It's a lot easier if you recognize a few other symetries. The symmetry you describe around 10:00 is across the diagonal plane x-y=0, which is the same plane as theta = pi/4. This symmetry works because you can switch any two axes and define an identical shape. Taking this one step further, we can also cut along the plane y-z=0. (For symmetry's sake, we'll also visiualize the plane x-z=0, which would cut slightly less.) The half-octant is now divided into 6 congruent regions that match the symmetries of our volume; which demands a 48 become the coefficient outside the integral. How is the integral simplified? Some neat geo/trigo-nometry shows that we have triangles whose height is sin(theta). These triangles are oriented awkwardly, but we can just redefine the bounds of r to follow the triangle, from z/sin(theta) to 1. The height that we'll integrate across will be 0 to sin(theta). Integration has to go dr, dz, d(theta) because of the interdependencies. Going through it gives only a few longer expressions that get neatly cancelled out. The factor of 3 gets cancelled out amidst the cancellations and from the thrice integrated z. A little visual intuition goes a long way.
No π again. Disappointed. But suddenly if you intersect an infinite amount of cilinders the volume becomes 4/3 π r³. Hmm I smell something fishy here... Maybe π could be written as a limit based on a succession of volumes of intersecting cilinders?
Incredible! Just back to home from Orientation yesterday and saw this today. Just an instant thought, maybe generalizing the result to n-dimension? But that might be way beyond my ability to understand😂
I've been curious about this shape for some time... With a bit of visualization you can get the volume in a super easy way. First you separate the solid into its internal cube (vertices at the point where all 3 surfaces of the cylinders meet) and the 6 curved pieces. Those curved pieces are made of square slices, so we make an integral running along the height (along the z axis for example) of one of those pieces (1-1/sqrt2) and express the side of the square as a function of the z coordinate: L=2sqrt(1-z^2). It's really easy because the square root cancels as you have to square the side to get the area of the section. This way we calculate the volume which is 8/3 - (5/3) sqrt2. The volume of the central square is easy and it's 2sqrt2. We multiply the curved parts area by 6 and sum everything to get 8(2-sqrt2). I like this system better because it's easier and makes you understand why there's no pi in the final formula.
It looks like the common intersection of an infinite number of cilinders(rotated in all directions) should be a cube, it would be very cool to prove that :D Awesome video
Andrej Janchevski I dont think so! The intersection of an infinite amount of cylinders (I suppose your intention was a cylinder for every line through the origin) would be continiously symmetric (invariant under any kind of rotation in ℝ³) with respect to rotation (as the construct of all cylinders on their own would be, because every line - and cylinder - would be mapped 1:1 to another). Because a cube does not satisfy this property (it may not seem like that but the coordinates of i.e. a corner of the cube would change under a rotation like π/4 around the z-axis). Therefore a cube can't be the intersection. Moreover (as far as I know) a sphere is the only continously symmetric object with respect to rotation in ℝ³ and would therefore be this intersection! :) Quite a nice thought experience! :D
Hey Dr Peyam, I'm from Colombia and I just can't find help with this situation; what if the ratio of x^2 + y^2 = r^2 is >1? I mean, it grows in the XY plane. will the integral take the "circle" volume (in double integrals: polar coordinates) or the new volume will be just the other two cylinders? Honestly it will be great if someone like you help us with this problem. (btw, there's no need to solve the integral, just set up).
Great video! however I cant seem to get the value for 8(2-2^0.5)r^3. if i let z = (r^2 - r^2cos^2(theta))^0.5? the calculation doesn't work. I solved the integrals using wolfram alpha. it comes out with 8/3*(2)^0.5*r^3
Great video Dr Peyam. I was wandering what about find the volume enclosed by 2 cylinders using cylindrical coordinates. I know you already did a video on that using Cartesian coordinates. My problem is that I get the same limits of integration for this two cylinder problem compared to 3 cylinders.
Wow. It has such a cool shape. Also I'm surprised that there's no pi in the answer. It's a part of the cylinder so my logic told me that pi somehow should be in the answer.
n cyclinders??each at a constant horizontal,vertical angle from each other say a and b,also at what angular configuration of a and b would the volume of the intersection be maximum??(spherical co-ordinate)
@Debraj banerjee nice observation,but we aren't tending 'n' to infinity we simply are looking at the maximization factors for a given value 'n' and (a,90-a,b)
And i can not find the volume when given 2 cylinders. Not talking about 3 😥Which Parallel to the z axis. When the first cylinder is in the origion and with height 2h and radius a. The second cylinder is with height h and radius 1 and its center is in (x,y,z)=(0,1,0) . I dont know how to find the volume trapped outside the first cylinder and inside the second cylinder as function of a ( known that: 0
i really liked that but ....do you need to spell out every small algebraic manipulation? You explain the computations at 19:18 with the same emphasis as mentioning the jacobian factor or the pi/4 symmetry.
But your explanations are totally complecated....wish u simply your approach to maths. The answer is real and true but complicated working at some levels. Otherwise thank u so much.
@@drpeyam thanks Dr so much.... actually finally have got it. The major prob we have here in UGANDA is lack of proper 3D demonstrations on screens to the extent that whatever we learn we take it theoretically; but to a few willing to learn, we research and we get to know the reality of what we are doing. However,in that number, the region of intersection needs a clear demonstration on a 3D screen for proper visualization of Octan.....and so on. Otherwise am greatiful and happy that I have read from u. I love your passion and the communication of the mathematics language. U so often humble me down. Thanks Doctor so much.
It's actually not that bad
blackpenredpen Isn't it.
blackpenredpen As long as you do it do it carefully
+blackpenredpen
i love you
The connection to the Reimann
2019: Finding the volume of the intersection of 11 cylinders
4/3 pi r^3 + epsilon, where epsilon is too small to care about
I suspect 6 then then 24
Nay, I demand it. 120 minutes for 120 cylinder challenge!
😅😂😂
Don't forget to mention in how many dimension you want the 11 cylinders to intersect ;-)
It's a bit hard to follow his accent, but the lecture is excellent. If you listen carefully, you will understand very well.
It's a lot easier if you recognize a few other symetries.
The symmetry you describe around 10:00 is across the diagonal plane x-y=0, which is the same plane as theta = pi/4.
This symmetry works because you can switch any two axes and define an identical shape.
Taking this one step further, we can also cut along the plane y-z=0.
(For symmetry's sake, we'll also visiualize the plane x-z=0, which would cut slightly less.)
The half-octant is now divided into 6 congruent regions that match the symmetries of our volume; which demands a 48 become the coefficient outside the integral.
How is the integral simplified?
Some neat geo/trigo-nometry shows that we have triangles whose height is sin(theta).
These triangles are oriented awkwardly, but we can just redefine the bounds of r to follow the triangle, from z/sin(theta) to 1.
The height that we'll integrate across will be 0 to sin(theta).
Integration has to go dr, dz, d(theta) because of the interdependencies. Going through it gives only a few longer expressions that get neatly cancelled out. The factor of 3 gets cancelled out amidst the cancellations and from the thrice integrated z.
A little visual intuition goes a long way.
Congratulations!
You're so close to 10k
No π again. Disappointed. But suddenly if you intersect an infinite amount of cilinders the volume becomes 4/3 π r³. Hmm I smell something fishy here...
Maybe π could be written as a limit based on a succession of volumes of intersecting cilinders?
You're so cool Peyam! A real maths hero.
Incredible! Just back to home from Orientation yesterday and saw this today. Just an instant thought, maybe generalizing the result to n-dimension? But that might be way beyond my ability to understand😂
Even if your drawing skills are not amazing, the math-content you are producing as well as the dumb puns are art as well :D
Great video, perfectly explained! Liked and subscribed
I've been curious about this shape for some time... With a bit of visualization you can get the volume in a super easy way. First you separate the solid into its internal cube (vertices at the point where all 3 surfaces of the cylinders meet) and the 6 curved pieces. Those curved pieces are made of square slices, so we make an integral running along the height (along the z axis for example) of one of those pieces (1-1/sqrt2) and express the side of the square as a function of the z coordinate: L=2sqrt(1-z^2). It's really easy because the square root cancels as you have to square the side to get the area of the section. This way we calculate the volume which is 8/3 - (5/3) sqrt2. The volume of the central square is easy and it's 2sqrt2. We multiply the curved parts area by 6 and sum everything to get 8(2-sqrt2). I like this system better because it's easier and makes you understand why there's no pi in the final formula.
thank you for the atention and answer !
Great video!
No pi on the answer? What? I would never imagine that!
It looks like the common intersection of an infinite number of cilinders(rotated in all directions) should be a cube, it would be very cool to prove that :D Awesome video
Andrej Janchevski I dont think so! The intersection of an infinite amount of cylinders (I suppose your intention was a cylinder for every line through the origin) would be continiously symmetric (invariant under any kind of rotation in ℝ³) with respect to rotation (as the construct of all cylinders on their own would be, because every line - and cylinder - would be mapped 1:1 to another). Because a cube does not satisfy this property (it may not seem like that but the coordinates of i.e. a corner of the cube would change under a rotation like π/4 around the z-axis). Therefore a cube can't be the intersection. Moreover (as far as I know) a sphere is the only continously symmetric object with respect to rotation in ℝ³ and would therefore be this intersection! :) Quite a nice thought experience! :D
Thank you, this vid is awesome.
These videos make me refresh 2D+3D integration which is #good
I laughed like crazy with the call me maybe reference 😂😂😂
Hey Dr Peyam, I'm from Colombia and I just can't find help with this situation; what if the ratio of x^2 + y^2 = r^2 is >1? I mean, it grows in the XY plane. will the integral take the "circle" volume (in double integrals: polar coordinates) or the new volume will be just the other two cylinders? Honestly it will be great if someone like you help us with this problem. (btw, there's no need to solve the integral, just set up).
Great video. How about in n dimensions (n intersecting cylinders)?
can you pls upload a video that explains how to calculate the area of the surface of this Steinmetz solid?
Volume of 4 4D spherinders?
Good Job 👍🏻
Great video! however I cant seem to get the value for 8(2-2^0.5)r^3. if i let z = (r^2 - r^2cos^2(theta))^0.5? the calculation doesn't work. I solved the integrals using wolfram alpha. it comes out with 8/3*(2)^0.5*r^3
Great video Dr Peyam. I was wandering what about find the volume enclosed by 2 cylinders using cylindrical coordinates. I know you already did a video on that using Cartesian coordinates. My problem is that I get the same limits of integration for this two cylinder problem compared to 3 cylinders.
same problem . when i go to two cylenders and search the limits . i found that it is same as 3 cylinders .
Thank you
Could you do that triple integral if f(x,y,z) =/= 1 ??
And now for something different: the surface area of this monstrosity!
awsome man
So can you do the xi²+xj²=1 for all pairs of i!=j ; i,j=1,...,n Volume?
Wow. It has such a cool shape. Also I'm surprised that there's no pi in the answer. It's a part of the cylinder so my logic told me that pi somehow should be in the answer.
n cyclinders??each at a constant horizontal,vertical angle from each other say a and b,also at what angular configuration of a and b would the volume of the intersection be maximum??(spherical co-ordinate)
Google Steinmetz solid
I guess we could try a rather brute force method writing out each individual axes in terms of (a,90-a,b) and then integrate and maximize....
It would be a sphere of radius 1
hari kishan cylinders
@Debraj banerjee nice observation,but we aren't tending 'n' to infinity we simply are looking at the maximization factors for a given value 'n' and (a,90-a,b)
Now can you do it using single integrals please answer
0:55 - 1:05
Perfect
Intersecting cylinders are so cruel ... killing pi :-O
excuse me, Can you write for full version which don't care the symmetry?
very interesting.
And i can not find the volume when given 2 cylinders.
Not talking about 3 😥Which Parallel to the z axis. When the first cylinder is in the origion and with height 2h and radius a. The second cylinder is with height h and radius 1 and its center is in (x,y,z)=(0,1,0) .
I dont know how to find the volume trapped outside the first cylinder and inside the second cylinder as function of a
( known that: 0
Can we solve it in cartesian coordinates?
I thought cylindrical coordinates would be with the distance, r, and two angles.
Forget about it. I was thinking about spherical coordinates!
i really liked that but ....do you need to spell out every small algebraic manipulation? You explain the computations at 19:18 with the same emphasis as mentioning the jacobian factor or the pi/4 symmetry.
But your explanations are totally complecated....wish u simply your approach to maths. The answer is real and true but complicated working at some levels. Otherwise thank u so much.
It’s the easiest approach there is
@@drpeyam thanks Dr so much.... actually finally have got it. The major prob we have here in UGANDA is lack of proper 3D demonstrations on screens to the extent that whatever we learn we take it theoretically; but to a few willing to learn, we research and we get to know the reality of what we are doing. However,in that number, the region of intersection needs a clear demonstration on a 3D screen for proper visualization of Octan.....and so on. Otherwise am greatiful and happy that I have read from u. I love your passion and the communication of the mathematics language. U so often humble me down. Thanks Doctor so much.
Great video!