How can I calculate the perimeter of an ellipse? What is the general method of finding out the perimeter of any closed curve?
Perimeter of an ellipse
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0For what it's worth, this is known as the [complete elliptic integral of the second kind](https://en.wikipedia.org/wiki/Elliptic_integral#Complete_elliptic_integral_of_the_second_kind). – 2012-09-05
3 Answers
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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0Perimeter = $\int r d \theta$? Shouldn’t it be $\int \sqrt{r^2 + r^2_\theta} d \theta$? – 2018-05-05
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
For any ellipse, its perimeter is given by $p=2πa(1-(\frac{1}{2})^2ε^2-{(\frac{1.3}{2.4})}^2\frac{ε^4}{3}-\cdots)$