Length of Lemniscate of Bernoulli

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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

$\blacksquare$


Sources

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