# Prove by the integral definition that total curvature of lemniscate is $0$.

Let for an arc-lenght parametrized curve $$\gamma$$ its total curvature is given by: $$T_\gamma=\int\kappa(s)ds$$ but, by the chain rule, we can see that: $$T_\gamma=\int\kappa(t)\frac{ds}{dt}dt$$ I choose to use the polar form of the lemniscate $$r^2=\cos(2t)$$ and I arrived to: $$\dot{r}^2=\tan(2t)\sin(2t)$$ $$r\ddot{r}=-\cos(2t)[\sec^2(2t)+1]$$ In the other hand, the curvature in polar coordinates is given by: $$\kappa(t)=\frac{r^2+2\dot{r}^2-r\ddot{r}}{(r^2+\dot{r}^2)^{3/2}}$$ so we have: $$T_\gamma=\int_{0}^{2\pi}\frac{r^2+2\dot{r}^2-r\ddot{r}}{(r^2+\dot{r}^2)^{3/2}}(r^2+\dot{r}^2)^{1/2}dt=\int_{0}^{2\pi}\frac{\cos(2t)+2\tan(2t)\sin(2t)+\cos(2t)[\sec^2(2t)+1]}{\cos(2t)+\tan(2t)\sin(2t)}dt$$ but when I perform the integral it results in $$6\pi$$, but that cannot be, because the rotation index of lemniscate is 0, so its total curvature must vanish. Where are my mistake? Can you please help me to derive this result. I've been thinking that maybe my definition of total curvature must change when I use polar coordinates but I cannot find information. Thanks in advantaged.

• Isn't the problem that the polar form of the lemniscate makes sense for those values of $t$ such that $\cos(2t)\geq0$ but your integral works for all values of $t$? – Raskolnikov Jan 21 at 16:24
• In other words, your polar form is simply not an arc-length parametrization as required by your initial definition of the curvature. – Raskolnikov Jan 21 at 16:31