Define an ellipse of eccentricity "e" as the real points {x,y} satisfying: (1-e^2)*(x+e)^2 + y^2 = (1-e^2), y[x] = sqrt((1-e^2)*(1-(x+e)^2)). Notice that (e^2-1)*d/de*y[x-e] - e*y[x-e] = 0, and that: (e^2-1)*d/de*y[x] - e*y[x] = d/dx*((e^2-1)*y[x]) = d/dx*Cy[x] . Action angle coordinates {p,q} satisfy {x,y}=sqrt(2*p)*{cos[q],sin[q]}, with y[x] replaced by p[q]=(1/2)*((1-e^2)/(1+e*cos[q]))^2, and then: (e^2-1)*d/de*p[q] - e*p[q] = d/dq*((2*sin[q]+e*cos[q]*sin[q])*p[q]) = d/dq*Cp[q] . Subtracting both equations, and integrating between points on the ellipse circumference, obtains Int_{P1}^{P2} e*( p[q]*dq - y[x]*dx ) = e*(Apq - Axy) = Cy[x_P2] - Cy[x_P1] - Cp[q_P2] + Cp[q_P1], which says that the difference of sectorial area Apq and Cartesian area Axy is a difference of certificates evaluated at P1 and P2. ( I tested this equation numerically and it works, no problem. ) The difference Apq - Axy is also the area difference of a triangle and a quadrilateral, so can alternatively be calculated by trigonometry. When this is done separately, it should then be possible to reverse the derivation and conclude that p[q] is the desired function "1/2*r^2". It satisfies a certain differential equation, with a known, unique certificate. --Brad PS. In case there are any typos in the above, here are the computes: Factor@Dot[ D[Sqrt[(1 - e^2) (1 - x^2)], {e, #}] & /@ {0, 1}, {-e, e^2 - 1}] With[{ye = Sqrt[(1 - e^2) (1 - (x + e)^2)]}, Factor@Plus[Factor@Dot[D[ye, {e, #}] & /@ {0, 1}, {-e, e^2 - 1}], D[(1 - e^2) ye, x]]] With[{r2e = (1/2)*((1 - e^2)/(1 + e Cos[q]))^2}, ReplaceAll[ Factor@Subtract[Dot[D[r2e, {e, #}] & /@ {0, 1}, {-e, e^2 - 1}], D[(2*Sin[q] + e*Cos[q] Sin[q]) r2e, q]], Sin[q]^2 -> 1 - Cos[q]^2]]