How to evaluate the length of the perimeter of a low eccentricity ellipse?

Question

Given that $ e= \frac{a^2-b^2}{b^2} $ , and $L$ is the length of the perimeter, which equals $4aE(e, \pi/2)$, find the length of the perimeter up to $e^2$ in terms of $a$ and $b$.

How does one begin this?

Answer 1

Note that the eccentricity of the ellipse is defined by $e:=\sqrt{a^2-b^2}/a$, whereby $a$ is the longer semiaxis.

A beginning: Use the parametrization $$x=a\cos \phi,\quad y=b\sin\phi\qquad(0\leq\phi\leq2\pi)$$ in order to set up the integral for $L$. In the resulting expression substitute $b^2:=a^2(1-e^2)$, whereupon $a$ can be factored out. Now expand the integrand into a power series with respect to $e$ and keep as many terms as needed for the desired accuracy.

Answer 2

If our ellipse is given by the equation $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$, its length is given by: $$ L(a,b)=4a\int_{0}^{\pi/2}\sqrt{1+e^2\sin^2\theta}\,d\theta. \tag{1}$$ Since: $$ \int_{0}^{\pi}\sqrt{1+x^2+2x\cos(2\theta)}\,d\theta = \pi\sum_{n\geq 0}\left(\frac{1}{(2n-1)4^n}\binom{2n}{n}\right)^2 x^{2n}\tag{2} $$ if we define the elliptic parameter $\lambda$ as $\frac{a-b}{a+b}$ it follows that: $$ L(a,b) = \pi(a+b)\left(1+\frac{1}{4}\sum_{n\geq 0}\left(\frac{1}{(n+1)4^n}\binom{2n}{n}\right)^2 \lambda^{2n+2}\right).\tag{3}$$ If we define the mean of order $\alpha>0$ between $a$ and $b$ as: $$ M_{\alpha}(a,b)=\left(\frac{a^\alpha+b^{\alpha}}{2}\right)^{1/\alpha} $$ two classical inequalities about the ellipse length are given by Muir (1883) and Alzer-Qiu (2004):

$$ M_{\frac{3}{2}}(a,b)\leq \frac{L(a,b)}{2\pi}\leq M_{\frac{\log 2}{\log(\pi/2)}}(a,b)\tag{4} $$

and since $\frac{\log 2}{\log(\pi/2)}=1.5349285\ldots$ is pretty close to $\frac{3}{2}$, for low-eccentricity ellipses $2\pi$ times the $\frac{3}{2}$-th mean of $a,b$ is an eccellent approximation for the length.