Martin Gardner, somewhere in the book Mathematical Carnival; talks about superellipses and their application in city designs and other areas. Superellipses(thanks for the link anorton) are defined by the points lying on the set of curves:

$$\left|\frac{x}{a} \right|^n + \left|\frac{y}{b} \right|^n = 1$$

After reading the chapter, I was wondering how to calculate the area of these shapes. So I started by the more simplistic version of supercircles' area:

$$\frac{A}{4}=\int_0^1 \sqrt[n]{1-x^n}dx$$

Although, it looks simple, but I wasn't able to evaluate the integral(except some simple cases, i.e. $n=1,2,\frac{1}{2},\frac{1}{3},\cdots$). So I asked Mathematica to see if its result can shed some light on the integration procedure, the result was:

$$\int_0^1\sqrt[n]{1-x^n}dx=\frac{\Gamma \left(1+\frac{1}{n}\right)^2}{\Gamma \left(\frac{n+2}{n}\right)}$$ where $\Re(n)>0$. But I still couldn't figure out the integration steps. So my question is: how should we do this integration?


It's easy to evaluate the integral in the limit of $n \rightarrow \infty$! One way to do it is using Taylor series expansion, and keeping the relevant terms(only first term in this case).

Some beautiful supercircles are shown in the image bellow:

supercircles beautiful supercircles

As one can see their limiting case is a square.

Also, it will be really nice, if one can calculate the volume of the natural generalization of the curve to 3(or $k$) dimensions:

$$\left|\frac{x}{a} \right|^n + \left|\frac{y}{b} \right|^n +\left|\frac{z}{c} \right|^n = 1$$

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    $\begingroup$ This page gives a parametrization for the bounding curve... it could be that Green's Thm would produce the area, but I haven't worked it out yet. $\endgroup$ – apnorton Aug 3 '13 at 13:39
  • $\begingroup$ @anorton That was an interesting link(with some beautiful curves)! $\endgroup$ – Ali Aug 3 '13 at 13:43
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    $\begingroup$ It might also be mentioned that these are the unit balls of the $L^p$ norms in $R^2$. $\endgroup$ – abnry Aug 3 '13 at 15:10

Let $t=x^n$, hence $dt = nx^{n-1}dx = nt^{1-\frac{1}{n}}dx$ \begin{align*} \int_0^1 \sqrt[n]{1-x^n}dx&=\frac{1}{n}\int_0^1t^{\frac{1}{n}-1}(1-t)^{\frac{1}{n}} dt\\ &=\frac{1}{n}\int_0^1t^{\frac{1}{n}-1}(1-t)^{1 + \frac{1}{n} - 1} dt\\ &=\frac{1}{n}\beta\biggr(\frac{1}{n}, 1+\frac{1}{n}\biggr)\\ &=\frac{1}{n}\frac{\Gamma(\frac{1}{n})\Gamma(1+\frac{1}{n})}{\Gamma(\frac{n+2}{n})}\\ &=\frac{\Gamma(1+\frac{1}{n})^2}{\Gamma(\frac{n+2}{n})} \end{align*}

Wonderful problem presentation by the way! I enjoyed waking up to this.


Hint: Use the change of variables $t=x^{n}$ and then use the $\beta$ function

$$ \mathrm{\beta}(u,v) = \int_0^1 t^{u-1}(1-t)^{v-1}\,dt=\frac{\Gamma(u)\Gamma(v)}{\Gamma(u+v)},\quad \textrm{Re}(u), \textrm{Re}(v) > 0.\, $$

  • $\begingroup$ Indeed! Why did I miss that? $\endgroup$ – Ali Aug 3 '13 at 13:45
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    $\begingroup$ @Ali: We miss things sometimes. We are humans. $\endgroup$ – Mhenni Benghorbal Aug 3 '13 at 13:48

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