Obtaining the Surface Area of a Superegg with a Given Volume

Question

I have been stuck trying to find an expression for the surface area of a superegg of a given volume. Specifically, the shape I'm looking at is the solid of revolution obtained by rotating a squircle (i.e. the shape graphed by $x^a+y^a=r^a$) around the $x$-axis.

Here's what I've done so far. Integrating to get the solid of revolution I've gotten an expression for volume:

$$V = 2\pi r^3 \frac{B(\frac{1}{a},\frac{2}{a}+1)}{a}$$

Where $B$ is the beta function. I rearranged it to get the radius in terms of $V$ and $a$. Thus, in order to answer my question, I just needed to get an expression for surface area in terms of $a$ which I tried to get by doing the classic $$\int_0^r 2\pi f(x) \sqrt{1 + \dot f(x)^2}dx$$

This is an integration that none of my tools seem to enjoy doing, neither numerically nor analytically. I've also tried to see if this problem has already been figured out in some obscure paper, with no luck. Part of me dreads that the integral for finding the surface area will not have a neat, closed form expression in terms of any combinations of functions, just like that of the perimeter of an ellipse. Is this true? If not, how would one figure out this integral? Or am I going about this whole problem wrong?

Answer 1

I think it can't be simpler than this:

The superegg you're talking about can be written in $\mathbf R^3$ as $$x^a+y^a+z^a=r^a.$$

The surface area can be calculated via

$$\mathcal A=\iint_{S}||\partial_u\mathbf x\times\partial_v\mathbf x||dudv.$$

If we parametrize the superegg equation in modified spherical coordinates choosing $u=\theta$ and $v=\Phi$, we'd get

$$ \mathbf x= \left\{ \begin{aligned} &x=r\sin^{\frac{2}{a}}\theta\cos^{\frac{2}{a}}\Phi\\ &y=r\sin^{\frac{2}{a}}\theta\sin^{\frac{2}{a}}\Phi\\ &z=r\cos^{\frac{2}{a}}\theta \end{aligned} \right. . $$

Since the cross product involves a lot of algebra, I'll provide the result I got which'll have to be computed numerically for sure.

$$ \begin{aligned} \mathcal A &=r^2(s-1)^2\int_0^{2\pi}\int_0^{\pi}\sqrt{\sin^{2s}\theta\cos^{2s-4}\theta(\sin^{2s-4}\Phi\cos^2\Phi + \cos^{2s-4}\sin^2\Phi)\\ + \sin^{4s-6}\theta\cos^2\theta(\sin^{s-2}\Phi\cos^s\Phi + \sin^s\Phi\cos^{s-2}\Phi)^2 }\ d\theta d\Phi\\ &=\sqrt[3]{\dfrac{V^2(s-1)^4}{\pi^2\mathcal{B}^2\left(\frac{s-1}{2},s\right)}}\int_0^{2\pi}\int_0^{\pi}\sqrt{\sin^{2s}\theta\cos^{2s-4}\theta(\sin^{2s-4}\Phi\cos^2\Phi + \cos^{2s-4}\sin^2\Phi)\\ + \sin^{4s-6}\theta\cos^2\theta(\sin^{s-2}\Phi\cos^s\Phi + \sin^s\Phi\cos^{s-2}\Phi)^2 }\ d\theta d\Phi \end{aligned} ,$$

where $s=\dfrac{2}{a}+1$.

Answer 2

I’m going to demonstrate how to get the volume and surface area of rotation for a general shape called supersonics, that I have previously described here.

To my mind, the volume and surface area of bodies of revolution are totally independent and neither one shows up in the calculation of the other. You can have many volumes with the same surface area and vice versa. The methods I use are the Pappus Centroid Theorems.

First, some basics. Superconics can be considered a generalization of the superparabola and superellipse. The squircle is a subset of superconics. The general equation is

$$ \bigg|\frac{X}{a} \bigg|^q+\bigg|\frac{Y}{b} \bigg|^{\frac{1}{p}}=1\\ f(X)=b(1-|X/a|^q)^p $$

in the region $X\in[-a,a],\ \ Y\ge 0$. Without any loss in generality, we can consider the canonical form

$$f(x)=(1-|x|^q)^p\quad x\in[-1,1]$$

We can calculate the general integral for areas, moments, and so on, analytically as follows

$$\int_{-1}^1x^n f^m(x)dx=\frac{2}{q}\text{B}\bigg(\frac{n+1}{q},mp+1\bigg)$$

where B is the beta function. Defining

$$\Psi(p,q)=\frac{2}{q}\text{B}\bigg(\frac{1}{q},p+1\bigg)=\frac{2\Gamma(p+1)\Gamma(1+\frac{1}{q})}{\Gamma(p+1+\frac{1}{q})}=\Psi\bigg(\frac{1}{q},\frac{1}{p}\bigg)$$

we can show that the area under the curve and centroid (relative to the $x$-axis) are given by

$$ A=\Psi(p,q)\\ R=\frac{\Psi(2p,q)}{2\Psi(p,q)} $$

We are now prepared to calculate the volume with Pappus’s $2^{nd}$ Centroid Theorem

$$V=2\pi RA=\pi \Psi(p,q)$$

This is in agreement with your result. Moreover, the centroid is seen to be the area of the square of the function divided by twice its area. Thus, the volume of revolution is seen to be $\pi$ times the area of the square of the function.

Now, the surface area is an entirely different matter. By Pappus’s $1^{st}$ Centroid Theorem,

$$S=2\pi CL$$

Where $C$ is the centroid of the curve (not of the area) and $L$ is the curve length. Neither the centroid or the length can be expressed in terms of $\Psi$ or B. And it is well known that there are very few closed form solutions for curve length. So we may as well acknowledge that we’ll need numerical analysis to get the surface area. However, there is a relatively simple equation for the surface area of revolution in the complex plane. From Pappus’s theorem we can show that

$$S=2\pi\int z|\dot z|du$$

Fortunately, we can express the superconics equation in the complex plane as follows,

$$z=|\cos^2(u)|^{1/q}\text{sgn}(\cos(u))+|\sin^2(u)|^{p}\text{sgn}(\sin(u)),\quad u\in[0,2\pi]$$

Notice that nowhere in the equation for $S$ can we see anything to do with the volume. The complex plane also offers the advantage that we can express closed curves without the ambiguities arising from double-valued functions. In the present calculation we use $u\in[0,\pi]$.

To give an example of how this is to be applied, look at the following Matlab code for such a calculation.

$$ \begin{align} &\text{p=.2; q=5; % a reasonable squiricle}\\ &\text{u=pi*linspace(0,1,1001)';}\\ &\text{z=abs(cos(u).^2).^(1/q).*sign(cos(u))+i*abs(sin(u).^2).^p.*sign(sin(u));}\\ &\text{zdot=gradient(z);}\\ &\text{S=2*pi*imag(trapz(z.*abs(zdot)));} \end{align} $$

Finally, going back to the original problem, you can multiply the volume by $r^3$ and the surface area by $r^2$. And, of course, you can get a much greater variety of shapes than just the squircle.