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Let S be a sphere of unit radius in three dimensional Euclidean space, R^3. Given a positive real number e, does there always exist a convex polyhedron P in R^3 such that: (1) S is a subset of P (2) The boundary of P is homeomorphic to the boundary of S (3) The volume of P does not exceed the volume of S by more than e? It is not required that S be tangent to any of the faces of P.

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(2) S has no boundary, so I assume you mean boundary of P is homeomorphic to S. Otherwise, the answer is YES, and this looks an awful lot like a homework problem. –  Igor Rivin Dec 11 '10 at 16:11
    
I interpreted "sphere" to mean "ball". I would not have answered except the OP gives his name and has asked some reasonable questions in the past. –  Bill Johnson Dec 11 '10 at 16:21
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2 Answers

up vote 3 down vote accepted

For small $\epsilon$ let $Q$ be the convex symmetric hull of a finite $\epsilon$ net for the boundary of $S$ and let $P=(1+\epsilon) Q$.

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To Bill Johnson: I should have written "ball" instead oF "sphere". Your answer beautifully clarified for me what should be the right approach to the problem. If we begin with a ball B concentric to S whose volume exceeds that of S by exactly e, then by taking a small enough positive number v, it should be possible to prove the existence of a finite v-net for B whose convex hull, H, contains S as a subset. Then H is the sought for polyhedron. –  Garabed Gulbenkian Dec 11 '10 at 20:25
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This is proposition 17 of book 12 of Euclid's elements.

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@Scott. Proof that you can make a trivial problem hard if you look at it the wrong way. :) –  Bill Johnson Dec 11 '10 at 18:26
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