Dividing concentric super ellipses to equal area slices How the concentric super ellipses as shown in the figure can be divided into parts containing equal area such that the total area of the superellipse A = A1 + A2 + ... An where n = 60 in the shown figure given its semi-major axis a and semi-minor axis b. The origin of the sectors lies at the center of the superellipse. Any help is appreciated.
 A: Solve the problem for concentric circles in the unit circle centered at the origin. Then stretch the $x$-axis by $a$ and the $y$-axis by $b$ to turn the circles into ellipses. All the areas will be scaled by the same stretch factor.
This is essentially the same question you asked here:  Dividing an ellipse into equal area
A: Following the same idea presented here Dividing an ellipse into equal area, it suffices consider the affinity by $X=ax$ and $Y=by$ such that
$$x^2+y^2=1 \to \frac{X^2}{a^2}+\frac{Y^2}{b^2}=1$$
and divide the unit circle in 60 equals parts by the lines


*

*$y=0$

*$y=\pm\frac{\sqrt 3}{3}x$

*$y=\pm{\sqrt 3}x$

*$x=0$


and by the circles


*

*$x^2+y^2=\frac{\sqrt 5}5$

*$x^2+y^2=\frac{\sqrt {10}}5$

*$x^2+y^2=\frac{\sqrt {15}}5$

*$x^2+y^2=\frac{2\sqrt {5}}5$



thus the equations for the lines and ellipses dividing the main ellipse into $60$ sectors of equal area $A=\frac{\pi ab}{60}$ are


*

*$Y=0$

*$Y=\pm\frac{\sqrt 3}{3}\frac b a X$

*$y=\pm{\sqrt 3}\frac b a X$

*$X=0$


for the lines and


*

*$\frac{X^2}{a^2}+\frac{Y^2}{b^2}=\frac{\sqrt 5}5$

*$\frac{X^2}{a^2}+\frac{Y^2}{b^2}=\frac{\sqrt {10}}5$

*$\frac{X^2}{a^2}+\frac{Y^2}{b^2}=\frac{\sqrt {15}}5$

*$\frac{X^2}{a^2}+\frac{Y^2}{b^2}=\frac{2\sqrt {5}}5$

