integrate. But all the functions which you are basically learning at school are usually continuous. And that includes, obviously, the functions of two arguments. And that's why we can really talk about integration as a valid procedure, because we all kind of have it in mind that there is a theorem which states that no matter how we break it, we will get it right. All right, thanks very much. But I would suggest you to try to derive the same formula for cylinder, whatever I just did, just by yourself in a clean sheet of paper. Do this integration completely. Do not forget the factor, which I have forgotten. And see if you get the right results. It's a very good exercise. And again, next lecture, maybe I will just come up with some other little example of that. Thanks a lot and good luck.