What fraction of a regular octahedron is closer to center than any surface

Calculus Level pending

The same method was used as was used in "What fraction of a regular tetrahedron is closer to center than any surface".

This problem's question is its title. Since the volumes of both the regular octahedron and the interior closer space scale with the cube of the edge the fraction remains the same regardless of the chosen edge length. The answer is in parts per billion rounded to an integer, i.e., Round[1000000000 value]. I used an edge length of 1 1 .

In the solution the integral and a closed form are given.


The answer is 94371862.

A Former Brilliant Member by A Former Brilliant Member

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1 solution

∫ 1 4 ( 1 − 3 ) 0 ∫ 1 3 − 1 2 2 − 4 x 2 x 3 ∫ 6 x 2 + 6 y 2 − 1 2 6 2 y 1 d z d y d x \int _{\frac{1}{4} \left(1-\sqrt{3}\right)}^0\int _{\frac{1}{\sqrt{3}}-\frac{1}{2} \sqrt{2-4 x^2}}^{\frac{x}{\sqrt{3}}} \int _{\frac{6 x^2+6 y^2-1}{2 \sqrt{6}}}^{\sqrt{2} y}1dzdydx

3 π 3 − 24 3 + 26 576 2 1 72 2 → 1 8 ( 3 π 3 − 24 3 + 26 ) ≈ 94371862 \frac{\frac{3 \pi \sqrt{3}-24 \sqrt{3}+26}{576 \sqrt{2}}}{\frac{1}{72 \sqrt{2}}}\to \frac{1}{8} \left(3 \pi \sqrt{3}-24 \sqrt{3}+26\right) \approx 94371862

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