Intersection of 3 cylinders volume

In this video, I calculate the volume of the Steinmetz 3-solid obtained by intersecting three cylinders, using a symmetry argument and cylindrical coordinates. It’s actually not as bad as you think!

[povilasdapsys]

2019: Finding the volume of the intersection of 11 cylinders

[TheAllen]

It's a bit hard to follow his accent, but the lecture is excellent. If you listen carefully, you will understand very well.

[PeterBarnes]

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.

[cosimobaldi]

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.

[shiina_mahiru_]

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😂

[Koisheep]

These videos make me refresh 2D+3D integration which is #good

[TheMauror]

I laughed like crazy with the call me maybe reference 😂😂😂

[fabiomoraiscartaxo]

thank you for the atention and answer !

[Bani]

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

[gnikola]

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?

[silasrodrigues]

No pi on the answer? What? I would never imagine that!

[cycklist]

You're so cool Peyam! A real maths hero.

[Rundas]

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

[rialtho_the_magnificent]

And now for something different: the surface area of this monstrosity!

[titoapen]

Congratulations!
You're so close to 10k

Intersecting cylinders are so cruel ... killing pi :-O

[xlgabriel]

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).

[ThuTa-dw]

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

[shayanmoosavi]

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.

[spiritgoldmember]

Great video. How about in n dimensions (n intersecting cylinders)?