I'm trying to calculate the arc length of the innerloop see picture of a cardioid-like $r=4a\cos^3\frac\theta3$ but don't know where to start. I know you have formulas like $L=\int_{a}^{b} \sqrt{r^{2}+\left(\frac{d r}{d \theta}\right)^{2}} d \theta$, but don't know how to do this for the innerloop.
thanks in advance
We can take out constants and multiply them back in at the end, so the curve is $r=\cos^3\frac{\theta}3$.
The required length is twice the length of the curve from $\theta=\pi$ to $\theta=\frac{3\pi}2$. Thus $$\frac L2=\int_\pi^{3\pi/2}\sqrt{\cos^6\frac\theta3+\cos^4\frac\theta3\sin^2\frac\theta3}\,d\theta$$ $$=\int_\pi^{3\pi/2}\sqrt{\cos^4\frac\theta3}\,d\theta=\int_\pi^{3\pi/2}\cos^2\frac\theta3\,d\theta=\frac34\left(\frac\pi3-\frac{\sqrt3}2\right)$$ Therefore the length of the inner loop for $r=4a\cos^3\frac\theta3$ is $a(2\pi-3\sqrt3)$.