# Perimeter of an ellipse

How can I calculate the perimeter of an ellipse? What is the general method of finding out the perimeter of any closed curve?

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For what it's worth, this is known as the complete elliptic integral of the second kind. – Rahul Sep 5 '12 at 9:09

For general closed curve(preferably loop), perimeter=$\int_0^{2\pi}rd\theta$ where (r,$\theta$) represents polar coordinates.

In ellipse, $r=\sqrt {a^2\cos^2\theta+b^2\sin^2\theta}$

So, perimeter of ellipse = $\int_0^{2\pi}\sqrt {a^2\cos^2\theta+b^2\sin^2\theta}d\theta$

I don't know if closed form for the above integral exists or not, but even if it doesn't have a closed form , you can use numerical methods to compute this definite integral.

Generally, people use an approximate formula for arc length of ellipse = $2\pi\sqrt{\frac{a^2+b^2}{2}}$

you can also visit this link : http://pages.pacificcoast.net/~cazelais/250a/ellipse-length.pdf

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To add, it's a non-integrable function, i.e., the integral cannot be expressed in terms of elementary functions so you have to resort to numerical methods to evaluate it. Their study led to a very important class of functions called elliptic functions. – ajay Sep 5 '12 at 8:29
@ajay I think the term "non-integrable" function is usually not used for this purpose. – Tunococ Sep 5 '12 at 9:18
By non-integrable, I mean the antiderivative of the function can't be expressed in terms of elementary functions even though the function itself is Riemann integrable. A classic example being the integral $\int e^{-x^2} dx$ – ajay Sep 5 '12 at 9:34
@ajay: Preferable term is 'closed' form – Aang Sep 5 '12 at 9:40

I do not know if that's what you wanted, but the only general method is to calculate the length of the curve. If we have a ellipse equation:

$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$

with parametric representation:

$x=a \cos t, \ \ y=b \sin t, \ \ \ t\in [0,2\pi]$

the length of the curve is calculated knowing:

$x'=-a \sin t, \ \ y'=b \cos t, \ \ \ t\in [0,2\pi]$

and is (see Arc length)

$\int_{0}^{2 \pi} \sqrt{a^{2}\sin^{2}t+b^{2}\cos^{2} t} dt$

this integral can not be solved in closed form. There are various approximations (they take advantage of the power series) that you can see in this link

ellipse

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For any ellipse, its perimeter is given by $p=2πa(1-(\frac{1}{2})^2ε^2-{(\frac{1.3}{2.4})}^2\frac{ε^4}{3}-\cdots)$

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