15
$\begingroup$

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?


SideNotes:

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:

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

$\endgroup$
  • 3
    $\begingroup$ 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
  • 1
    $\begingroup$ 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
12
$\begingroup$

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.

$\endgroup$
7
$\begingroup$

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

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

$\endgroup$
  • $\begingroup$ Indeed! Why did I miss that? $\endgroup$ – Ali Aug 3 '13 at 13:45
  • 6
    $\begingroup$ @Ali: We miss things sometimes. We are humans. $\endgroup$ – Mhenni Benghorbal Aug 3 '13 at 13:48

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.