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

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

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

\begin{equation}\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)} \end{equation}

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

\begin{equation} \Gamma(n) = (n-1)! \end{equation}

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 $r$:

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

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

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

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

  • $\begingroup$ Is "superellipsoid" your terminology, or is it standard? $\endgroup$ – Igor Rivin Nov 6 '11 at 21:46
  • $\begingroup$ The results you have on volume were done first by Dirichlet and are in a book from about 100 years ago by Whitaker and Watson, spelling uncertain. I do not expect any closed form for surface area, although, on second glance, I see that all the three letters $r$ are the same, so can be taken to be 1 without penalty. $\endgroup$ – Will Jagy Nov 6 '11 at 21:47
  • $\begingroup$ Igor, the name superellipse was probably coined by Piet Hein in Sweden, he was commissioned to design a traffic oval somewhat between an ellipse and a rectangle. en.wikipedia.org/wiki/Superellipse#History Note that, in three dimensions, similar (and real analytic!) figures can be constructed by taking $g(x,y,z)$ to be a homogeneous symmetric positive polynomial of degree 6, every exponent even, then setting $g=1.$ $\endgroup$ – Will Jagy Nov 6 '11 at 21:53
  • $\begingroup$ @Igor Rivin: That is not my terminology. I read about it for the first time in a journal for chemistry; later on Wolfram Mathworld. mathworld.wolfram.com/Ellipsoid.html mathworld.wolfram.com/Superellipse.html I should note that where I write $x/r$, $y/r$, $z/r$ originally I read $x/a$, $y/b$, $z/c$. However, my particles are centrally symmetric: $r=a=b=c$. $\endgroup$ – Aldo Nov 7 '11 at 14:28
  • $\begingroup$ Here is a figure explaining the influence of varying $m$. Note that I am only interested in $2 \leq m \leq \infty$. Mostly $m=3.5$. cl.ly/0E370V2K333c192K0X3m $\endgroup$ – Aldo Nov 7 '11 at 14:56

You can't expect a closed formula for this surface area. The perimeter of an ellipse, much less the perimeter of a superellipse or the surface area of an ellipsoid or a superellipsoid, is already an integral that doesn't have a formula in the usual sense of an elementary formula. Instead, people did what they always do when an integral is meaningful but has no formula: They named it. That one is called an elliptic integral, but it's just for the case $m=2$. For the $m=2$ case, there is a way out in 3 dimensions: After you define elliptic functions to be the answer in 2 dimensions (for not just the full perimeter but the perimeter of arcs), you can express the 3-dimensional surface area in terms of them without naming any more functions. However, I don't expect that to happen for general $m$, which in any case only reduces the dimension of the problem by 1 instead of solving it.

It's true that your superellipsoids are in the special case that the three semiradii are equal, but I don't think that that rescues you for general values of $m$. Obviously it does rescue you when $m=2$.

What you really want to do for your problem is integrate numerically. You might want a convenient initial change of variables such as $$(a,b,c) = (|x/r|^m,|y/r|^m,|z/r|^m),$$ so that you then get an integral over the triangle $a+b+c=1$. After that there are a variety of numerical integration methods that converge quickly. For instance, you can do the double integral by doing Gaussian integration twice. This particular strategy is not robust as $m \to \infty$, but you didn't say that you needed that limit. An alternative which I guess is robust in that limit, in the case that the semiradii are all equal, is to integrate over one 1/24 region of the spherical angle with the change of coordinates: $$(x,y,z) \propto (1,u,v)$$ with $0 \le u \le v \le 1$. That is also an integral over a triangle.

  • $\begingroup$ I am sure the OP knows how to numerically integrate. The question is: why does he want a closed form? He might want to find the functional dependence on $m,$ for example... $\endgroup$ – Igor Rivin Nov 7 '11 at 8:12
  • $\begingroup$ Thank you for your answer. What I am most interested in are 3 separate cases: $m=2$ (sphere), $m=3.5$ (a representative deformation parameter) and $m=\infty$ (cube). You understand that I'm interested in the sphere and cube cases as a check (extreme case scenario), since of course I already know the volume and surface area of a sphere and a cube. Essentially: I want to know the surface area for the case $m=3.5$ and $r = 80\times10^{-9}$. Later, also other physical cases ($2.5 \leq m\leq 10$ and varying $r$). $\endgroup$ – Aldo Nov 7 '11 at 15:01
  • $\begingroup$ Unfortunately, my skills in numeric integration are poor. I do code in Mathematica. I do really appreciate an (extensive) example calculation. Also, I have difficulties understanding by what you mean with $(x,y,z)\propto (1,u,v)$. My maths schooling doesn't go much further than secondary school. $\endgroup$ – Aldo Nov 7 '11 at 15:07
  • $\begingroup$ The symbol $\propto$ (proportional to) means that you should divide the right side by $(1+u^m+v^m)^{1/m}/r$ to get it to equal the left side. As for numerical integration, Mathematica can do it for you! In any case MathOverflow is for graduate-level and research-level questions in mathematics. You should go to math.stackexchange.com if the issue is schooling and skills. I will point out that you might as well set $r=1$ for your question. $\endgroup$ – Greg Kuperberg Nov 7 '11 at 23:56
  • $\begingroup$ Haha, I get it. Thanks for all the answers all. This helps a lot. Cheers. $\endgroup$ – Aldo Nov 8 '11 at 12:48

This is hardly what you asked, but you might consider a different description for colloidal dice, constants $A,B,C,D$ and $$ A(x^6 + y^6 + z^6) + B ( y^4 z^2 + y^2 z^4 + z^4 x^2 + z^2 x^4 + x^4 y^2 + x^2 y^4) + Cx^2 y^2 z^2 = D.$$ If you take $A=2, B=-1, C=2, D=2$ the result is a slightly non-convex cuboid shape that passes through 26 standard integer lattice points.

For comparison, in two dimensions graph $$ x^4 + 2x^2 y^2 + y^4 = 1, \; x^4 + x^2 y^2 + y^4 = 1, \; x^4 + y^4 = 1 $$ on the same axes, then consider $ x^4 + B x^2 y^2 + y^4 = 1$ for $B > -2.$ Taking $B=-1$ gives $x^4 - x^2 y^2 + y^4 = 1,$ slightly non-convex and passing through eight lattice points,


Sometimes people use the notion "supersphere" for the shape you mention, instead of superellipsoid which is more general.

You can also start from the volume of supersphere

$$\mid x/a\mid^p + \mid y/a\mid^p +\mid z/a\mid^p \leq 1$$

given by: $\ V=\frac{8}{3}\frac{a^3}{p^2}\frac{[\Gamma(1/p)]^3}{\Gamma(3/p)}$. It gives $V=(4\pi a^3)/3$ when $p=2$, and $V=8a^3$ when $p\rightarrow \infty$.

Then we observe the perimeter $\mu$ describing the cuboidability of a supersphere:

$$\mu =\sqrt{2} \cdot \frac{1}{2^{1/p}}$$

giving: $\mu=1$ for $p=2$, and $\mu = \sqrt{2}$ when $p \rightarrow \infty$.

Knowing the volume $V$ of the supersphere, we observe the characteristic radius $r$ satisfying: $$V=(4\pi r^3) / 3$$. Then the surface area S of the supersphere with the volume V is:

$S=4\pi r^2 \cdot f(\mu)$, where $f(\mu)$ increases with increasing of $\mu$, between $2$ and $\sqrt{2}$, and probably can't be written with elementary functions. One can simplify $f$ by approximate function $g$:

$g(\mu)=K\cdot(\mu-1)^2$ where $K=[(6/{\pi})^{1/3}-1] / (3-2\sqrt(2))$.

Hope this helps a bit. Few more details in Prof. Susumu Onaka's papers mentioning superspherical shapes.


Here is an image of Will Jagy's colloidal dice shape:
      alt text

  • $\begingroup$ Thanks for the picture, Joseph. I once drew this as a parametrized surface using spherical coordinates, which works as it is also star-shaped around the origin. I do wonder if Aldo has really found the best description for his dice. I would expect some sort of calculus of variations minimizer, with an unusual combination of spherical and cubical terms in the functional. $\endgroup$ – Will Jagy Nov 7 '11 at 21:45

If we were lucky, the formula for surface area with fixed $m$ and $r$ variable would just be the derivative of the formula for volume (with fixed $m$ and $r$ variable). This is the case for the sphere and the cube.

Have you tried this with any computer approximations?

[edit: computations -> computer approximations]

  • $\begingroup$ this is obviously wrong as the level sets are not equidistant unless $m=\infty$(cube) or $m=2$ (sphere). $\endgroup$ – Vitali Kapovitch Nov 6 '11 at 23:32
  • $\begingroup$ The level sets are also equidistant at least for $m=1$ (octahedron). $\endgroup$ – Noam D. Elkies Nov 7 '11 at 0:22
  • $\begingroup$ true but the Op restricted the range to $m\ge 2$ so I didn't mention the case $m=1$. $\endgroup$ – Vitali Kapovitch Nov 7 '11 at 0:47
  • 1
    $\begingroup$ Yes, removing unfortunate posts is a good idea. You may be able to see the word "delete," if so click on that, and then when it asks you whether you wish to vote to delete say yes. If you cannot see any such option, you can still edit your post down to the minimum 15 characters, such as "Well, never mind." $\endgroup$ – Will Jagy Nov 7 '11 at 2:27
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    $\begingroup$ Given a function $f$ we have that $\frac{d}{dt}vol\{f\le t\}=\int_{f=t}\frac{1}{|\nabla f|}$. For $f(x)=||x||_p=(|x_1|^p+\ldots+|x_n|^p)^{1/p}$ on $\mathbb R^n$ with $n\ge 2,p\ge 1$ it's only true that $|\nabla f|=const$ for $p=1,2$ and $\infty$. That's what makes the formula jbsmoove was suggesting valid for those $p$ and no others. $\endgroup$ – Vitali Kapovitch Nov 7 '11 at 14:46

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