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Frabjous

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- #1

Frabjous

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- #2

anuttarasammyak

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I have no idea but I suppose area of polygon may be easier problem to challenge first.

- #3

Frabjous

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- #4

Frabjous

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I guess I could break it into pyramids with a face for each base. D’oh.

- #5

jim mcnamara

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The general formula is

## {\displaystyle {\frac {1}{3}}\left|\sum _{F}(Q_{F}\cdot N_{F})\operatorname {area} (F)\right|,} ##

... where the sum is over faces F of the polyhedron, QF is an arbitrary point on face F, NF is the unit vector perpendicular to F pointing outside the solid, and the multiplication dot is the dot product

Do not trust my latex here. Use this:

See:

Goldman, Ronald N. (1991), "Chapter IV.1: Area of planar polygons and volume of polyhedra", in Arvo, James (ed.), Graphic Gems Package: Graphics Gems II, Academic Press, pp. 170–171

See for higher dimensions: Büeler, B.; Enge, A.; Fukuda, K. (2000), "Exact Volume Computation for Polytopes: A Practical Study", Polytopes — Combinatorics and Computation, p. 131, CiteSeerX 10.1.1.39.7700, doi:10.1007/978-3-0348-8438-9_6, ISBN 978-3-7643-6351-2

- #6

Frabjous

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Thanks. I think I can find my way now.

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