Efficient approach for calculating Perimeter Of Ellipse

Question

There are many formulas for calculating the perimeter of an ellipse but the most accurate ones are very lengthy with big infinite series.

I want to ask if there is any simpler proof. Any help is appreciated

Answer 1

Let us define, for some $p>0$, the order-$p$ mean between two positive numbers $a,b$ as $$ M_p(a,b) = \sqrt[p]{\frac{a^p+b^p}{2}}. $$ If $a,b$ are the lengths of the semiaxis of an ellipse, the perimeter $L$ of the ellipse fulfills $$2\pi\cdot M_{\frac{\log 2}{\log(\pi/2)}}(a,b) \geq L \geq 2\pi\cdot M_{\frac{3}{2}}(a,b)\qquad (\text{Muir,Alzer,Qiu}) $$

A couple of paragraphs of my notes are devoted to this interesting problem, related to complete elliptic integrals of the second kind and continued fractions. We also have $$ L\approx \pi\left[3(a+b)-\sqrt{(a+3b)(b+3a)}\right]\qquad (\text{Ramanujan}) $$ which is pretty accurate.

Answer 2

consider the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$,$\implies$ upper half is given by the formula, $y=\frac{b}{a}\sqrt{a^2-x^2}.$

Then $$Perimeter=2\int_{-a}^{a}\sqrt{1+\frac{dy}{dx}^2}dx$$