2 edited tags
1

# Surface area of superellipsoid (dice)

I'm a physical chemist and I am involved in “colloidal dice”. These are small, cube-like particles with a really nice, regular shape. These particles are not really cubic, but more rounded, much like a dice. I've got a neat way to quantify their size and "roundness" and I'm interested in their volume and surface area. I've already found an expression for the former, but I'm still looking for the latter.

The (three-dimensional) particles have a diameter $r$ and can be described as being a special case of superellipsoids. My superellipsiod is a centrally symmetric particle defined by

$$\label{eq:superellipsoid} \left| \frac{x}{r} \right|^{m} +\left| \frac{y}{r} \right|^{m} + \left| \frac{z}{r} \right|^{m} \leq 1$$

where $x$, $y$ and $z$ are Cartesian coordinates, $r$ is the radius and $m$ is the deformation parameter. Here, $2 \leq m \leq \infty$, where $m=2$ represents a sphere and $m=\infty$ a sharp cube. Most of my dice have a roundness $m=3.5$ and a radius $r\sim80$ nm.

The question: What is the surface are of a centrally symmetric superellipsoid, given the radius $r$ and deformation parameter $m$. A complete proof would be appreciated.

Additional information: The volume of such a superellipsoid is given by

$$\label{eq:volumecuboid} V(r,m) = 8 r^{3} \frac{\displaystyle\left[ \Gamma \left(1+\frac{1}{m}\right) \right]^{3}}{\displaystyle\Gamma \left(1+\frac{3}{m}\right)}$$

and the $\Gamma$-function has the property that

$$\Gamma(n) = (n-1)!$$

where $n$ must be a positive integer.

The two important properties of the gamma function which will be used here are $x \Gamma(x)=\Gamma(x+1)$ and $\Gamma(\frac{1}{2})=\sqrt{\pi}$. Using these properties, it can be shown that for a sphere ($m=2$) with radius $r$:

$$V_{\textrm{sphere}} (r,2) = \frac{4}{3}\pi r^{3}$$

and for a cube ($m = \infty$) with edges $2r$

$$V_{\textrm{cube}} (r,\infty) = (2r)^{3} .$$

But now for the surface area. Thank you for reading.