 Hi, I'm Zor. Welcome to Unizor Education. Let's solve a few problems related to cylinders. I do suggest you to try to solve these problems just by yourself. They are really easy. They are presented on Unizor.com as usual. That's where I suggest you to watch this lecture from. But first try to solve the problems they are in the notes for this lecture on this website and then which to lecture my solutions actually to these problems, which you might actually find some other solutions, whatever. Alright, so cylinders and the problems. Problem number one. Prove that the section of the cylinder which is parallel to the base is a circle. Well, what does it mean that it's a circle? It means that this particular curve contains some kind of a point equally distance from all other points of this curve. So this is a flat curve. It's an intersection of some kind of a plane with the cylinder. Alright, so how can I prove that? Well, I know that the bottom and the top are cylinders of the cylinder are circles, right? So let's just take the center of let's say bottom circle and draw a perpendicular to the plane. Well, this perpendicular intersects my the plane, let's call it gamma which cuts the cylinder. It intersects at some point. Alright, now what I would like to do is I would like to prove that the distance from this point to any other point on that curve is the same. So let's just take one particular point. Let's let's call this is A, B, C. Now from C, I will draw perpendicular down on the surface of a cylinder. This is basically a generatrix, one of the positions of the generatrix as it's moving parallel to itself perpendicular to the basis around this circle. And let's consider this quadrilateral. Well, first of all, it's a flat figure. Why? Because C, G is as I was saying, this is the perpendicular to the plane and A, B also is perpendicular. That's how I constructed A, B. So they are both perpendicular to the plane of the base, which means they are parallel to each other, which means there is a plane which contains them both. So A, B, C, D is a flat figure. Now A, B is parallel to C, D because they're both perpendicular by construction. Now B, C and A, G. Now, they're also parallel. Why? Because you consider the two planes, the gamma plane, which is my section, and the base plane, they are parallel among themselves because that's exactly how I said I will draw the gamma in this particular case. It's parallel to the base. So they are parallel to each other, the section and the bottom base. Now, A, B, C, D is a plane which cuts the bulbs. So A, G and C, and A, B, A, sorry, A, G and B, C, okay, and B, C are intersections of the plane which cuts two parallel planes. And there is a theorem about this that they are parallel to themselves. So B, C and A, G are parallel. A, B and C, G are parallel. And that's why this is parallelogram. And since this is the parallelogram, B, C is equal to A, D. Now, let's take any other point, C prime, and drop down this perpendicular along the surface, side surface of a cylinder. Then I will have some other point, D prime, right? And again, B, C prime would be equal to A, D prime. Same thing, exactly. But now the bottom is a circle, which means A, G and A, D prime are equal to each other. Therefore, B, C and B, C prime are equal to each other. And these are two points which I have randomly chosen, which means that the distance from any point on this curve to B is exactly the same as is equal to radius of the bottom. And that's why it's a circle and that point is, the point B is the center of the circle. Okay. Now, the second problem is, well, in some way similar, but we cut differently. We cut our cylinder instead of cutting it horizontally. We will cut it vertically. By the way, why the center line of a cylinder is also perpendicular to the bottom? Well, the previous theorem actually can be applied in this particular case. So, the top is a plane, which we have drawn parallel to the bottom, right? Which means that its center would be parallel to any other generator. So, I will just use the fact that these two are, the center line is also perpendicular to the base, as well as every line on the side surface, which is basically positions of the generator. So, now we cut it vertically, which means we do something like this. So, in this case, what actually it means is, I'm cutting with a plane, which is parallel to the the center line. So, I'm cutting the cylinder with a plane parallel to the center line, which is not far from the center line. It's not outside of the cylinder. It's inside of the cylinder. Okay. Well, let's call it A, B, C, D. Okay. Now, I would like to prove that this is a rectangle. So, if you cut vertically, the cylinder, you will get a rectangle. Horizontally, you get the circle. Vertically, you get a rectangle. All right. So, first of all, I would like to prove that this is parallel to the ground. And to prove it, I have to prove the parallelism of the opposite sides. Well, A, B and C, G, these are easy. They are parallel because we have two parallel planes, two bases, and you have a cutting plane, A, B, C, G, right? So, the result, the intersection of the cutting plane and two parallel planes are A, B and C, G, which must be parallel because of some theory which we have proven long time ago. Now, let's consider the parallelism between these lines, B, D and A, C. Okay. How can we do that? Well, let's take line A, B. And we know the theorem. Again, we proved it in some other lecture that if you have a line and two other points and you would like to draw, if you would like to draw a plane parallel to the line which goes through these two points in space. Well, the plane can be in some position and in only one position, it would be parallel to this particular line. Like in this particular case, if you take all the different planes, there is only one plane which is parallel to this one. So, let me, using this, prove the parallelism of B, D and A, C to the center, to the center line. Let's draw a perpendicular from B down to this bottom and let's consider it's not the point D. Let's say it's D prime and correspondingly from A. We draw a perpendicular and let's say it's not C, it's some kind of other line, a point C prime. Now, let's consider A, B, C prime, D prime. A, B, C prime, D prime. Now, these are two perpendiculars which means they're parallel to this guy, to the center line. Which means that the whole plane A, B, C prime, D prime is parallel to the center line. But I said that A, B, C, D is parallel. We cannot have two different planes parallel to a line and going through two given points. As I was just saying. So, there is no way C prime or D prime would be different from correspondingly C and D. They must be exactly the same points. Otherwise, I will have two different planes parallel to center line. So, that's why both B, D and A, C are parallel to center line. Which means they're parallel to themselves. Which means this is parallelogram. And also, I know that any generatrix and B, D now becomes a generatrix, right? Since it's parallel to the center line and correspondingly perpendicular to the bottom base. So, since it's perpendicular, these are right angles and this one is also right angles. So, it's parallelogram with right angles, which means it's rectangle. Okay. Next. Okay. Again, we have a cylinder. No, that's not a good cylinder. I need a shorter cylinder because I have a cube which is... Yeah. Not such an easy thing. Let's start from the cube. So, I have a cube which is inscribed into cylinder. Now, the cylinder is described, circumscribed around the cube in such a way that the bottom base, which is a square, is inscribed into a bottom circle. This is the bottom circle. And the top base is also... And this is my cylinder. So, cylinder is around the cube or cube is inscribed into a circle in such a way that both squares are inscribed into both circles on a circle. Now, what I do know about this is I know that the cube side, the edge of the cube is equal to G. What I have to determine is the full surface area and the volume of the cylinder. Now, let me go back and let's recall what is the area and the volume of the cylinder. You remember that the area contains two bases. Each of them is pi R square, where R is a radius. And I also have a side surface. Now, side surface, if you remember, I just cut it along one particular generatrix and open up into a flat rectangle. My width of this rectangle would be a circumference of the circle, which is 2 pi R. And the height of the cylinder will be the height of the rectangle, will be the height of the cylinder. So, this is basically my area, right? So, what I have to determine, knowing G, I have to determine R and H. Well, the H is easy. H is equal to D. Right? Because my cylinder completely has exactly the same height as the cube, right? Because I said that the bottom base of the cylinder contains the bottom face of the cube and the top base of the cylinder contains the top face of the cube. So, the distance between these two planes is exactly the height of the cylinder. Now, the radius might be slightly more complex, but that's actually very easy as well. Let's just consider a circle which is surrounding a square of the side D. Now, what would be the radius of this square? Well, let's consider this right triangle. See, this is a square, right? So, this is a right triangle, OAB, this is the right angle. So, I know the hypotenuse and I do have to determine the catatose, right? And the catatose, which is OA, is equal to D square root of 2 over 2, right? Because this square plus this square should be equal to D. And this square will be D square over 2 and D square over 2 would be D, right? So, this is D square root of 2 over 2. That's the radius. Radius is equal to D square root of 2 over 2. So, now, and knowing this formula, I can determine that my area equals to pi R square, which is, well, actually 2 pi R square, I'm sorry, because we have two bases, right? Two pi R squares. And R square is D square over 2, plus 2 pi R, which is D square root of 2 over 2, and H, which is D equals to 2 pi D square, I can factor out D square, D square, and 1 half would be out. So, I will have 1 plus square root of 2 divided by 2. That's my formula. That's the area. Okay. Now, how about volume? Well, the volume of the cylinder is, as we know, the area of the base, which is pi R square times the altitude, the height. Okay. In this particular case is pi R square is D square over 2 and D. So, it's pi D cube over 2. That's the volume. So, everything is determined by one particular parameter D, based on the fact that the cube is inscribed into a circle. So, all we have to do is determine the radius and altitude of the cylinder. Next. Okay. If you have a cylinder, prove that this particular line, which connects the centers of the bases, is axis of symmetry. Now, what is axis of symmetry? Well, if you remember, the symmetry relative to an axis is, if you have a point, you drop a perpendicular to this line from this point and extend it by the same length. Right? Okay. So, let's just take any point on the cylindrical surface. Well, this point can be either on top base, on bottom base, or at side surface. Now, speaking about top and bottom bases, obviously, the situation is very easy because it's a circle, right? And if you take any point here, then you can drop a perpendicular to the center line, which will be within this plane, which will be basically a connection between our point and the sector, extend it, and it will be another point within this particular circle in the same distance from the center. So, every point on any base has a symmetrical point on the same base. So, that's why the bottom and the top are completely symmetrical relatively to this axis because this plane is perpendicular, which means when I drop the perpendicular to the axis, I will basically connect it with the center and then I extend by the same length. Now, what about any point on the surface? Well, let's take any point on the surface, let's say here. And now, when you drop a perpendicular, the easiest way is, let me just draw a plane through this point perpendicular to the center line and connect the intersection of this line, center line and this plane, connect with my point. Now, obviously, since my center line is perpendicular to the plane, that would be perpendicular to this line as well, right? Now, but you remember, I was just proving it before, that the intersection of the plane with the side surface would be a circle. And for every circle from this point, if I extend it to the same, I will get the other point, the opposite point on the same circle, which is actually lying on the surface of this particular cylinder. So, the symmetrical point to this point is the point on this circle, which also belongs to the intersection of this plane with the cylinder. So, for every point on the cylinder side surface, there is a corresponding symmetrical point relatively to this center line. So, that's why the center line is axis of symmetry. For every point on the cylinder, there is another point symmetrical relative to this axis on the cylinder. And by the way, incidentally, if you remember, the symmetry relative to the axis is equivalent to the rotation by 180 degree. So, basically, what I'm just saying is that if you will turn the cylinder by 180 degree, it will convert basically into itself. It will transform into itself. Every point would be in a different location, but within the same cylinder. Okay, next. Okay, we have a cylinder and we cut it vertically on a distance d from the center line. So, we cut it vertically on a distance d from the center line. So, if you drop a perpendicular from the center to this chord, well, actually it's a perpendicular to the whole plane, but obviously since this is the plane which is perpendicular to this plane, then this perpendicular would be exactly a perpendicular to the intersection of these two planes. There was a theorem about this. If you have two intersecting planes and you drop, and they are perpendicular to each other, if you drop a perpendicular to the intersection line, it will be perpendicular to another plane for obvious reasons. Okay, now. So, I know this, this distance. I also know the parameters of the cylinder, r and h, radius and the height. Now, what I have to determine is the area of this as I know by now a rectangle, right? Now, how can I determine the area of the rectangle? I need two parameters, widths and height, right? Now, how can I determine, well, the height is easy. The height is exactly the same as the height of the cylinder. But the widths might be a little bit more calculations involved. So, you have a circle of radius r, and you have a chord which is on the distance g from the center. So, what's the length of the chord? Obviously, it's double this catechus which has a hypotenuse r and another catechus g. So, the half of this is square root of r square minus g square by Pythagorean theorem, but the whole chord is double. And now I have to multiply it by the height to get the area. That's it. Very small calculation based on Pythagorean theorem. Now, I was saying that all these problems are relatively easy. Next, okay, next is the following. You remember we were inscribing a cube into the cylinder. Now we will inscribe a regular tetrahedron. So, let me start again from a tetrahedron. It might be easier to draw. So, this is a regular tetrahedron. Now, I'm inscribing this tetrahedron into a cylinder in such a way that my base triangle would be inscribed into the bottom of the cylinder. And my top, my apex of my triangular pyramid would coincide with the center of the top. So, that's how it looks. So, this is the center of the top. And now, all faces of my pyramid are equilateral triangles, right? Because it's a regular tetrahedron, as I said. Now, what do I know about this? I know that the h is equal to d. And I have to calculate again my area, total area of the cylinder and this volume. So, again, my total area is 2 pi r square, right? Pi r square is one base, another base. And the surface, if I will open it up, I will have 2 pi r circumference times h. Okay, so I have to determine r and h. Now, this is area. I have to determine r and h knowing only d. Well, first of all, let's talk about the radius of the circle. Now, if you have a circle, this is my bottom base. And you have an equilateral triangle inscribed into this with side is equal to d. What is the radius of the circle? Well, that's easy. So, a, b is equal to g. Now, this midpoint, called d. Now, triangle a, d, o is a right triangle. This is obviously a 30 degrees angle, right? So, what do we know about this triangle? How can they determine r? Now, we know from this triangle that a, o is equal to r over 2, right? So, a, d is equal to square root of r square minus r square divided by 4, which is r square root of 3 over 2. So, that's my a, d. Now, I know that a, d is actually half of the d, right? So, this is half of the d. From which follows that r is equal to d divided by square root of 3 or d multiplied by square root of 3 divided by 3. Fine. This is done. r is determined. Now, let's determine the height of this pyramid, because it's exactly the same as the height of the cylinder, right? So, let's drop the perpendicular here and let's connect. Now, obviously this is perpendicular, this is the center, so this is the center. So, this is a, b, c. Now, oa is a catechus, which I do know, that's r, right? Top sa, I also know this is d. So, I have a right triangle, so a, where I know the hypotenuse and one catechus. Now, I have to determine the second catechus. So, the second catechus would obviously be equal to d square, hypotenuse minus this square, which is d square divided by 3, which is d square root of 2 divided by square root of 3, which is d square root of 6 divided by 3. So, that's my h, d square root of 6 divided by 3. So, I know both, radius of the base and the height, so I can determine what's my area. Okay, this is just a simple algebraic exercise. Now, the area is equal to 2 pi r square, which is d square divided by 3, plus 2 pi r and h multiplied. So, it's d square divided by 9 multiplied by square root of 2, square root of 3 and square root of 3, right? Square root of 6 is square root of 2 times square root of 3. So, this is 3, this is 3, which is equal to 2 pi d square, 1 plus square root of 2 divided by 3. That's my formula. That's the total surface. The volume is equal to area of the base, right? Which is p pi r square times h times the height, area of the base times h, which is equal to pi d square divided by 3 times d square root of 6 divided by 3, which is pi d cube square root of 6 divided by 9. That's the answer. Well, that's it. That's the last problem. I hope you will try to do exactly the same thing as I did just by yourself. I strongly suggest you to do it. That's a very good exercise. And good luck. That's it. Thank you very much.