The following Mma code uses integral certificates to derive that "A_s=Sqrt(1-e^2)*(A-A_t)". This is not the same as the previous result, but equivalent when measuring sectorial area from the opposed focus. I have not seen the functions rzCert, xyCert elsewhere, but believe that L. EULER could already have known them. --Brad ========== Mma. Code ========== rad = (1 - e^2)/(1 + e Cos[z]); rzCert = (Sin[z] + (e/2)*Cos[z] Sin[z]) ((1 - e^2)/(1 + e Cos[z]))^2; xyCert = (1 - e^2) (1 - e^2)/(1 + e Cos[z]) Sin[z]; ADiff = Cancel[ Times[Factor[xyCert - rzCert], x y /(rad^2 Cos[z] Sin[z])]/e] xyAInt = Cancel[ ReplaceRepeated[ Sqrt[(1 - e^2)] Integrate[Sqrt[(1 - x^2)], {x, 1, t}][[1]], {t -> x + e, Sqrt[1 - (e + x)^2] -> y/Sqrt[1 - e^2]}]] Factor[xyAInt - ADiff] Factor[% /. {x -> Cos[t] - e, y -> Sqrt[1 - e^2] Sin[t]}] Out[]= (x*y)/2 (1/2)*(e*y + x*y - Sqrt[1 - e^2]*ArcCos[e + x]) (1/2)*(e*y - Sqrt[1 - e^2]*ArcCos[e + x]) (1/2)*Sqrt[1 - e^2]*(-ArcCos[Cos[t]] + e*Sin[t]) On Thu, Dec 26, 2019 at 8:53 AM Brad Klee <bradklee@gmail.com> wrote:
This solution is pretty well recognized: Let “t” be the angle around a unit circle, then A=1/2*t gives circular area, and A_t=1/2*e*sin(t) gives the area of a triangle such that the sectorial area can be written: A_s=Sqrt(1-e^2)*(A+A_t). Invert this equation to get the correct time parameter. I doubt that your pals Julian and Neil are the first to have noticed this.
Your preferred calculation is very similar to the certificate method I was discussing recently. One area integral is easy to calculate and equals the sectorial area by the addition of a polygon volume. I have already shown *on this list* that the same tactic is capable of defining sectorial area relative to Cartesian area, which is calculated by a relatively simple trigonometric integral.
I don’t have my computer with me at the moment, but later I will try to copy your Mathematica code with my alternative approach.
Merry Ellipse-Mas, and a Happy New-Circle,
—Brad
On Dec 25, 2019, at 6:03 PM, Bill Gosper <billgosper@gmail.com> wrote:
Yes! How could we possibly have gone this long without a universally recognized Kepler(a,b,t) function? —Bill