Say $K$ is a $n$-dimensional convex subset of $\mathbb R^n$ around the origin. Say we know $Vol_{n - 1}(\pi_{\theta^{\bot}}) K)$ where $\theta^\bot$ is the linear subspace of $\mathbb{R}^n$ orthogonal to the point $\theta$ of $S^{n - 1}$, and $\pi_{\theta^{\bot}}$ is orthogonal projection onto that subspace; also, $Vol_{n - 1}$ is Lebesgue measure on $\mathbb{R}^{n - 1}$. Can we write $Vol_{n}(K)$ as an integral over $Vol_{n - 1}(\pi_{\theta^{\bot}}) K)$ (and say its derivatives with respect to $\theta$ etc). It is ok to include some kind of diameter function in the formula of $Vol_{n}(K)$.
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$\begingroup$ One should be aware that for hyperplanes $H$, the measures of the intersection $K \cap H$ and the image of the orthogonal projection $\proj_H K$ can vary significantly (depending on the body $K$). I think one can expect an integral formula relating $vol_n(K)$ to something like $\int_{H hyperplane} \vol_{n-1} K\cap H$, but not so easily a precise integral formula $\int_{H hyperplane} \vol_{n-1} \proj_H K$. But maybe one just needs to soul-search through the estimates available in Burago-Zalgaller's "Geometric inequalities", e.g. the Bieberbach inequality (thm 11.2.1) or Loomis-Whitney(11.3.1). $\endgroup$– JHMCommented Apr 24, 2013 at 18:39
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2 Answers
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If I understand the question correctly, no. In particular, for a curve of constant width, the width does not determine the area.
answered Apr 24, 2013 at 18:12
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$\begingroup$ We are not talking about arbitrary curve. We are taking about convex sets. In two dimensional case such a formula exists and known as Cauchy's formula. $\endgroup$ Commented Apr 25, 2013 at 5:22
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$\begingroup$ @pallab: can you offer a reference for this cauchy's formula (which must certainly be different than the usual complex analysis formula)? $\endgroup$– JHMCommented Apr 25, 2013 at 6:19
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$\begingroup$ Please have a look: math.utah.edu/~treiberg/IntGeomSlides.pdf. Page 19 $\endgroup$ Commented Apr 25, 2013 at 7:17
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$\begingroup$ I see a formula for the perimeter there, not for the area. $\endgroup$ Commented Apr 25, 2013 at 10:55
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$\begingroup$ $A(\Omega)=\int h (h+\ddot{h}} d\theta$. This is the formula for area not perimeter. It is at the end of page. 19. $\endgroup$ Commented Apr 25, 2013 at 11:11
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Have a look at:
[JFM 34.0649.01 Minkowski, H.
Volumen und Oberfläche. (German)
Math. Ann. 57, 447-495 (1903)] and
[MR0478079 (57 #17572)
Pogorelov, Aleksey Vasilʹyevich.
The Minkowski multidimensional problem.
Translated from the Russian by Vladimir Oliker. Introduction by Louis Nirenberg. Scripta Series in Mathematics. V. H. Winston & Sons, Washington, D.C.; Halsted Press [John Wiley & Sons], New York-Toronto-London, 1978. 106 pp.]
answered Apr 25, 2013 at 6:10