How to calculate the area of a generalized ellipse in form $\frac{|x|^c}{a^c} + \frac{|y|^c}{b^c} = 1$? I have a generalized ellipse (ref. Eq(1) in Athanassoula, et al. "The shape of bars in early-type barred galaxies." (via harvard.edu) (1990) )
$$
\frac{|x|^c}{a^c} + \frac{|y|^c}{b^c} = 1
$$
where $a$, $b$ are semi-axis, and $c$ a free parameter in a range $(0,4)$, which is enough for me in my problem; when $c=2$, it returns to normal ellipse.
How to calculate its area?
Thanks!
 A: The cartesian version of the equation of your function is
$$y=\pm b\left(1-\dfrac{|x|^c}{a^c}\right)^{1/c}$$
which represents in fact two functions (resp. the upper and lower parts of the curve according to the choice between signs $+$ or $-$).
But, due to the symmetry of the curve with respect to $x$ and $y$ axes (see figure below), it is sufficient to consider the $+$ case on its $[0,a]$ part, and compute the quarter of this area (situated in the first quadrant) by the following integral :
$$b\int_{x=0}^{a}\left(1-\dfrac{x^c}{a^c}\right)^{1/c}dx=ab\int_{X=0}^{1}\left(1-X^c\right)^{1/c}dX=ab\dfrac{\Gamma(1+\tfrac1c)^2}{\Gamma(1+\tfrac2c)}\tag{1}$$
the second part resulting from the change of variable $X:=\dfrac{x}{a}$, the last part being obtained using Wolfram Alpha (in fact, @Will Jagy has given in a comment below a reference where this result is proved).
with, of course, a final multiplication by $4$.
In the particular case $c=2$ (ordinary ellipse), (1) becomes:
$$ab\dfrac{\Gamma(\tfrac32)^2}{\Gamma(2)}=ab\dfrac{(\tfrac12\Gamma(\tfrac12))^2}{1}=ab\tfrac14 (\sqrt{\pi})^2=ab\tfrac{\pi}{4}$$
as awaited for the quarter of the area of an ellipse with semiaxes $a$ and $b$.

Fig. 1: For $a=3$ and $b=2$. Cases $c=0.5$ (starlike), $c=1$ (lozenge) ... to $c=4$ (almost rectangular).
