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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?

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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.

$$ \begin{aligned} x_C & =0 \\ y_C & = \frac{17\pi\,\rho}{4 M} = \frac{17}{18} \end{aligned}$$

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  • $\begingroup$ Wow cheers, very coherent. So to find $\bar{r}$ and $\bar{\theta}$ do I just plug in $\bar{r}=\sqrt{\bar{x}^2+\bar{y}^2}$ and $\bar{\theta}=\arctan{\bar{y}/\bar{x}}$? $\endgroup$ – Sam Spedding May 15 '18 at 12:44
  • $\begingroup$ Also, what is the point in the inertia? Surely you could substitute in $x=r\cos\theta$ and $y=r\sin\theta$ into the integrals for $x_C$ and $y_C$ to get the results at the bottom straight away? $\endgroup$ – Sam Spedding May 15 '18 at 12:51
  • $\begingroup$ The center of mass integrals produces results in terms of $\rho$, which you need to eliminate using the mass equation. MMOI is defined about the origin. If you need to transfer it to the center of mass, use the parallel axis theorem. Finally, yes, convert the Cartesian center of mass $(x_C,y_C)$ into polar form $r_C \angle \theta_C$ as you described. $\endgroup$ – John Alexiou May 15 '18 at 13:15

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