This is what I have so far:
Using the formula $\mathrm ds = \sqrt{r^2 + \left(\frac{\mathrm dr}{\mathrm dθ}\right)^2}$
$\frac{\mathrm dr}{\mathrm d\theta} = 3\sin\;\theta $
$r^2 = 9 - 18\cos\;\theta + 9\cos^2\theta$
$\mathrm ds = \sqrt{9 - 18\cos\;\theta + 9\cos^2 \theta + 9\sin^2 \theta}$
using $\cos^2 \theta + \sin^2 \theta = 1$,
$\mathrm ds = \sqrt{18(1-\cos\;\theta)}$
Then I have
$\int_0^{2\pi} \sqrt{18(1-\cos\;\theta)}\mathrm d\theta$
But I'm not sure how to integrate this.