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Is it possible to integrate a function that would give the perimeter of an ellipse?

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  • $\begingroup$ Something like ArcLength[Circle[{0, 0}, {1, 2}]] or RegionMeasure[Circle[{0,0}, {1, 2}]]? $\endgroup$ – Carl Woll Feb 6 '17 at 23:50
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    $\begingroup$ How to determine the arc length of ellipse? $\endgroup$ – C. E. Feb 6 '17 at 23:52
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    $\begingroup$ Why using numerical methods when there are complete elliptical integrals implemented in Mathematica? en.wikipedia.org/wiki/Ellipse#Circumference $\endgroup$ – Felix Feb 7 '17 at 1:26
  • $\begingroup$ @CarlWoll Do you know why I get two difference answer? $\endgroup$ – yode Jun 15 '17 at 18:20
  • $\begingroup$ I think they are equal under the assumption a>0 && b>0. $\endgroup$ – Carl Woll Jun 15 '17 at 18:41
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$c = 4 a \int\limits_{\theta = 0}^{\pi/2} \sqrt{1 - e^2 \sin^2(\theta)} d\theta = a \pi (2 + e^2)$,

where

$a$ is the semi-major axis length and $e$ the eccentricity.

c[a_, e_] := 4 a Integrate[1 + e^2 Sin[θ]^2, {θ, 0, π/2}]

You can visualize the ellipse by:

ellipseplotter[a_, e_] := 
 ParametricPlot[
  a {Cos[θ], e Sin[θ]}, {θ, 0, 2π}]

ellipseplotter[1, .5]

enter image description here

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To illustrate parametric approach and ArcLength (I acknowledge this was mentioned by @Carl Woll):

r[a_, b_, u_] := {a Cos[u], b Sin[u]}
perimeter[a_, b_] := 
 NIntegrate[
  Sqrt[FullSimplify[D[r[a, b, u], u].D[r[a, b, u], u]]], {u, 0, 2 Pi}]
Manipulate[
 ParametricPlot[r[a, b, t], {t, 0, 2 Pi}, 
  PlotLabel -> 
   Grid[{{"perimeter:", perimeter[a, b]}, {Style["ArcLength", Bold], 
      ArcLength[r[a, b, u], {u, 0, 2 Pi}]}}], 
  PlotRange -> Table[{-3, 3}, 2]], {a, 1, 3}, {b, 1, 3}]

enter image description here

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