I was browsing through pages on Wolfram Mathworld, and I came through this interesting page about the 'Rectellipse.' The page had formulas for the area and perimeter of the shape in two dimensions with the equation:
$x^4+y^4=1$
I was wondering how one could calculate the volume and surface area of the same shape in 3D (3D Rectellipse), with the equation:
$x^4+y^4+z^4=1$
I couldn't come up with much so I asked. Thanks!
2026-03-27 16:53:54.1774630434
Volume and surface area of a fourth degree superellipse
359 Views Asked by Bumbble Comm https://math.techqa.club/user/bumbble-comm/detail At
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There are 1 best solutions below
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don't know about surface area, the volume in the first octant is $$ \frac{ \Gamma \left( \frac{5}{4} \right)^3}{\Gamma \left( \frac{7}{4} \right)} $$ so the whole figure has volume $$ \frac{ 8 \; \Gamma \left( \frac{5}{4} \right)^3}{\Gamma \left( \frac{7}{4} \right)}. $$
My calculator says this is about $6.481987,$ smaller than the cube of edge two $8,$ but bigger than the sphere of radius one $4.18879.$
Method was published by Dirichlet, article title: Über eine neue Methode zur Bestimmung vielfacher Integrale