 Hi, I'm Zor. Welcome to Inezor Education. Well, today is a very special day, and there are two major reasons which kind of coincide. The one is, today is a total solar eclipse, which is a very remarkable event by itself, obviously, and also to commemorate that particular event. This is my last lecture of the course of Advanced Maths for Teenagers and High School Students. Most likely I will add to the course certain exams where they don't exist and maybe some other little details. But generally speaking, I consider the theoretical material presented in the course to be relatively complete and my plans are to go to start Physics 14. That's another advanced course of physics for high school students, primarily, and teenagers. Alright, so back to business. This last topic which we have discussed before is about double integrals, double integrals, and I have exemplified this particular topic in the previous lecture with the volume of the cylindrical cylinder, basically. Now, in this particular lecture, I would like to apply exactly the same kind of technique to have calculated the volume of the pyramid. Again, we know the volume of the pyramid from the geometry course, so we will just check if we will get exactly the same results using integration. Now, I have to add actually that the volume of the pyramid, as it was explained, presented in the course of geometry was, how should I say it, it was a sneaky way to introduce integration, basically. Because I remember where we were talking about this, we sliced the parallel planes, the pyramid, and basically went to a limit whenever the difference between the different planes is going down to zero, the distance between them. So it's kind of a, I would say, sneaky way to introduce the same concepts which integration puts on a solid mathematical foundation. So I'm going to do straight integration in this particular case using, as I was saying, the same approach as in the case of the cylinder. So again, what's complicated in this particular case, that our base is not rectangular. So whenever we are dividing it, we are supposed to choose one particular dimension, let's say x as a primary, and have x actually varying from the leftmost to the rightmost. But for each particular x, my y has certain limits which depend on the x, and here is y. Now, this is the pyramid, which is basically inscribed into the coordinate system. So I just chose three points, a, b, and c, on x's, x, y, and z. And the lengths from the origin, all point a is lowercase a, point b lowercase b, and point c is lowercase c. So we have to calculate the volume of this pyramid, and as we know, it's one-third of the area of the base times height, right? So area of the base is, it's a right triangle, let me just write it down here. So this is x, this is y, this is my point a, this is my point b, this is a, and this is b, and this is my triangle. So this is my base. So the area of the base is a times b divided by 2, and then we should multiply by height. Since all these angles are right angles, obviously, the height is c. So my total, the volume will be one-third of one-half of a, b, c. So the volume is equal to one-sixth a, b, c. This is six. All right. So that's what we have to get. Now, how are we going to integrate? Well, the same way as before, we will divide our base into small rectangles. Obviously, not exactly, because this is a triangle. So the edges are obviously not coinciding with the edge, with the size of rectangles. However, as we are increasing the number of these dividing lines and decreasing the dimension of the largest of them, obviously, the sum of these rectangles will be closer and closer to the area of the base, right triangle. And whenever we are on each particular rectangle, we will build a parallel pipette to the intersection with the plane going through a, b, c. Obviously, the sum of these parallel pipettes will be closer and closer to the sum, to the volume of the pyramid. So let's just calculate what exactly are the volumes of a particular rectangle, which is based on x and y. Here is x and here is y. And we will integrate it. So what's the process of integration? Well, we have to summarize from 0 to a, right, by x from 0 to a. This is all. So my integration by x will be from 0 to a. Now, for each particular x, what is the y? It depends on where exactly I am intersecting my perpendicular to the x-axis with this line. Now, obviously, what I have to do, I have to find out, to find out y, I have to basically find out the equation which describes this line. Now, equation which describes line which intersects x and y axis at points a and b is very simple. It's x divided by a plus y divided by b is equal to 1. Now, let me just explain why. Because if x is equal to a and y is equal to 0, so this point, point a, point a has coordinates a, 0, right? If you will substitute a, 0, x is equal to a and y is equal to 0. This would be 1 and this would be 0. So 1 is equal to 1. So the point satisfies. Now, b, b has coordinates 0, b, right? 0 by x and b by y. So if I will substitute coordinates 0, b as x and y, this would be 0 and this would be b divided by b1. So, again, corresponds. So both points a and b have coordinates which satisfy this equation. Well, two points determine the line on the plane. So this is a correct representation. Now, from this correct representation, I can find very easily y. So what is y? It's b times 1 minus x divided by a, right? So what my point right now is that I have to integrate by x from 0 to a, but for integration by y, I have to integrate only until from the point x, I integrate only to the point where it intersects this line. So from this point to this point, which means I have to integrate from 0 to 1, no, b times 1, x divided by a. So that would be my integration by y. Okay, great. Now, what exactly I'm integrating? Well, I'm integrating, it would be dx times dy, right? That's my base times the height. What's the height? Well, we also have to determine the equation which determines the plane, which connects these three points, right? And it's very similar to this one because if you will write an equation x divided by a plus y divided by b plus z divided by c equals 1, you will see with exactly the same clarity that all three points, now, this one has coordinates 0, 0, a. This one has coordinates 0, b, 0, and this one 0, 0, c, right? So if you will substitute any of these three points, let's take points c, for instance. So the x is equal to 0, y is equal to 0, and z is equal to c. 0, 0, that will be 1. So point satisfies. Similarly with b, in this case, this will be 1 and this will be 0, etc. So this is an equation of the plane. And if you define x and y somewhere in the base, then the height of the, how is it called? Well, no, it's not a prison. It's parallel to Pippet. Yes, forgot the word. So the height of this parallel to Pippet would be exactly determined by this particular equation. So what is this for z? That's the height. Well, it's c times 1 minus x divided by a minus y divided by b. So that's my function, which describes the height. So what is the element of the value? It's dx, this little piece, dy, this little piece, that will be my base of this parallel to Pippet, times height, which is this. So dx and dy would be here. So this would be dx, this would be dy, and here I will have c times 1 minus xa minus yb. So that's an integral, which I'm supposed to calculate, double integral obviously, which I'm supposed to calculate, which will give me the answer to the question about the volume of the pyramid. All right, let's take this integral. So how do we do it? Well, first of all, we have to take the inner integral, considering x is a constant. So it's only dependent on y. So it's basically a linear function of y. So this integral would be equal to, so let's preserve the outer integral, and inside I will have, well, in this particular case I can do it the following way. I can break it into integrals, from 0 to b times 1 minus xa of c times 1 minus xa, because this is the constant, right? c times 1 minus x dy. Plus another integral, well, actually minus, minus, this minus. So minus the same limits of integration of cy divided by b, dy. Equals. Okay, now again, this is retained outside. Now the first integral, now this is a constant, because x is a constant. So it goes outside of the integration, and I have integral from 0 to this of basically function which is equal to 1. And as we know, the M2i derivative is y in this case, which should be substituted using the Newton-Libnitz formula, the top minus bottom. Now the top is, so c times 1 minus x divided by a is a multiplier goes outside. Now inside would be y in the limits from this to this. Well, this is 0 anyway, so just this one remains, so it's b 1 minus x divided by a. So that's the first integral. Now the second integral minus cb, that's a factor. Integral of y is y squared divided by 2, right? Because the derivative of y squared divided by 2 is y. So I have to substitute again, top and bottom, bottom is 0, so it disappears. So the only thing which remains is top, which is b squared 1 minus x over a squared divided by 2. So that's my single now integral, which depends only on x. Okay, let me just wipe this out. I hope I did not make any arithmetic mistake. And if I didn't, and the result should be this. Well, let's see. Now this is just a plain quadratic polynomial. I think it would be easier if I would substitute t is equal to 1 minus x divided by a. You see here, here, and here it's all the same, right? Which means x is equal to 1 minus t a 1 minus t, right? Which means dx is equal to the derivative of this, which is minus a times dt, right? Am I right? Right. Now, if x is equal to 0, t is equal to 1, right? And if x is equal to a, t is equal to 0, right? From here. If x is equal to 1, I have 0. I mean, if x is equal to a, I have 0. If x is equal to 0, I have 1. So my limits of integration now will be, bottom limit is x is equal to 0, so it's from 1 to the top limit, which is 0. Inside, I will have c times b times t square, right? t square minus c divided by 2, also b square and b, so only 1b remains, and t square again. And instead of dt, I have to put minus a, I mean, instead of dx, I have to put minus dt. Okay, so how should I take this integral? Okay, now you see it's much easier, it's much more convenient thing. Well, first of all, I should put plus here equals, I'll change this minus to plus, and I will change the limits of integration. As you remember, from the course of integration, exchanging the limits will cause negative sign. So it's a times b and c. By the way, this minus this, this is one half of this, right? So what's remained is a bc divided by 2 integral from 1 to 0 of t square dt. Am I right? Now this minus this is one half of bc times a, and the t square goes from 0 to 1 because I've changed the signs. Now the derivative of t cube divided by 3, a, b, c divided by 2, so it's t cube divided by 3 from 0 to 1, right? That's what it is. The derivative of t cube by 3 is t square. So if you will substitute, so this is my application of Newton-Leibnitz formula. And as a result, I have this formula, which is if I will substitute 1, I will have a, b, c divided by 6, right? And 0 will give me 0. So this is the total result of the integration, which corresponds to this one from geometry. Now it's exactly the same type of process of dividing my three-dimensional figure into small pieces where we know how to calculate the volume of, because this is the parallel pippets. All parallel pippets with dx times dy square base, and the height determined by the formula for z, which we have done earlier. So again, the integration is actually dividing your three-dimensional piece into infinite number of infinitely small pieces, so to speak. As long as you know how to calculate each piece, then you integrate by x and by y in this particular case to get the final result. And as we see, the final result is the same. Now, can I say that this is a proof of this formula? Well, in some way, yes. Because whenever we derived the same formula in the course of geometry, we were really did exactly the same thing. We were dividing with planes, the whole height, and it's exactly the same. We basically calculated the volume of each piece, and then we summed up the result, and we went to a limit, which is actually the derivation of everything, which I have basically assumed known, and which basically is the foundation of the Newton-Leibniz formula whenever you're integrating. So, yes, it can be considered in some way as just more rigorous proof of the same formula. Okay, so that's it for today. As I was saying, this is, at least I think, this is my last lecture of the course of mathematics for teenagers. And well, I don't know when exactly I will start the physics, but I think it's very important. I have to really tell you that the course of physics will be really very much oriented on the mathematics which we have learned here. It would be a theoretical physics, if you wish, but only on a classical level. And I invite you, obviously, to go to these lectures as they are, as they will come into the Unison.com. Everything will be there. I have already prepared the contents. And gradually, I will start filling up with lectures. So, thanks very much for your patience, your participation, and good luck in your endeavors. Thank you.