 Hi, I'm Zor. Welcome to Unisor Education. This lecture is part of the Advanced Mathematics course for high school students, which presented on website unisor.com. That's where I suggest you to watch this lecture from, because the site contains not only the video part, but also some notes. And in this case, notes are important because there are a lot of calculations here, which I will try to do without any errors. On the board, and you can definitely review it all in the notes. So this is about the volume of the pyramid. Now, the volume of the pyramid can be approached in many different cases. Unfortunately, none of these cases can be 100% rigors on the level of high school. However, I have made all the preparations for this lecture to be relatively rigorous, as much as possible. And the preparations were related to three-dimensional similarity concepts. And also there was a lecture in that particular topic related to two-dimensional area calculation as a limit. The calculation of the area of triangle, actually, as a limit. Because right now we are going to approach the volume of the pyramid in exactly the same fashion as I did for area of the triangle, as a limit of some other more elementary objects, which we can approximate our complex object, which is a pyramid. Now, pyramid is a complex object versus prism, which is a more elementary object. So the right prism is something which we have already studied before, and we know the volume of the right prism. So my idea is to explain how the volume of the pyramid can be approximated with some of the volumes of little prisms, which are basically combined together, make a shape close to pyramid. And the smaller the height of these prisms, the better the approximation is. So that's the whole approach. So let me just go on this particular road, and I'll try to explain the details. So, first of all, let's consider we have a triangular pyramid, and the base lies in some kind of a plane beta. Now, I will drop the perpendicular from the apex S down to the plane beta. Let's say it's point H. And now what I'm going to do is I will build an object which contains small prisms, one on the top of another, stacked one on the top of another, which together approximate the shape of this particular pyramid. Now, how can I do it? Well, let's divide SH, which is the height, into n equal parts, where n is some number. And the greater this number is, the more precise my approximation will be. And now, so let's say these are division points. So somewhere I will use H lowercase n as an index. Now, at each point of division, I draw a plane parallel to beta, which basically intersects my pyramid at triangle An, Bn, and Cn. So, at each division point, Hn from H1, which is here, to Hn minus 1, and then H itself can be treated as Hn with capital N. So I draw a plane and the plane intersects our pyramid at these points. I mean, it's obviously a triangle, very simple proof. Now, how can I build a prism? Now, the prism I will build based on this triangle An, Bn, and Cn, I will build the right prism, so I will go with the perpendicular here, and that will be my prism. Now, the height would be exactly to the next plane, to beta n minus 1. If this is beta n, that would be beta n minus 1. In between these two planes, I build a right prism with bottom base at points An, B, and Cn. Now, if I will do it at every level, my pyramid would be kind of converted into some kind of a stack of the prisms which make steps like Egyptian pyramid, for instance, or pyramids in Mexico, for instance. They have these steps, right? So I will approximate the volume of the pyramid as the sum of the volume of these prisms, and then if I will use the larger and larger N, which makes actually taking the limit as capital N goes to infinity, my sum of the volumes of these prisms will be the approximation of the volume of the pyramid. So, let's take a look at this particular prism, which is number N. What is its height and what is its base? Because that's all we need to calculate the volume of the prism. As we know, it's just multiplication of area of the base times the height. All right, the height is easy because it's the distance between these two planes, and these two planes are in the distance which is equal to H divided by capital N, right? Since I have divided the very beginning into N equal parts, this perpendicular. And it's obviously perpendicular to every plane because they're all parallel among themselves. So that's the height. How about the base? How about this triangle? Well, first of all, let me point out that this triangle, A and B and C and N, is similar to A, B, C, and the similarity, it's a 3G similarity, with a center of this particular scaling at point S at the apex of the pyramid. And the scaling factor is basically the scaling factor between SH and divided by SH, which is N divided by capital N. Because there are N small segments up to the plane called beta Ns, and there are capital N, these segments, to the beta. So SHN relates to SH as lowercase N to uppercase N. Now, obviously, the same scaling factor exists with SA and over SA. Why? Because triangles S, A, N, H, N, and SAH, these two triangles, are obviously similar in two-dimensional sense. Because these planes are parallel, so the plane SAH cuts two parallel lines. So these are parallel, obviously. And since they are parallel, then all the angles are equal. So obviously, they are similar. And since they are similar, the same proportionality between SHN and SH exists between SA, N, and SA. And the same as SBN and SCN, etc. So all these sides of these triangles are all proportional. And this is the coefficient of proportionality. Now, if you remember from the similarity lecture, if you have two flat figures, flat objects, which are similar to each other, and there is a scaling factor. Now, the area of these two flat objects is related as F squared, where F is a scaling factor. So all linear dimensions are scaled by the factor of F. But all area dimensions, like area of a triangle in this particular case, are related as F squared. So the area of triangle A, N, B, N, C, N divided by the area of triangle ABC as N squared over capital N squared. Because this is the factor, and this is the factor squared. So that's basically enough for us to determine the area of A, N, B, N, C, which is the base of the Nth prism. So let me just put here that the area of triangle ABC is lowercase S. Then I can say that the area of triangle A, N, B, N, C, N is equal to S times this factor. So I know the area of the base of this prism, and I know the height of this prism. So I know the value. So the volume of Nth prism is equal to product of area of the base, which is S N squared over N squared times H over N. Now, this is prism number N. Let me just put it slightly different. Why? Because S and H and capital N are constants in my case, and lowercase N is an index. Because what I have to do right now, I have to summarize. That would give me the expression for the sum of this summarized volume of the stacked prisms. And then, if I will have a formula for this, I can use some limit theory, put N to infinity, and that would give me the formula for my pyramid. So let's find out what this actually is, if I will summarize this. Well, obviously, constant can be moved out, and here I will have 1 plus 2 squared plus 3 squared plus etc. plus N squared. That's what I have to calculate. Now, this is algebra now. It's not geometry, so I don't really need this. So let me go to the algebraic aspect of this. How can I calculate this sum? Well, there is an easy way. You can just find out the formula. And if you want to prove it, you can prove it by induction, which is very, very easy. Now, if you don't know the formula, which I obviously don't remember, I will try to derive it, but I do know the methodology. Here is the methodology. Let's go and put 1 cubed plus 2 cubed plus etc. plus N cubed. That's equal to 2 cubed plus 3 cubed plus etc. plus N cubed plus N plus 1 cubed plus 1 minus N plus 1 cubed. Now, this is my sigma N squared from 1 to N. This is sigma of N plus 1 cubed. And then plus 1 minus N plus 1 cubed. Now, why did I do it? Here is why. Let me open it up. N cubed plus 3 N squared plus 3 N plus 1 plus 1 minus N plus 1 plus sigma N cubed. I will not use from 1 to N, it's obvious. Plus 3 sigma N squared plus 3 sigma N plus N actually, right? Because I'm summarizing 1 N times plus 1 and minus N plus 1 cubed. Let me put it this way. Now, why did I do this? Well, I did it for a very, very simple reason. Because now, you see, this can be reduced. And I have basically an equation for this thing. So it's 0 equals 3 sigma N squared plus 3 sigma N actually, we know what it is. I did it when I was calculating the area of a triangle. It's sum of arithmetic progression from 1 to N and obviously it's N times N plus 1 divided by 2. Plus N plus 1 plus minus N plus 1 cubed. So now we have equation for this guy. And we can determine this sum of N squared and this is basically the formula. So I can just simplify it as much as possible. So let me multiply everything by 2. So I will have 0 equals 6 sigma N squared plus 3 N N plus 1 plus 2 N plus 1 minus 2 N plus 1 cubed, right? So 6 sigma N squared equals 2 N plus 1 cubed minus 3 N N plus 1 and minus 2 N plus 1 equals... I can factor out N plus 1 and here I will have 2 N plus 1 squared minus 3 N minus 2, okay? Which is N plus 1. And let me open this up. 2 N squared plus 4 N plus 2 minus 3 N minus 2 N plus 1 2 N squared plus N equals N plus 1 N and 2 N plus 1. So that's 6 sigma. So 1 sigma of N squared from 1 to N equals to N N plus 1 2 N plus 1 divided by 6. That's my final formula for sum of the squares of the integer numbers from 1 to N. And I can use it here. So this equals to SH divided by N cubed times N N plus 1. Sorry, it was 2 N, 2 N cubed. My mistake. 2 N cubed. Is that right? No, this is 1, but I lost 2 somewhere. SH divided by N. Okay, let me finish this divided by 6. I don't know, everything is fine. I don't know, everything is fine. Right. Because this goes to square, right? So it's equal to... Let me multiply this. SH divided by N squared divided by 6. And here I will have 2 N squared plus 2 N plus N plus 3 N plus 1 and 1 1. That's my formula. Equals. Alright. So it's SH. I will divide by N squared 1 by 1 divided by 6. 2 N squared, this will be 2. 3 N plus 3 divided by N. This is N, this is N squared and 1 over N squared. That's it. Now I'm ready to increase N to infinity. What happens with this formula? Well, obviously these two members will disappear and I will have only this one. So my total volume would be SH divided by 3. And this is the final formula for volume of the pyramid. We kind of assumed that this particular process of constructing this step-like figure is approximating my pyramid and the volume approximately is equal to the volume of the pyramid. And as this capital N goes to infinity, it's kind of intuitively obvious, although it's not rigorously proven, that the volume of this figure would go to the volume of the pyramid. So there are certain assumptions which I allowed myself just to relate to your intuitive feelings without actually proof, and this is one of those. In any case, this assumption seems to be reasonable and in higher levels of mathematics, this assumption is actually proven that this is really true. I just don't want to spend right now much time because, again, my purpose was not to derive the formula per se, but to explain the direction, the methodology, how this formula can be derived. So you basically approximate a complex figure with something simpler which you already know how to deal with, in this case, prisms. So I would suggest you to go again through the notes of this lecture. Again, it's very educational, I believe, and what's important is, if you can, try to do whatever I just did yourself, just on a piece of paper. Try to derive whatever the formula is because there is nothing which I'm using here in this particular lecture which I had to remember something. Well, maybe accept formula for rhythmic progression. But anyway, it's just, again, the methodology is important and I do encourage you to try to go through all these calculations and to get similar results. That would be a very good exercise. Well, other than that, that's it for today. Thank you very much and good luck.