Timeline for Advanced view of the napkin ring problem?
Current License: CC BY-SA 2.5
9 events
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| Jun 29, 2018 at 3:37 | comment | added | Đào Thanh Oai | @MichaelHardy Dear Sir, I know You four year ago on wiki, now this theorem have 9 cite publish in 5 Journals. Can You convert to en.wiki See also: mathoverflow.net/questions/234053/… | |
| Jun 29, 2018 at 3:37 | comment | added | Đào Thanh Oai | @MichaelHardy Dear Sir, I know You four year ago on wiki, now this theorem have 9 cite publish in 5 Journals. Can You convert to en.wiki See also: mathoverflow.net/questions/234053/… | |
| Sep 17, 2010 at 12:38 | comment | added | Michael Hardy | I've neglected this thread for a while; maybe I look it over again today or tomorrow....... | |
| Sep 1, 2010 at 2:13 | comment | added | Steve Huntsman | OK, the volume of the complementary part of the unit sphere is $V_c(z) = 2S(z)+2\pi(1-z^2)z-2\pi(1-z^2)z/3$, where the terms correspond to the two spherical cones, the cylinder, and the "negative" cones, respectively. The solid angle of each spherical cone is $\Omega = 2\pi(1-z)$, so $2S(z) = (4\pi/3)(1-z)$ and $V_c(z) = 4\pi(1-z^3)/3$. Now $V(1,z) = 4\pi/3 - V_c(z) = 4\pi z^3/3$. As to how advanced this is, well I don't think it's advanced at all. But it is a manifest expression of scaling symmetry. | |
| Sep 1, 2010 at 0:32 | comment | added | Michael Hardy | I'm not following your proposed decomposition of the complementary part of the sphere. Can you explain that a bit more long-windedly? | |
| Sep 1, 2010 at 0:31 | comment | added | Michael Hardy | Your identities are certainly correct, but it doesn't seem to address the second paragraph of my question, any more than the argument from Cavalieri's principle does. | |
| Sep 1, 2010 at 0:23 | history | edited | Steve Huntsman | CC BY-SA 2.5 |
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| Aug 31, 2010 at 21:50 | history | edited | Steve Huntsman | CC BY-SA 2.5 |
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| Aug 31, 2010 at 21:09 | history | answered | Steve Huntsman | CC BY-SA 2.5 |