Timeline for The Isoperimetric problem for domains constrained to lie between two parallel planes
Current License: CC BY-SA 4.0
13 events
| when toggle format | what | by | license | comment | |
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| Sep 21, 2020 at 21:41 | vote | accept | Jules | ||
| S May 8, 2019 at 21:58 | history | suggested | Glorfindel | CC BY-SA 4.0 |
broken image fixed (click 'rendered output' or 'side-by-side' to see the difference; image retrieved via Wayback Machine); for more info, see https://meta.mathoverflow.net/a/4058/70594
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| May 8, 2019 at 20:03 | review | Suggested edits | |||
| S May 8, 2019 at 21:58 | |||||
| May 13, 2013 at 19:32 | history | edited | user9072 |
edited tags
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| May 13, 2013 at 13:32 | comment | added | Benoît Kloeckner | I took the liberty to improve your title and retag your question. | |
| May 13, 2013 at 13:31 | history | edited | Benoît Kloeckner | CC BY-SA 3.0 |
changed the title and retaged
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| May 12, 2013 at 18:10 | answer | added | Rbega | timeline score: 4 | |
| May 12, 2013 at 8:42 | comment | added | Douglas Zare | I believe the curve you want to rotate is a piece of a nodary between two vertical tangencies. en.wikipedia.org/wiki/Nodary | |
| May 12, 2013 at 5:29 | comment | added | user5810 | ... um, he didn't say that he wants "to count the area where" his "set touches $x = \pm a$ planes" differently from the rest of the area, so ... $\:$ | |
| May 12, 2013 at 4:49 | comment | added | Anton Petrunin | You say "Since the problem is completely symmetric in the y,z directions, the solution can be represented as the rotation of the graph of a function r(x) around the x axis." This is true and if you want to prove it you may use Schwarz symmetrization in the directions of yz-plane. $$ $$ Once you get to this point the remaining part is ODE. It is a simple problem, but I can not help since you do not specify how you want to count the area where your set touches $x=\pm a$ planes. | |
| May 12, 2013 at 2:42 | comment | added | Otis Chodosh | fyi your title is very misleading: in common mathematical terminology a "minimal surface" minimizes the surface area with no constraints on volume (actually "minimal" only means a critical point of area). You are discussing "isoperimetric surfaces" . Also, could you clarify exactly what your question is, it seems like you've found an answer, up to integrating an ODE.... | |
| May 12, 2013 at 1:37 | answer | added | Will Jagy | timeline score: 2 | |
| May 12, 2013 at 1:17 | history | asked | Jules | CC BY-SA 3.0 |