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  "body": "\n## Statement\n\nLet $P_t$, for $t \\in [0, 1]$, be a continuous flex of a closed orientable polyhedral surface in $\\R^3$.\nThat is, each face of $P_t$ stays congruent to the same face of $P_0$, and the combinatorics stay the same.\nThen the oriented volume enclosed by $P_t$ does not depend on $t$.\n\n## Remarks\n\nThe statement was proved in 1997 by Connelly, Sabitov and Walz, so it is now a theorem.\nThe proof rests on a result of Sabitov: the volume of a polyhedron is a root of a monic polynomial whose coefficients are polynomials in the squared edge lengths.\nAlong a flex the edge lengths are fixed, so the volume takes only finitely many values.\nSince it varies continuously in $t$, it is constant.\n\nExample: every [[flexible polyhedron|ZGD3TQX0K6]], such as Steffen's polyhedron, keeps its volume while it moves.\nSo such a polyhedron cannot work as a bellows that pumps air by changing its volume.\n\n## Explanations\n\n### L1 human\n\nSome polyhedra made of rigid faces with hinged edges can still wobble.\nOne might hope to use such a shape as a bellows, squeezing air in and out.\nThe theorem says this is impossible: the volume inside stays the same during the whole motion.\n\n### L2 human\n\nThe result complements [[Cauchy's rigidity theorem|CJ7KCDDF31]].\nCauchy rules out any flex of a convex polyhedron, while the bellows theorem allows flexes of nonconvex surfaces but fixes their volume.\nIts algebraic proof treats the volume as an algebraic function of the edge lengths, a three-dimensional relative of [[Heron's formula|0FRBZ2X4HJ]].\n",
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    "name": "Bellows conjecture",
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  "msc": [
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  "name": "Bellows conjecture",
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