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Suppose I have a thing such as an ellipse:

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

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?

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    If there would be a simple formula for the perimeter of an ellipse you would have met it in high school $\ldots$2012-02-12

1 Answers 1

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I would write the following

$ds=\sqrt{\left(\frac{dx}{d\theta}\right)^2+\left(\frac{dy}{d\theta}\right)^2}d\theta$

and so

$P=\int_0^{2\pi}\sqrt{a^2\sin^2\theta+b^2\cos^2\theta}d\theta$

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

$P=4bE(e)$

being

$E(e)=\int_0^{\frac{\pi}{2}}\sqrt{1-e^2\sin^2\theta}d\theta$

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

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    @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.2012-02-12