 Hi, I'm Zor. Welcome to Unisor education. I would like to present to you an interesting problem related to the geometry of sphere. I would say it's kind of unexpected result. Let's put it this way. At least unexpected for me. I do suggest you to watch this lecture from the Unisor.com website because there are notes for any lecture and also problems are usually given with answers and sometimes with solutions. In this case I do have both answer and solution, but if you try to solve this problem yourself and just check against the answer, that would be much much more useful, obviously. And that's exactly what I'm suggesting you to do. Now if you cannot solve it, that's okay. The solution is on the website and I will present it in the lecture. But then try it maybe afterwards to do it just by yourself. All right, so the problem. Consider we have a sphere and a cylindrical hole is actually drilled inside that sphere. So the cylinder has the axis of symmetry going exactly through the center of a sphere. So it's a very symmetrical kind of thing. So the sphere was drilled in cylinder. Well, it's not actually a cylinder. It's a cylinder and two caps. That's what we are touching off from this from this field. Now what I would like to know is the volume of the remaining part. The remaining part it's like some kind of a ring without tops. Actually, you know what when you go to a good restaurant and they give you napkins in some kind of rings, the rings usually have shape of this type of thing. So Now to evaluate the volume of the remaining part, one would think that you really need lots of information. For instance, the radius of a sphere, right? Then the radius of a cylinder you are cutting through, the height of the cylinder. I mean there are certain parameters which must be specified. Now to our surprise, at least to my surprise, there is only one parameter which is actually sufficient to determine the volume of the remaining part after we drill this cylinder in. It's the H, the height of that cylinder. Apparently, if just this one parameter is given, without the radius of the sphere, without the radius of the cylinder, whatever, everything else seems to be just, you know, going, putting together a nice formula for the remaining volume after the drilling process. So just one parameter H is given, nothing else. Well, in theory, you might kind of understand it, that if you have a bigger sphere and this parameter is given, it probably means it should be a wider cylinder, something like this. If you have a bigger sphere, if you have just this length of the height, then it should be somewhere here. And the remaining part, although of a bigger radius, would have probably thinner walls, something like this. And the volume of the remaining part would be the same. So only one parameter H is given. All right, now let's go to the calculations. We have to get the formula and hopefully the formula will depend only on one parameter H. Now, I would like actually to specify radius of the sphere as capital R, radius of the cylinder as lowercase R. And since I needed the caps here, let's say the height of each cap is capital H here and there. Now, what do I know just looking at this picture? Well, first of all, obviously, I can relate capital R, lowercase R and H using the Pythagorean theorem, right? So it's R squared is equal to lowercase R squared plus half of the H squared. That's one equation, right? Another equation is, well, if you just drill through, well, if you imagine the axis, now, the axis of the sphere is basically two radiuses, right? And each radius is equal to half of the height of the cylinder plus height of the cap, right? So each radius is equal to half of the height of the cylinder, which is this one, or on the bottom it's this one, plus height of the cap. That's what I know. So if my unknown variables are radius of the sphere, radius of a cylinder and height of the cap, three unknown variables. I have only two equations. So obviously, I cannot determine them, knowing only the height of the cylinder H. But maybe it will not be necessary, because maybe in the formula, in the final formula, everything will just, you know, reduce itself and cancel itself, and only dependency on the lowercase H will remain in the formula. Alright, so now let's go to calculations. Well, the V, which is the volume of the remaining part after the drilling, is equal to volume of the sphere minus volume of the cylinder and minus two volumes of the caps. Right? Volume of the cylinder and two caps we take out. Okay, in terms of these variables, it's equal to volume of the sphere is four-third pi r cube minus volume of the cylinder is, if radius is lowercase r and the height is lowercase H, it's pi r square H and minus two volume of the cap is pi H square 3r minus H. That was in one of the previous lectures when I was talking about caps. Somehow, I remember this formula. Quite frankly, it's not an easy thing to remember, but I think that's what it is. So, alright, so we have all these dependencies on r, capital R, lowercase r and capital H, which we don't really know. However, I can use these dependencies, which I know about to basically use it here. So, what's the result if I will substitute? So, let's say I will leave just the capital H as unknown and r and r will be in terms of lowercase H and uppercase H. So, four-third pi, instead of r, I will put H plus H over 2 cube minus pi r square. Instead of r square, I will put capital R square minus H over 2 square and instead of r, I will put this. So, that would be r square, which is this, H plus H over 2 square minus, so that's r square, minus H over 2 square. That's r square and times H and minus 2 pi H square 3r minus H. So, 3r, it will be 3 capital H plus this. So, it would be minus H. So, it's 2 H plus 3 H over 2. Is that right? Well, we'll see. I mean, hopefully I didn't make a mistake. All right. Now, let me wipe out this and we will go off. All right, let's just, you know, open all the parentheses. Let me put it this way. Pi will go out and I will probably have... Oh, wait a moment, wait a moment. I think I have to divide it by three. Yeah, that's right. I said I remember the formula, but actually I did not. Now I remember because I have to have one-third, actually, also out. So, it's not just pi, it would be pi over 3, but I will have to multiply by 3 this middle part. So, I don't have 3 here. I don't have 3 there and what's remaining? Okay. 4 and this cube, right? So, it would be 4 H cube plus 3 H square times this H over 2 times 4, that's 12 divided by 2, 6. So, it's 6 H square H plus 3 H and this square. That would be 4. Okay. So, it would be 3 H H square and the cube would be H cube over 8. Now, this is 4. So, it's H cube over 2 minus minus. Now, this should be multiplied by 3 and divided by 3 because I put 3 in the denominator, right? So this is, well, obviously, H over 2 square will be reduced. So, I will have only H square plus 2 times H times H over 2 plus H H. So, it will be 3 H square minus 3 H, or K is H, right? And then H square over 4 and H square over 4 will be reduced. Minus. The last one. 4, okay, pi goes out and 3 goes out. So, it's 4 H cube and 3 H square H and I close. Now, it would be nice if everything would be just cancelling out. Okay. 4 H cube, 4 H cube. Okay. 6 H square H and 3 H square H. Now, I remember it was reduced. I missed something. Okay. How about this? H H square plus something is wrong here. I think I missed H here. Here. Yeah, this is the H. Of course. My fault. So, that would be H and would be H square. Of course. I forgot this H to multiply by this. Now, everything is cancelling out. 6 H square H minus 3 and minus 3. And 3 H H square and 3 H H square. You see, everything is cancelling out. And the only remaining part is pi H cube divided by 2 and 3, 6. So, the volume of the remaining part depends only on the H. And this is a very interesting formula actually. If instead of H, you would put, let's say, 2 X. What this formula would look like. It would be 8 over 6. So, it's 4 third pi X cube. Now, this is a formula for the volume of a sphere. So, basically what it says is that the remaining part would be exactly equal in volume to the volume of a sphere with the radius equal to half of the height of the cylinder. Strange. I don't know how it happens and why I cannot explain it, but I think it's an interesting observation. All right. Well, that's it. Now, I suggest you to do the following. If you did not do it yourself before the lecture, do it right now. Just go through all these calculations. It's very useful because it basically involves all the elements of the sphere. The volume of a sphere, the volume of the cap plus a cylinder also involved. And a couple of accurate calculations which you have to do, hopefully without making any mistakes. I made a little mistake here, but that's okay. And see if you will get the same result. Well, other than that, that's it for today. Thank you very much and good luck.