Suppose I have a thing such as an ellipse:


now we can define it so that $\frac{x}{a}=cos(\theta)$ and $\frac{y}{b}=sin(\theta)$. I know the perimeter formula

$$\mu(S)=\int\sqrt{1+\left(f'(x)\right)^{2}} dx.$$

It is easy to paramerize the ellipse but how can I parametrize the perimeter formula so that I can easily calculate the perimeter?

I find that I am doing things the hard way like this:

$$y=\pm b \sqrt{1-\left(\frac{x}{a}\right)^{2}}$$

now if I plug in the y into the formula of perimeter, it is messy. Can I do it elegantly with parametric form somehow?


I would write the following


and so


and this is just an elliptic integral. The final result takes the form ($b>a$)




and $e^2=1-\frac{a^2}{b^2}$ the eccentricity.

  • $\begingroup$ ...I have never really understoond what the $S$ here means? In physics, it sometimes denotes some surface $S$ but here it is? Do you change the coordinates somehow and use the Jacobian matrix or something like that, sorry thinking alound (maybe totally skewed wrong ideas)... $\endgroup$ – hhh Feb 12 '12 at 15:23
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    $\begingroup$ @hhh: it's the differential corresponding to arclength. $\endgroup$ – J. M. is a poor mathematician Feb 12 '12 at 15:27
  • $\begingroup$ @J.M.: yes but what is is $S$? It is a partial derivative apparently, there is something I cannot grap in the very beginning of this answer -- chain rule ...have to calculate... thinking still aloud... $\endgroup$ – hhh Feb 12 '12 at 15:32
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    $\begingroup$ If we have $x=a\cos\theta$ and $y=b\sin\theta$ as in OP and $a\geq b$ (major axis on $x$ axis, length $2a$), then $e^2=1-\left(\frac{b}{a}\right)^2$ and $P=4aE(e)$; cf. en.wikipedia.org/wiki/Ellipse#Eccentricity & en.wikipedia.org/wiki/Ellipse#Circumference. So in general, $e^2=\frac{|a^2-b^2|}{\max(a,b)^2}$ and $P=4\max(a,b)E(e)$. @hhh: $s$ is arc length, i.e. the length traversed along the portion of curve parametrized in the integral. $\endgroup$ – bgins Feb 12 '12 at 17:34
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    $\begingroup$ @hhh: Indeed, the initial integral goes from $0$ to $2\pi$ but the elliptic integral is given between $0$ and $\frac{\pi}{2}$. The factor 4 comes out as you need four times the elliptic integral. $\endgroup$ – Jon Feb 12 '12 at 20:25

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