Center of Mass double integral A lamina occupies the region which is the intersection of $x^2+y^2-2y \leq 0$ and the first quadrant of the $xy$-plane. Find the center of mass if the density at a point of the lamina is twice the point's distance from the origin.
Does the setup of this look like this:$$\int_0^{\frac{\pi}{2}}\int_0^{2\sin\theta}2r^2drd\theta?$$
 A: Actually, the factor of $2$ is unimportant.  The center of mass in the $y$ direction is given by
$$\bar{y} = \frac{\displaystyle \int_0^{\pi/2} d\theta \: \int_0^{2 \sin{\theta}} dr \, r  (2 r)(r \sin{\theta})}{\displaystyle \int_0^{\pi/2} d\theta \: \int_0^{2 \sin{\theta}} dr \, r  (2 r)}$$
Note that the weight of the $y$ coordinate is expressed in the $r \sin{\theta}$ term in the numerator.  Note also the boundary of the lamina is expressed in its polar form, $r=\sin{\theta}$.  Evaluating the radial integrals, we get
$$\bar{y} = \frac{\frac{16}{4} \displaystyle \int_0^{\pi/2} d\theta \: \sin^5{\theta}}{\frac{8}{3} \displaystyle \int_0^{\pi/2} d\theta \: \sin^3{\theta}}$$
or
$$\bar{y} = \frac{6}{5}$$
For $\bar{x}$, we do a similar calculation:
$$\begin{align}\bar{x} &= \frac{\displaystyle \int_0^{\pi/2} d\theta \: \int_0^{2 \sin{\theta}} dr \, r  (2 r)(r \cos{\theta})}{\displaystyle \int_0^{\pi/2} d\theta \: \int_0^{2 \sin{\theta}} dr \, r  (2 r)}\\ &= \frac{\frac{16}{4} \displaystyle \int_0^{\pi/2} d\theta \: \cos{\theta} \sin^4{\theta}}{\frac{8}{3} \displaystyle \int_0^{\pi/2} d\theta \: \sin^3{\theta}}\\ &= \frac{9}{20}\end{align}$$
