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?
 A: 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
A: 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
A: For any ellipse, its perimeter is given by
$p=2πa(1-(\frac{1}{2})^2ε^2-{(\frac{1.3}{2.4})}^2\frac{ε^4}{3}-\cdots)$
A: If the semi-major axis has length $a$ and the eccentricity is $e$, then the perimeter is
$$
2πa \left(\frac{1}{1} \left(\frac{(-1)!!}{0!!} e^0\right)^2 + \frac{1}{-1} \left(\frac{1!!}{2!!} e^1\right)^2 + \frac{1}{-3} \left(\frac{3!!}{4!!} e^2\right)^2 + \frac{1}{-5} \left(\frac{5!!}{6!!} e^3\right)^2 + ⋯\right)
$$
where $(-1)!! = 1 = 0!!$, and for all $n ∈ \{⋯, -7, -5, -3, -1, 0, 1, 2, 3, 4, 5, ⋯ \}$: $(n+2)!! = (n+2) n!!$. The sum extends to negative powers of $e$, if you take $n!! = ∞$ for $n ∈ \{⋯, -8, -6, -4, -2\}$; and if you work out the double-factorials, using:

*

*$(2n)!! = 2^n n!$, for $n ≥ 0$;

*$(2n-1)!! = \frac{(2n)!}{2^n n!}$, for $n ≥ 0$;

*$(2n-1)!! = \frac{(-1)^n}{(-2n - 1)!!} = (-2)^{-n} \frac{(-n)!}{(-2n)!}$, for $n ≤ 0$;

then it comes out to the following:
$$
2πa \left(1 - \left(\frac{1}{2} e\right)^2 - \frac{1}{3} \left(\frac{1}{2} \frac{3}{4} e^2\right)^2 - \frac{1}{5} \left(\frac{1}{2} \frac{3}{4} \frac{5}{6} e^3\right)^2 + ⋯\right).
$$
Im my previous life oops I said that out loud I offered the following estimate:
$$\mbox{Ramanujan's Formula}:
π(3(a+b) - \sqrt{(3a+b)(a+3b)}) = π(a+b) (3 - \sqrt{4 - h}),$$
where $b = a \sqrt{1 - e^2}$ is the length of the semi-minor axis and $h = ((a - b)/(a + b))^2$, while secretly holding onto the other, much better, estimate:
$$π(a+b) \left(\frac{12 + h}{8} - \sqrt{\frac{2 - h}{8}}\right).$$
