A lamina (two–dimensional plate) occupies the region inside the circle $x^2 + y^2 = 2y$, but outside the circle $x^2 +y^2 = 1$. Find the centre of mass if the density = $\frac{k}{r}$ (inversely proportional to its distance from the origin). The formulae for centre of mass for $x,y$ are below: 
the graph of the two circles is below ($x^2 + y^2 = 2y$ is $x^2 + (y – 1)^2 = 1$ ):
I don't understand how to find the boundaries of region D. I know that it has to be in the form of polar co-ords $(r, \theta)$ but I can't figure it out. Plz help!
Best Answer
For $m$, we have that
$$\begin{align} m&=\int_{\pi/6}^{5\pi/6} \int_1^{2\sin \phi} \frac{k}{\rho}\rho d\rho d\phi\\\\ &=k\int_{\pi/6}^{5\pi/6} (2\sin \phi -1)d\phi\\\\ & =(2\sqrt{3}-2\pi/3)k \end{align}$$
For the term $M_x$, we have that
$$\begin{align} M_x&=\int_{\pi/6}^{5\pi/6} \int_1^{2\sin \phi} \frac{k\rho \cos \phi}{\rho}\rho d\rho d\phi\\\\ &=k\int_{\pi/6}^{5\pi/6} \cos \phi \int_1^{2\sin \phi} \rho d\rho \\\\ &=k\int_{\pi/6}^{5\pi/6}\cos \phi \left(\frac12 (4\sin^2\phi -1)\right)d\phi\\\\ &=0 \end{align}$$
For the term $M_y$, we have that
$$\begin{align} M_y&=\int_{\pi/6}^{5\pi/6} \int_1^{2\sin \phi} \frac{k\rho \sin \phi}{\rho}\rho d\rho d\phi\\\\ &=k\int_{\pi/6}^{5\pi/6} \sin \phi \int_1^{2\sin \phi} \rho d\rho \\\\ &=k\int_{\pi/6}^{5\pi/6}\sin \phi \left(\frac12 (4\sin^2\phi -1)\right)d\phi\\\\ &=\sqrt{3}k \end{align}$$