# 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:

 $\ds L$ $=$ $\ds 4 a \map F {\sqrt 2, \dfrac \pi 4}$ $\ds$ $=$ $\ds \dfrac 1 {\sqrt {2 \pi} } \paren {\map \Gamma {\dfrac 1 4} }^2$

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

## Proof

The arc length of a small length increment $\d s$ is given in polar co-ordinates by:

$\paren {\d s}^2 = \paren {r \rd \theta}^2 + \paren {\d r}^2$

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:

$\ds L = 4 \int_0^{\pi/4} \sqrt {r^2 + \paren {\dfrac {\d r} {\d \theta} }^2} \rd \theta$

First we show:

 $\ds r^2$ $=$ $\ds a^2 \cos 2 \theta$ $\ds \leadsto \ \$ $\ds 2 r \frac {\d r} {\d \theta}$ $=$ $\ds -2 a^2 \sin 2 \theta$ $\ds \leadsto \ \$ $\ds \frac {\d r} {\d \theta}$ $=$ $\ds -\frac {a^2 \sin 2 \theta} {a \sqrt {\cos 2 \theta} }$ $\ds \leadsto \ \$ $\ds \paren {\dfrac {\d r} {\d \theta} }^2$ $=$ $\ds \frac {a^2 \sin^2 2 \theta} {\cos 2 \theta}$

So:

 $\ds r^2 + \paren {\dfrac {\d r} {\d \theta} }^2$ $=$ $\ds a^2 \cos 2 \theta + \frac {a^2 \sin^2 2 \theta} {\cos 2 \theta}$ $\ds$ $=$ $\ds a^2 \frac {\cos^2 2 \theta + \sin^2 2 \theta} {\cos 2 \theta}$ $\ds$ $=$ $\ds \frac {a^2} {\cos 2 \theta}$ Sum of Squares of Sine and Cosine

Thus:

 $\ds L$ $=$ $\ds 4 a \int_0^{\pi / 4} \frac {\d \theta} {\sqrt {\cos 2 \theta} }$ $\ds$ $=$ $\ds 4 a \int_0^{\pi / 4} \frac {\d \theta} {\sqrt {1 - 2 \sin^2 \theta} }$ $\ds$ $=$ $\ds 4 a \map F {\sqrt 2, \dfrac \pi 4}$ Definition of Incomplete Elliptic Integral of the First Kind

$\blacksquare$