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I'm having a weird issue with the EllipticE function, and just wanted to make sure I've understood it correctly. In most sources (Wikipedia, MathWorld), the perimeter of an ellipse of semi-major axis unity is given by

$P = 4E(e)$

where $e$ is the eccentricity of the ellipse and $E$ the complete elliptical integral of the second kind, where $E(e) = \int_{0}^{\pi/2} \sqrt{1 - e^2\sin^2\theta} d \theta$. In mathematica, the indefinite integral is given by

eqn = Sqrt[1 - e^2 (Sin[x]^2)];
ans = Integrate[eqn, x]

Which gives EllipticE[x, e^2] - If I evaluate this between $\pi/2$ and $0$, I get EllipticE[e^2], but this is odd, as it would imply the perimeter of the ellipse is $P = 4E(e^2)$. To test this, I imagined an ellipse with $e=0.6$, $a = 1$ and $b =0.8$. Then you'd get the following values:

$P_{Texts} = 4E(e) = 5.19371$

$P_{Mathematica} = 4E(e^2) = 5.67233$.

Google has an ellipse perimeter calculator too, and entering the axis values gave me approximately 5.67, in line with Mathematica's definition. Am I missing something really obvious?

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  • $\begingroup$ Compare 4 (ans /. {e -> 0.6, x -> [Pi]/2}) // N and 4 (ans /. {e -> Sqrt[0.6], x -> [Pi]/2}) // N and you will catch the problem $\endgroup$
    – Acus
    Commented Jun 1, 2018 at 13:03
  • $\begingroup$ @user18792 I think I see your point to an extent by using this substitution : eqn = Sqrt[1 - m (Sin[x]^2)]; ans = Integrate[eqn, x] 4 (ans /. {m -> 0.6, x -> Pi/2}) // N 4 (ans /. {m -> 0.36, x -> Pi/2}) // N Problem is, the definition problem 4aE(e) versus 4aE(e^2) still exists. But I'm starting to suspect this is due to some authors being sloppy with their definition of e! Found a 4aE(e^2) source too... curious... mathematica-journal.com/2009/11/on-the-perimeter-of-an-ellipse $\endgroup$
    – DRG
    Commented Jun 1, 2018 at 13:22

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Mathematica has some odd definitions of elliptic functions that are not the same as in Wikipedia. From the Mathematica documentation:

$E(\phi,m)=\int_0^\phi \sqrt{1-m\sin^2\theta}d\theta$

In Wikipedia's definition, the $m$ in the integral is replaced by $m^2$.

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