Length of Lemniscate of Bernoulli

Theorem
The total length of the lemniscate of Bernoulli given in polar coordinates as:
 * $r^2 = a^2 \cos 2 \theta$

is given by:
 * $L = 4 a \map F {\sqrt 2, \dfrac \pi 4}$

where $F$ denotes the incomplete elliptic integral of the first kind.

Proof
The arc length of a small length increment $\mathrm d s$ is given in polar co-ordinates by:
 * $\paren {\d s}^2 = \paren {r \d \theta}^2 + \paren {\d r}^2$

from which:
 * $\dfrac {\d s} {\d \theta} = \sqrt {r^2 + \paren {\dfrac {\d r} {\d \theta} }^2}$

Half of one lobe of the lemniscate is achieved when $\theta$ goes from $0$ to $\pi / 4$.

Therefore the total length of the lemniscate of Bernoulli is given by:
 * $\displaystyle L = 4 \int_0^{\pi/4} \sqrt {r^2 + \paren {\dfrac {\d r} {\d \theta} }^2} \rd \theta$

First we show:

So:

Thus: