 Hi, I'm Zor. Welcome to Unisor Education. Let's talk about spheres, primarily about surface area and the volume of the sphere. This lecture is part of the course for high school students in advanced mathematics. It's presented on Unisor.com. And not only the video you can get if you go to this particular site, but also notes for each lecture. And sometimes you have tests, exams, etc. So I recommend you to watch this lecture from that site and use it as much as you can. Okay, spheres, area and volume of the sphere. All right. Well, first of all, we probably have to just spend a couple of seconds to talk about what is a sphere. It has been defined in one of the previous lectures where I introduced all the solid geometry objects. So it's basically all points in the space, in three-dimensional space, which are equidistant from a center at fixed point. So now we have to talk about volume and surface area of the sphere. Well, first of all, sphere is defined by one and only one parameter, which is radius. So volume and surface area will be defined in terms of radius. So we need some formulas, basically. So that's what I'm going to do. I'll grab a couple of formulas for volume and the surface area of the sphere. And I will start with volume. When I will explain whatever I want to talk about as far as the surface area is concerned, you will understand why I decided to have the volume first. All right. Now, it would be easier for me to evaluate the volume of half a sphere. Let's say this one. So we have certain plane which goes through the center of the sphere. It divides it basically in half. And I will evaluate the volume of this upper half. And then I'll multiply it by two like this. Now, to do that, I have to, unfortunately, resort to certain intuitive understanding of certain things. I mean, there is some limit theory involved in this. And I would probably not like to do it in all rigorousness. However, whatever I will talk about will be relatively well understood intuitively. And that's pretty sufficient for this particular lecture. Now, how can I evaluate the volume of this half a sphere? Well, I will have a stack of very short cylinders which fill up this particular thing. So the way how I do it is the following. I will divide this perpendicular from the center perpendicular to this plane into n equal parts. And basically, I will draw parallel lines here, like here. They will divide the area into slices. Alright, and then I will build a cylinder from each upper part down. So it will be like this, something like this. So it's cylinders which are sitting on the top of each other. They have the height equals to, obviously, r divided by n, right? And as you understand, I will then increase n to infinity and the sum of the volumes of these cylinders is supposed to go to the volume of the sphere, half a sphere in this case, because the cylinders will be narrower and narrower, and that's why it will be more and more tightly filling my half a sphere. So what I'm going to do is I'm going to evaluate first k's, number k, cylinder here, and then I will summarize it by k from 1 to n. And then I will increase n to infinity and take the limit. So that's the plan, alright? Alright, so the case cylinder. Let's call this point a 1, a 2, a 3, etc. So a k's is somewhere here. Well, first of all, why am I saying that these are cylinders? Well, it's actually very easy. If you draw a plane parallel to the base of this half a sphere, you obviously will get in a section a circle. Now, y, well, that's kind of simple. So if this is my circle, main circle, well, this is invisible, and I have a perpendicular and draw another plane parallel to my base plane here. Now, why is this a circle? Well, if you take any two points, let's say this point and this point, or any other point actually. Now, these distances are supposed to be the same because if you consider o, a, b, c, d, if you consider triangle o, a, b, for instance, or o, a, c, or o, a, d, all these triangles are right triangles because this plane is parallel to this one and this is the perpendicular to both of them, right? Now, also o, b and o, c and o, d, since b, c and d are on this sphere, they're all radiuses of the same sphere. So they're all equal to each other. So o, a, it's a catheters which they all share. That's why a, b and a, c and a, g are all equal to each other, so this is a circle. So when I draw these planes parallel to the base plane, in this section I always have circles. And whenever, if I have two circles parallel to each other, and then another, oops, we take another. If I have two circles parallel to each other and I draw basically perpendicular lines through each of them, I obviously have a cylinder because this is a cylindrical surface with this as a director's, right? And you have two parallel planes which are bounding this particular cylindrical surface. All right, so these are cylinders. We know their height, which is radius divided by n. And what is the radius of a circle? The radius of a circle which is somewhere, this is a point, a k, for instance, a k. So what's the radius of this cylinder which is formed by the plane on the k's place? Well, that's very easy, actually. Because the distance from o to ak, oak, is equal to what? The whole thing is r, and this is k out of r. So it's k divided by n radius, right? Now, the radius of the circle is the distance to this point. So you can always consider the right triangles. Now, the right triangle would have this as a quadratus, this is the quadratus, and this is hypotenuse which is equal to r, the radius, right? Because this point is on the surface of a circle. So we have that lower case r square of this particular circle formed by the plane going through ak is equal to r square minus this k divided by nr square. So that's the radius square of the k's cylinder. And this is the height. All cylinders have the same height. So what is my volume of this particular cylinder? Volume vk's of the k's cylinder is equal to the area of the base. Now the base is a circle of this radius. So it's pi r square times height equals 2. Now we can substitute for r square, we can substitute this. And we can have pi r square minus k square n square r square times r divided by n. So this is my volume of the k's cylinder. Well, let me just do something. Pi r to the third divided by n, that's this one, minus pi r to the third k square n cube. Second and then cube, right? I just opened the parenthesis. Now what I have to do now is I have to summarize it. That's the volume of half a sphere. Now if I summarize it by k, now this doesn't depend on k. So if I summarize this, it would be n times whatever p r cube divided by n, which is pi r cube. Now this would be minus pi r cube n cube sigma k square of 1 to n. So this is something which I, the only thing which I have to know. Now actually I did already summarization of all integer numbers from 1 to n and I have a formula ready for this. In the topic of volumes, volume of the pyramids, I actually in quite some details explain how this can be summarized. Now for those who know the formula, obviously the easiest way is to prove it by induction. Well in any case since number one I know the formula number two is also derived in a relatively logical way in another chapter. I'll just use the formula I wrote for myself here. So the formula would be pi r cube minus pi r cube divided by n cube. And here I will put the formula which I wrote here. It's n times n plus 1 times 2 n plus 1 divided by 6. That's what it is. Now okay so this is the formula for some of our volumes of the cylinders. Now if n increases to infinity, if number of these cylinders increases to infinity and they are tighter and tighter inscribed into half a sphere, we intuitively feel that the volume, this sum is supposed to tend to a limit which is the volume of the half a sphere. Now what is the limit of this? Well think about it this way. Basically I would like to consider this part which is only one which depends on n when n goes to infinity. If I will multiply these, what will I get? I will get 2 n cube plus something related to n square plus something related to n and plus something free member. This will be a polynomial of the third degree, right? 1, 2, 3 n's. And the coefficient at n cube will be 2. Now here is also n cube and there is a 6 here. So if I will divide this into this, what will be? Well first of all I will have 1 third, right? Plus. Now this is n square and this is n cube which means it will be something divided by n and then something else divided by n square and something else divided by n cube. But these are as n goes to infinity they all go to zero. And 1 third is the only thing which is left, right? 2, 6, 1, 3. So this piece will go to 1 third as n goes to infinity. And what I have left is I have pi r cube minus 1 third of pi r cube which is 2 pi r cube divided by n divided by 3 and this is half a sphere and now I have to multiply it by 2 to get the sphere which is 4 pi r cube divided by 3. This is the formula for the volume of the sphere, the entire sphere. Well that's it, formula is ready. The volume is this and we can use it wherever we need it. Now why did I start with volume and not with a surface area? Here is why. Now I will use the formula for volume to derive the formula for surface area because if I will start from the surface area and let's say I will use exactly the same technique. I will inscribe these cylinders and I will try to evaluate their side surface. Well that's not an easy thing because if you remember the radius, radius squared actually was equal to r squared minus k over nr squared, right? That's the radius of the k's cylinder, or k we can use, right? Now so if I have a cylinder, this is the radius, now my surface is equal to what? The circumference of the circle times height, right? If I will cut my cylinder and roll it out on a flat plate, I will have a rectangle, right? If I will cut it here. I will have a rectangle and lower side would be equal to a circumference which is 2 pi rk. Doesn't look like k, this is k. And the height would be equal to r divided by n, right? Because this cylinder is 1 nth in height than the radius of the sphere. Now but this thing it contains this square root of this expression which doesn't look right. And to deal with this square root would be kind of difficult I would say, right? So you have to summarize certain things for different cylinders, for different k. So it's not easy. I mean just purely technically I have decided that I don't want it to go this way. Whatever I suggest right now would be much easier if you already have the volume. And by the way, in case of a volume I need the square of the radius of each cylinder, right? Because it's the area of the base of the cylinder which is important, which is pi rk squared. Okay. So now let's go back to my sphere. Okay. This is my sphere. This is the center. Consider the following. Choose a few points on the sphere. Let's say here, that looks more like a sphere, right? So you choose this point for instance, this point, this point, and this point or whatever. And you connect, for each point you connect a couple of nearest ones. Let's say this would be connected to this and to this. And this would be connected to this and this to this. And then each of these points you connect to a center. Now what happens in this case? Well, if you imagine if this is a sphere and you have let's say three points here, let's put it closer and connect it to a sphere. To a center and in between. What is this? It's a pyramid, right? So you put another point here, another point here. It's all pyramids here. So you can actually fill your sphere with pyramids. In a very analogous situation in two-dimensional space it's visible much better. If you for instance take these few points on a circle and you connect them and connect with the center you are dividing a circle into triangles which are inscribed into this particular circle. Now in as much as the area of the circle can be approximated with the sum of these triangles and then the more points you put in between the better the approximation. The volume of the sphere can be approximated as the sum of the volumes of these pyramids, right? Now what happens if I will put my points denser and denser so they will put more points and the distance between points will be smaller and smaller then these pyramids will be with their bases will be closer and closer to the surface of the sphere, right? So now let's talk about the volume. Volume of each pyramid is what? Volume of one-third area of the base times height, right? Now what is the height of these pyramids? Well, the smaller they are the closer height becomes to the radius of the circle, right? Same thing is here. The smaller these triangles are the altitude of these triangles will be closer and closer to the radius. Same thing with the pyramid here. So this thing will go to the radius as the number of points is increasing infinitely and the distance between these points is also decreasing to zero. Now, what is the area of... Now, I will summarize that thing, right? So it will be sum of all these but this will be very close to R which means that when I'm summing I can take R out of... if I will do sigma ii, okay? i is the number of the pyramid number and summing is by i. So this will be very close to R so it's approximately... so it's approximately R times sigma one-third si. Well, one s also can be outside of the sigma. So one-third goes here and this is sigma by i. Now, what is sigma of s i where s i is the area of the base? Well, it's very closely... it will be closer and closer to the total surface of the area, right? Because we are covering... with bases of these pyramids we are covering an entire sphere. So basically what I would like to say is that the volume of the sphere would be equal to one-third R and area of the surface of the sphere. That's my kind of, again, half-intuitive, half-obvious consideration. And now, since I know the volume I can very easily determine the area, right? So s is equal to 3 v divided by R equals to 3 4 pi R cubed divided by 3 divided by R equals to 4 pi R squared. So that's the total area, the surface area of the sphere. Well, I think, again, it's better understood all this thing if you start from the volume and consider this equivalency between the surface area and the volume of the sphere of radius R. I think it's a very important consideration which goes, again, not very rigorously proven by whatever I was just talking about, but, again, intuitively very obvious. It's much obvious on the plane where you have a circle and you have different polygons inscribed and the sum of all these triangles is basically almost equal to the area of the circle and the better approximation would be the better if the number of edges goes to infinity and each edge is getting smaller and smaller. So this is just a three-dimensional equivalent of that. Instead of triangles, we are talking about little pyramids here, triangular pyramids. Okay, that's it. I do suggest you to read the notes for this lecture. I think it's very interesting to read after you have listened to whatever I was just talking about, and I don't think you should be really upset for the better word about non-strictly, rigorously defined certain things which I was just talking about. I did use certain considerations which seem to be intuitively obvious and don't be afraid to use your intuition. Your intuition should actually be developed because whenever you're trying to work on something new, it's not the calculations which drive you forward. It's your intuition first. First you have to really have a concept of whatever you're trying to create and then using the calculation you can support this concept with certain, more or less, rigorously obtained numbers. All right, that's it for today. Thank you very much and good luck.