Past exam of the mathematics course of the University of Cambridge / 2001 / iii / Paper 8 / 1 / i

Use polar coordinates $x=r\cos \theta$, $y=r\sin \theta$. Away from the crossing, the lemniscate of Bernoulli has $r^2=\cos 2\theta$. An explicit parametrization of its four quarter-arcs iswhere the two signs are independent. Indeed $x^2+y^2=t^2$ and $x^2-y^2=t^4$. The four arcs meet at the origin and at the two lobe endpoints, covering the entire curve.

For the upper-right quarter, take $0\le \theta \le \pi /4$, $r=\sqrt{\cos 2\theta }$. Differentiating gives $dr/d\theta =-\sin 2\theta /r$. Thus the arc length element satisfiesSet $t=r$, which decreases from one to zero. Since $|dt/d\theta |=\sqrt{1-t^4}/t$, it follows that $ds=|dt|/\sqrt{1-t^4}$. All four quarter-arcs have the same length, soThe endpoint singularity is proportional to $(1-t{)}^{-1/2}$ and is integrable. This is the geometric origin of the lemniscatic integral.