 Hi, I'm Zor. Welcome to Unisor Education. I would like to solve a few problems related to spheres. This lecture is part of the Advanced Mathematics course on Unisor.com. I suggest you actually to use the website because it has a complete curriculum of this course and every lecture has notes and then there are some exams. There is certain functionality related to the self-study, etc. Now, back to the problems related to spheres. Alright, I have four problems and let's just do one by one. They're not very difficult at all. But it's good exercise as far as the volumes and the surface of the sphere is. Okay. The first is extremely easy. Well, there are three planets which I would like to consider. Mars, Earth and Jupiter. Well, they're not in the order of their distance from Sun, they are in the order of size. So, I know that the diameter of Jupiter is 11 times greater than the diameter of Earth. And the diameter of Earth is twice as big as the diameter of Mars. So, the question is how much bigger the surface and the volume of Jupiter are than that of Mars, considering that these planets are ideal spheres. Well, that's actually very easy. Since Earth is twice as big as Mars and Jupiter is 11 times bigger than Earth, the diameter of Jupiter is 11 times 2 diameter of Mars, which is 22 diameter of Mars. Now, if diameter is 22 times greater or radius is 22 times greater, that doesn't really matter. What happens with the surface area and the volume? Well, we know that the surface area is proportional to a square of the radius. Well, actually it's 4 pi r square if you need the formula, but it doesn't really matter what the formula is in this case. What matters is that obviously the surface is proportional to r square or d square. Diameter doesn't matter, right? So, if my diameter of Jupiter is 22 times greater than the diameter of Mars, then the area of Jupiter is 22 times 22 times 22 square greater than the diameter of Mars, which is 484. Sorry, not the diameter. It should be the area, surface area, then surface area of Mars. Now, the volume of Jupiter, the formula again, is 4 3 pi r cube. That's the volume expressed in terms of the radius. But again, what's important is that the volume is proportional to the third degree of the radius or the third degree of the diameter, right? So, that means that the volume of Jupiter is 22 cube volume of Mars, which is 10648. So, that's very important actually. You see, volume grows in cubicle proportion to the radius. So, if you see the planet, for instance, and what you see is basically, you see the diameter. And another planet, let's say, is slightly bigger in diameter as you see it. Well, considering they are on the same distance. It's maybe only twice as big, but its volume is 8 times as big. And if you go back to Sun, for instance, Sun is visible from the Earth at the same angle, basically, as the Moon. So, they do look kind of the same. But considering that the difference to Sun is like 150 million in the difference kilometers and the difference to Moon is something like less than 500,000 kilometers. So, the difference in the lengths, considering the angle is the same, implies the difference in diameter in certain number of times. And the volume of the Sun is, well, you have to cube that number, which is huge. The Earth is still circling around the Sun. So, it doesn't really matter, it's really far, but it's really big. And that's why the gravitation, which is proportional to mass and mass proportional to volume, that's why gravitation is so big. Alright, anyway, it was just an exercise that the surface area of the sphere is proportional to the square of the radius and the volume is proportional to the third degree of the radius, the cube. Simple. Okay, all other problems are as simple as this one. So, next one. Okay, we have a glass of water of cylindrical form and it's filled up with water, let's say, to half of it. And the radius is r. Then we have a ball, metal ball, of the radius r and we drop it in, into the glass. Well, otherwise we will not be able to drop it, right? So, the radius of the ball is less than the radius of the cylinder. The question is, by how much the level of water will rise if I will drop this ball? Okay, well, since it's metal, it goes completely down, right? So, the amount of water which it will push out would be equal to the volume of this sphere, which is pi for third pi r cube. That's the volume. Now, this volume is the volume of water which is supposed to be pushed out by this ball. In this case, it's actually pushed up. And now, my question is, by how much it will raise the height of the water. So, now, this is a cylinder because the water is rising in a cylindrical glass, so the amount of water which is above the previous level will be a cylinder. It's volume, the volume of the cylinder should be equal to this because that's the volume of the ball, right? So, what's the volume of the cylinder? It's pi r square, the area of the base, times h. Now, we don't know h, but we do know that these are equal to each other, right? So, from this, you can have that h is equal to 4 pi is reduced r cube divided by 3 r square. So, that's the height of the water coming up, rising up. Obviously, if r, capital R is very big, then this h would be smaller. And if r, lowercase r, is getting greater, obviously, with a fixed radius of the cylinder, obviously this r will go, will increase. Well, now we can actually think about, depending on the existing height of the cylinder, of this glass, how big the radius of the ball might be to completely fill up the glass. But that's another problem, just came up to me. Another easy one. Alright, so that's my second problem. The next one is, okay, let's say you're playing with a rubber ball filled with air. So, there is a wall of rubber and inside the air is popped. Well, air is actually weightless. You can just consider it's weightless and disregard its weight. So, it's only the rubber itself from which the ball is made has the weight. We dropped it into the water and it's actually going down exactly by half. Alright? So, let's say we have the ball of radius r. That's what we know. And we also know that when it's dropped into the water, it goes down exactly half its own size. Now, my question is, I would like to calculate the width of the wall of this rubber ball if I know that the unit weight of the rubber ball, let's say, is B and water is, well, let's say, okay. Now, I was actually thinking about this problem before and it's not very easy to find an exact solution without some kind of numerical apparatus. But you can just basically make an equation which can be solved. It's not easy to solve the cubicle equation because it's supposed to be the volume, so it would be r cubed somewhere. But anyway, if you will come up with an equation, at least, it's very easy to solve. In this case, we have computers to solve equations if we cannot solve it in a regular way. So, we have unit weights for the rubber ball and for the water. And we know that the radius of the ball is r and it's floating exactly half of it is underwater and half another. Now, let's go back to physics a little bit. I think it's called the law of Archimedes which says that the weight of the floating object is equal to the weight of the water it displaced. So, how much water we displaced here? Well, the volume of the ball divided by 2, right? So, the volume is equal to 4 third p r cubed and we have to divide it by 2. So, it would be 2 third p r cubed. So, that's the amount of water which we have actually pushed out, displaced, so to speak. Now, if I will multiply it by the unit weight of the water, that would be the weight of that water, right? So, this is the volume, this is the unit weight, so this is the weight of the water which we have displaced. Okay, now, what's the weight of the ball? So, that's basically the weight of the balls of this ball. This ball, these balls, the volume of these balls is the difference in the volume of the ball itself minus the volume of the air inside the ball, right? So, if we have x as the width of the ball, then the outer radius is r and the inner radius, that's where the air is, would be r minus x, right? So, the difference in volumes would be, obviously, 4 third pi r cubed minus r minus x cubed. So, that's the difference in volume. And if I will multiply it by the unit weight of the rubber from which the ball is made, that would be the weight of the material from which the ball is made. Again, considering that the air is weightless inside. And they must be equal, that's the law of arches. Well, basically, that's it. All we can do right now is just make a little simplification to third pi. So, this, this, this, this, this, and we have only two, four and two. And that's the equation, two r cubed minus r minus x cubed b equals to r cubed w. That's the equation from which we can find x. And this is the third degree equation, so, unfortunately, I cannot easily give you the formula. I mean, there is a formula for the third degree equation, but that's too much. It's basically, it's very simple to solve these equations numerically nowadays with computers. So, whenever you want to solve the problem like this, first you have to come up with an equation, and that's what requires this problem. Just come up with an equation, no solution. And then, obviously, you can solve it. All right, that's my third problem. And the fourth one is coming next. All right. Okay, let's say you have a regular tetrahedron. Okay? And inscribed into this tetrahedron is a sphere. Now, inscribed, it means it's tangential to each phase. Now, we have four phases, and the sphere is inside. So, it's somehow something like this. It touches this, this, this and the front. Now, what I have to prove is that the radius of the sphere is equal to one quarter of the altitude of the tetrahedron. All right. How can I prove it? That's not easy, right? But actually, it is. Here is my suggestion. Let's take this center of the sphere and connect it with all vertices. One, two, three and four. Now, thereby, we are dividing our tetrahedron into one, two, three, four pyramids, right? Now, every pyramid has a volume equal to one-third surface area of the base times its height. But what's its height? If we connect this center with, let's say, these three, so the bottom pyramid, what's the height of this? Well, that's the perpendicular. And this is, since we are talking about tangential sphere, well, it's very easy to prove that the radius into the point of tangency is actually perpendicular. Well, to prove it is basically very easy. I mean, you can just cut the whole thing with a plane, basically, you'll reduce the problem to a known theorem that the radius to a tangent line is perpendicular to this line. That's very easy. By the way, it's a good exercise, actually, for you if you want to. Alright, so area of the base, which is area of this triangle times the height, and the height is the radius, right? Now, the same thing happens with other three pyramids, the one on the right, on the back, and on the front. All of them have, since this is a regular tetrahedron, all of them have exactly the same base, which is a triangle, equilateral triangle, actually, and the height, the altitude of each of these pyramids is the radius from the center to a point of tangency. So, I have to multiply it by four to get the volume of the entire tetrahedron. But on the other hand, the same volume is actually one-third s times h, where s is the surface area of this triangle and h is the altitude. Well, from which we see that 4r equals h, or r is equal to one-quarter h. That's it. I do suggest you to go back to theunisor.com and try to solve all these problems just by yourself. There are answers provided there. And just as a good exercise, and don't forget, basically the site contains lots of other material, and it's good for self-study of an entire course of mathematics. Thank you very much, and good luck.