Let a nephroid $H$ be embedded in a cartesian plane with its center at the origin and its cusps positioned at $\tuple {\pm a, 0}$. :600px 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: {{begin-eqn}} {{eqn | l = \frac {\d x} {\d \theta}       | r = -3 b \sin \theta + 3 b \sin 3 \theta       | c =  }} {{eqn | l = \frac {\d y} {\d \theta}       | r = 3 b \cos \theta - 3 b \cos 3 \theta       | c =  }} {{end-eqn}} Thus: {{begin-eqn}} {{eqn | o =        | r = \paren {\frac {\d x} {\d \theta} }^2 + \paren {\frac {\d y} {\d \theta} }^2       | c =  }} {{eqn | r = \paren {-3 b \sin \theta + 3 b \sin 3 \theta}^2 + \paren {3 b \cos \theta - 3 b \cos 3 \theta}^2       | c =  }} {{eqn | r = 9 b^2 \paren {\paren {-\sin \theta + \sin 3 \theta}^2 + \paren {\cos \theta - \cos 3 \theta}^2}       | c =  }} {{eqn | r = 9 b^2 \paren {\sin^2 \theta - 2 \sin \theta \sin 3 \theta + \sin^2 3 \theta + \cos^2 \theta - 2 \cos \theta \cos 3 \theta + \cos^2 3 \theta}       | c =  }} {{eqn | r = 9 b^2 \paren {2 - 2 \sin \theta \sin 3 \theta - 2 \cos \theta \cos 3 \theta}       | c = Sum of Squares of Sine and Cosine }} {{eqn | r = 18 b^2 \paren {1 - \sin \theta \sin 3 \theta - \cos \theta \cos 3 \theta}       | c =  }} {{eqn | r = 18 b^2 \paren {1 - \sin \theta \paren {3 \sin \theta - 4 \sin^3 \theta} - \cos \theta \cos 2 \theta}       | c = Triple Angle Formula for Sine }} {{eqn | r = 18 b^2 \paren {1 - \sin \theta \paren {3 \sin \theta - 4 \sin^3 \theta} - \cos \theta \paren {4 \cos^3 \theta - 3 \cos \theta} }       | c = Triple Angle Formula for Cosine }} {{eqn | r = 18 b^2 \paren {1 - 3 \sin^2 \theta + 4 \sin^4 \theta - 4 \cos^4 \theta + 3 \cos^2 \theta}       | c = simplifying }} {{eqn | r = 18 b^2 \paren {1 + 3 \cos 2 \theta + 4 \sin^4 \theta - 4 \cos^4 \theta}       | c = Double Angle Formula for Cosine }} {{eqn | r = 18 b^2 \paren {1 + 3 \cos 2 \theta + \dfrac {3 - 4 \cos 2 x + \cos 4 x} 2 - 4 \cos^4 \theta}       | c = Power Reduction Formula for $\sin^4$ }} {{eqn | r = 18 b^2 \paren {1 + 3 \cos 2 \theta + \dfrac {3 - 4 \cos 2 \theta + \cos 4 \theta} 2 - \dfrac {3 + 4 \cos 2 \theta + \cos 4 \theta} 2}       | c = Power Reduction Formula for $\cos^4$ }} {{eqn | r = 18 b^2 \paren {1 - \cos 2 \theta}       | c = simplifying }} {{eqn | r = 18 b^2 \paren {2 \sin^2 \theta}       | c = Square of Sine }} {{eqn | r = 36 b^2 \sin^2 \theta       | c = simplifying }} {{end-eqn}} Thus: :$\sqrt {\paren {\dfrac {\d x} {\d \theta} }^2 + \paren {\dfrac {\d y} {\d \theta} }^2} = 6 b \sin \theta$ So: {{begin-eqn}} {{eqn | l = \LL       | r = 2 \int_0^\pi 6 b \sin \theta \rd \theta       | c =  }} {{eqn | r = 12 b \int_0^\pi \sin \theta \rd \theta       | c =  }} {{eqn | r = 12 b \bigintlimits {-\cos \theta} 0 \pi       | c =  }} {{eqn | r = 12 b \paren {-\cos \pi - \paren {-\cos 0} }       | c =  }} {{eqn | r = 12 b \paren {-\paren {-1} - \paren {-1} }       | c =  }} {{eqn | r = 24 b       | c =  }} {{eqn | r = 12 a       | c =  }} {{end-eqn}} {{qed}}