A superellipse is a generalization of an ellipse. It is characterized by the equation \LARGE \left | \frac{x}{a} \right |^p + \left | \frac{y}{b} \right |^p = 1, p > 0. When p = 2, the superellipse becomes a regular ellipse. If p > 2, the superellipse approaches a rectangular shape. It is a diamond when p = 1 and takes on a star-like shape (called an astroid) when p < 1.

In the figure, p = 11, 2, 1, 1/2 going from outside to inside.

The superellipse was first studied by Gabriel Lame in the early 19th century in the midst of his exploration of curvilinear coordinates.

Sergels torg (Sergel’s square) in Stockholm is a superellipse with p=5/2.

One of the classic problems in elementary calculus is to compute the area of an ellipse via integration by trigonometric substitution to derive the formula area = \LARGE ab\pi. Wolfram math world states the formula for the area of a superellipse as area = \LARGE \frac{4^{1-\frac{1}{p}}ab\sqrt{\pi}\Gamma\left(1+\frac{1}{p}\right)}{\Gamma\left(\frac{1}{2}+\frac{1}{p} \right)}. We can obtain this formula in a straightforward manner via integration and manipulating beta and gamma functions.

Let us begin by solving the superellipse equation for \LARGE y and then writing the area integral as \LARGE 4b\int_0^a \left[ 1-\left(\frac{x}{a}\right)^p\right]^\frac{1}{p}dx. We evaluate the integral by first making the substitution, \LARGE y=\frac{x}{a}, this transforms the integral into \LARGE a\int_0^1 \left( 1-y^p \right)^\frac{1}{p}dy. Let \LARGE z=1-y^p so that the integral becomes \LARGE \frac{a}{p} \int_0^1 z^\frac{1}{p} \left( 1-z \right) ^{\frac{1}{p}-1}dz = \frac{a}{p}B\left( \frac{1}{p} + 1,\frac{1}{p} \right). Where B is the beta function. At this point, we can state that the area of a superellipse is equal to \LARGE \frac{4ab}{p}B\left( \frac{1}{p} + 1,\frac{1}{p} \right). Despite this being a nice, compact formula, let us continue to see how we can derive the formula used by Wolfram math world.

We can set \LARGE \frac{a}{p}B\left( \frac{1}{p} + 1,\frac{1}{p} \right) = \frac{a\Gamma\left(\frac{1}{p}+1\right)\Gamma\left(\frac{1}{p}\right)}{p\Gamma\left(\frac{2}{p}+1\right)} using the first equation here.

We now expand the Gamma function in the denominator: \LARGE \Gamma\left(\frac{2}{p}+1\right) = \frac{4^\frac{1}{p}\Gamma\left(\frac{1}{p}+\frac{1}{2}\right)\Gamma\left(\frac{1}{p}+1\right)}{\sqrt{\pi}}.

Substituting this result into the last expression for the integral and using \LARGE \frac{\Gamma\left(\frac{1}{p}\right)}{p} = \Gamma\left(1+\frac{1}{p}\right), yields the desired result.

Thus we have the area of a superellipse = \LARGE \frac{4ab}{p}B\left( \frac{1}{p} + 1,\frac{1}{p} \right) = \frac{4^{1-\frac{1}{p}}ab\sqrt{\pi}\Gamma\left(1+\frac{1}{p}\right)}{\Gamma\left(\frac{1}{2}+\frac{1}{p} \right)}.

Let us see if we can recover the formula for the area of a regular ellipse from that of the superellipse. Let p = 2, then area = \LARGE \frac{4^{1-\frac{1}{2}}ab\sqrt{\pi}\Gamma\left(1+\frac{1}{2} \right)}{\Gamma\left( 1 \right)} = ab\pi, using these formulas to compute values of the Gamma function.

We can also use the Beta function formula. For p = 2, we have area = \LARGE \frac{4ab}{2}B\left(\frac{1}{2}+1,\frac{1}{2} \right) = ab\frac{\Gamma\left(\frac{1}{2}\right)\Gamma\left(\frac{1}{2}\right)}{\Gamma\left(1\right)} = ab\pi.

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