Suppost the region $D$ is defined by $D=\{ (r,\theta)\in\mathbb{R}\times[0,2\pi)\;| \; 0\le r \le 2+\sin(\theta)\}$.
How do I find the centre of mass of this region $(\bar{r}, \bar{\theta})$ assuming uniform density?
I know that to find the mass $M$ you use the double integral with the polar Jacobian $r$: $$ M=\int_0^{2\pi} \int_0^{2+\sin\theta} \rho r\;\textrm{d}r\;\textrm{d}\theta = \frac{9\pi\rho}{2} $$ where $\rho$ is the density of the lamina per unit area.
But then to find the moments w.r.t $r$ and $\theta$, I'm stumped. Any thoughts?
You have total mass
$$M = \int \limits_{0}^{2\pi} \int \limits_{0}^{2+\sin \theta} \rho\,r{\rm d}r \,{\rm d}\theta $$
and you construct the center of mass with
$$ \begin{aligned} x_C & =\frac{1}{M} \int \limits_{0}^{2\pi} \int \limits_{0}^{2+\sin \theta} \rho\,x\,r{\rm d}r \,{\rm d}\theta \\ y_C & =\frac{1}{M} \int \limits_{0}^{2\pi} \int \limits_{0}^{2+\sin \theta} \rho\,y\,r{\rm d}r \,{\rm d}\theta \end{aligned}$$
where $x = r \cos\theta$ and $y=r \sin\theta$.
Finally, the mass moment of inertia is
$$I = \int \limits_{0}^{2\pi} \int \limits_{0}^{2+\sin \theta} \rho\,r^2\,r{\rm d}r \,{\rm d}\theta $$
From the mass integral you find that $\rho = \frac{2 m}{9\pi}$ which you plug into the center of mass and MMOI to get them in terms of mass.
$$ \begin{aligned} x_C & =0 \\ y_C & = \frac{17\pi\,\rho}{4 M} = \frac{17}{18} \end{aligned}$$