= Solution
Use <polar coordinates> $x=r\cos\theta$, $y=r\sin\theta$. Away from the crossing, the <lemniscate of Bernoulli> has $r^2=\cos2\theta$. An explicit parametrization of its four quarter-arcs is
$$
\boxed{x=\pm t\sqrt{\frac{1+t^2}{2}},\qquad y=\pm t\sqrt{\frac{1-t^2}{2}},\qquad 0\leq t\leq1},
$$
where 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 $0\leq\theta\leq\pi/4$, $r=\sqrt{\cos2\theta}$. Differentiating gives $dr/d\theta=-\sin2\theta/r$. Thus the <arc length> element satisfies
$$
\left(\frac{ds}{d\theta}\right)^2=r^2+\left(\frac{dr}{d\theta}\right)^2
=\frac{\cos^22\theta+\sin^22\theta}{\cos2\theta}=\frac1{r^2}.
$$
Set $t=r$, 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, so
$$
\boxed{L=4\int_0^1\frac{dt}{\sqrt{1-t^4}}}.
$$
The endpoint singularity is proportional to $(1-t)^{-1/2}$ and is integrable. This is the geometric origin of the <lemniscatic integral>.
Back to article page
