3 ms·

Yes it is. It feels like cosmic coincidence because I lost my father 2 weeks ago.

marcodiego 5y ago
Sorry for your loss. Lost mine when I was 17, I know the feeling. Hope the "Easter egg" brings you some comfort. A long time ago, during a night I couldn't sleep, I started thinking on the simplest way to calculate the volume of a torus without resorting to integrals. After a time I found a way to "turn it into a cylinder without changing its volume". Then went to my father's copy of "Simmons' Calculus with analytic geometry" to check my idea. How surprised was me to see that my father had had EXACTLY the same idea and left it written on the margin of the book! That moment I saw how careful he was to model my way of thinking to be so similar to his. I'll be eternally grateful for that.
- akkartik 5y ago
  Thank you for that lovely story!
- dmitryminkovsky 5y ago
  Thank you for sharing.
- thaumasiotes 5y ago
  Hmm. Thinking about it as an integral of cross sections, the torus is just a cylinder with height equal to the circumference of the circle that defines the center of its tube. And I see that that is correct, at least for non-weird tori. https://en.wikipedia.org/wiki/Torus https://en.wikipedia.org/wiki/Torus goes on to say this: > [For a non-weird torus,] The volume of this solid torus and the surface area of its torus are easily computed using Pappus's centroid theorem > These formulas are the same as for a cylinder of length 2πR and radius r, obtained from cutting the tube along the plane of a small circle, and unrolling it by straightening out (rectifying) the line running around the center of the tube. > The losses in surface area and volume on the inner side of the tube exactly cancel out the gains on the outer side. Emphasis mine; that third sentence is weirding me out. Still thinking about it as an integral of cross sections, I can't see why the surface area or volume of the inner half of the torus should be different from the surface area / volume of the outer half. Are they?

Sorry for your loss. Lost mine when I was 17, I know the feeling. Hope the "Easter egg" brings you some comfort. A long time ago, during a night I couldn't sleep, I started thinking on the simplest way to calculate the volume of a torus without resorting to integrals. After a time I found a way to "turn it into a cylinder without changing its volume". Then went to my father's copy of "Simmons' Calculus with analytic geometry" to check my idea. How surprised was me to see that my father had had EXACTLY the same idea and left it written on the margin of the book! That moment I saw how careful he was to model my way of thinking to be so similar to his. I'll be eternally grateful for that.

akkartik 5y ago
Thank you for that lovely story!
dmitryminkovsky 5y ago
Thank you for sharing.
thaumasiotes 5y ago
Hmm. Thinking about it as an integral of cross sections, the torus is just a cylinder with height equal to the circumference of the circle that defines the center of its tube. And I see that that is correct, at least for non-weird tori. https://en.wikipedia.org/wiki/Torus https://en.wikipedia.org/wiki/Torus goes on to say this: > [For a non-weird torus,] The volume of this solid torus and the surface area of its torus are easily computed using Pappus's centroid theorem > These formulas are the same as for a cylinder of length 2πR and radius r, obtained from cutting the tube along the plane of a small circle, and unrolling it by straightening out (rectifying) the line running around the center of the tube. > The losses in surface area and volume on the inner side of the tube exactly cancel out the gains on the outer side. Emphasis mine; that third sentence is weirding me out. Still thinking about it as an integral of cross sections, I can't see why the surface area or volume of the inner half of the torus should be different from the surface area / volume of the outer half. Are they?