# The area of the superellipse

I'm watching this video, where D. Knuth explains the connection of $\pi$ and factorials, and other matters (it is very interesting). Almost at the end of the talk he says the area of the superellipse

$$x^{\frac{1}{\alpha}}+y^{\frac{1}{\alpha}}=1$$

is given by

$$A(\alpha) = \frac{2 \alpha \cdot\Gamma{(\alpha)}^2}{\Gamma{(2 \alpha)}}$$ whic would be

$$A(\alpha) = 2 \alpha B(\alpha,\alpha) = 2 \alpha\int_0^1(1-u)^{\alpha-1}u^{\alpha-1}du$$

I was trying to check this so I put

$$A\left( \alpha \right) = \int\limits_0^1 {{{\left( {1 - {x^{1/\alpha }}} \right)}^\alpha }dx}$$

Now let $x = {u^\alpha }$

$$A\left( \alpha \right) = \alpha \int\limits_0^1 {{{\left( {1 - u} \right)}^\alpha }{u^{\alpha - 1}}du}$$

What's going on?

The $2$ in Knuth's formula probably comes from the fact he considers the full figure and not only a fourth, as I am, but I don't know what I'm doing wrong here. If you want to check, it is at $1:21:00$ aproximately.

PS: Just as a curiosity, does Knuth have a stutter or is it he is just thinking about too many things in too little time?

So it was just OK:

$$A\left( \alpha \right) = \alpha \int\limits_0^1 {{{\left( {1 - u} \right)}^\alpha }{u^{\alpha - 1}}du} = \frac{{\alpha \Gamma \left( {\alpha + 1} \right)\Gamma \left( \alpha \right)}}{{\Gamma \left( {2\alpha + 1} \right)}} = \frac{{\Gamma {{\left( {\alpha + 1} \right)}^2}}}{{\Gamma \left( {2\alpha + 1} \right)}}$$

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 I did the case $\alpha = 2,$ a sort of nonconvex star shape. In the first quadrant it is just $y = 1 + x - 2 \sqrt x,$ with area $1/6,$ so the whole figure has area $2/3.$ Agrees with Dirichlet. – Will Jagy Feb 25 '12 at 20:37 @WillJagy It does. Thanks for you help. Did you receive my response to your mail? – Peter Tamaroff Feb 25 '12 at 21:05

For integrating the constant 1 on $$x \geq 0, \; y \geq 0, \; x^{1/\alpha} + y^{1/\alpha} \leq 1,$$ the result is $$\frac{\alpha^2 \Gamma(\alpha)^2}{\Gamma(1 + 2 \alpha)} = \frac{ \Gamma(1 + \alpha)^2}{\Gamma(1 + 2 \alpha)},$$ multiply by 4 to get the whole thing, $$\frac{ 4 \; \Gamma(1 + \alpha)^2}{\Gamma(1 + 2 \alpha)}.$$ Sample points, $\alpha = 1/2$ is the circle, $\Gamma(3/2) = (1/2) \sqrt \pi$ and $\Gamma(2) = 1,$ so we get $\pi.$ With $\alpha = 1,$ we have a tilted square, $\Gamma(3) = 2,$ area is indeed $2.$ As $\alpha \rightarrow \infty,$ area goes to $0,$ it is not necessary to quote Stirling's to believe that $(\alpha!)^2 / (2 \alpha)! \rightarrow 0.$ Finally, as $\alpha \rightarrow 0,$ we approach the entire square, and the area approaches $4.$