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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0To 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. – 2012-09-05
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0@ajay I think the term "non-integrable" function is usually not used for this purpose. – 2012-09-05
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0By 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 $\int e^{-x^2} dx$ – 2012-09-05
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0@ajay: Preferable term is 'closed' form – 2012-09-05
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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 $p=2πa(1-(\frac{1}{2})^2ε^2-{(\frac{1.3}{2.4})}^2\frac{ε^4}{3}-\cdots)$