Centre of mass of region using density (multivariable calculus) 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!
 A: 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
$$\begin{align}
M_y&=\int_{\pi/6}^{5\pi/6} \int_1^{2\sin \phi} \frac{k\rho \sin \phi}{\rho}\rho d\rho d\phi\\\\
&=k\int_{\pi/6}^{5\pi/6} \sin \phi \int_1^{2\sin \phi} \rho d\rho \\\\
&=k\int_{\pi/6}^{5\pi/6}\sin \phi \left(\frac12 (4\sin^2\phi -1)\right)d\phi\\\\
&=\sqrt{3}k
\end{align}$$
A: You drew a good picture of the region of integration, though it might
be helpful to you if you make your drawing larger.
On a larger drawing it's easier to add details like these:

The possible values of $\theta$ for integration will range from the angle
of the ray $\overrightarrow{OA}$ to the angle of $\overrightarrow{OB}$.
We know that the length $OA = OB = 1$, and it should be clear by symmetry
that the $y$-coordinate of $A$ is $\frac12,$ and the same for $B$.
The ray $\overrightarrow{OA}$ therefore has $\theta = \arcsin \frac12,$
and you should recognize that $\arcsin \frac12 = \frac\pi6.$
The ray $\overrightarrow{OB}$ is in a symmetric position
with $\theta = \pi - \frac\pi6.$
Within that range of $\theta$ values,
consider an arbitrary line $\overleftrightarrow{OP}$.
The radius of any point on that line within in the region
of integration is at least $1$,
but no greater than the distance $OP$.
That distance is $r$ for a point with polar coordinates
$(r,\theta)$ on the circle $x^2 + y^2 = 2y.$
Subsituting $x^2 + y^2 = r^2$ and $y = r\sin\theta,$ 
the equation of the circle is
$r^2 = 2 r\sin\theta,$
or more simply $r = 2 \sin\theta.$
You should now have all you need to set up the boundaries of the
double integral.
