Length of Arc of Nephroid

Theorem
The total length of the arcs of a nephroid constructed around a deferent of radius $a$ is given by:
 * $\LL = 12 a$

Proof
Let a nephroid $H$ be embedded in a cartesian plane with its center at the origin and its cusps positioned at $\tuple {\pm a, 0}$.


 * Nephroid.png

We have that $\LL$ is $2$ times the length of one arc of the nephroid.

From Arc Length for Parametric Equations:


 * $\ds \LL = 2 \int_{\theta \mathop = 0}^{\theta \mathop = \pi} \sqrt {\paren {\frac {\d x} {\d \theta} }^2 + \paren {\frac {\d y} {\d \theta} }^2} \rd \theta$

where, from Equation of Nephroid:
 * $\begin{cases}

x & = 3 b \cos \theta - b \cos 3 \theta \\ y & = 3 b \sin \theta - b \sin 3 \theta \end{cases}$

We have:

Thus:

Thus:
 * $\sqrt {\paren {\dfrac {\d x} {\d \theta} }^2 + \paren {\dfrac {\d y} {\d \theta} }^2} = 6 b \sin \theta$

So: