Center of mass (COM) of a uniform quarter disk I was trying to calculate the coordinates of the COM for a uniform quarter disk. By symmetry, it must lie on the angular bisector of the central angle. Also,
$A=\frac{\pi R^2}{4}$, thus $dA=\frac{\pi}{2}xdx$, for an elemental strip ring of width $dx$. Thus, as the disk is uniform, $\frac{dm}{dA} = \frac{4M}{\pi R^2}$, which gives 
$$dm= \frac{4M}{\pi R^2} \frac{\pi}{2}xdx=\frac{2Mx dx}{r^2}.$$
This is pretty much it. To integrate $\int{rdm}$, I need $r$. I have no idea how to get that. I have a hunch that I may have to convert to polar coordinates, but I have nothing else to start with. Please help.
 A: This calculation should be done totally in polar coordinates, not half in cartesian and then in polar. It will give rise to complications then. 
We locate the quarter disk with its vertex at the origin and so that the polar
axis bisects the disk into two congruent figures. This gives us immediately
that $\bar y = 0$. Since the surface mass density is constant, we may assume for the calculation of the center of mass that the density is $\sigma$.
Letting $Q$ be the quarter disk, setting
$$M = \int_Q \sigma \,\ {dA} =\int dm$$
where $M$ is mass of the quarter disk and $N$ is the area of the quarter disk = $\frac{\pi R^2}{4}$, and
$$\bar x = \frac{1}{M}\int x dm = \frac{1}{M} \int_Q x \sigma \,\ dA =\frac{\sigma}{M} \int_Q x  \,\ dA$$
Writing this new integral in polar coordinates we have
$$\int_Q x  \,\ dA = \int^{-\pi/4}_{\pi/4} (\int^R_0 r \cos \theta)\,\ r \,\ dr \,\ d\theta$$
$$=  \int^{-\pi/4}_{\pi/4} \cos \theta (\int^R_0 r^2 \,\ dr )\,\ d\theta$$
$$=  \int^{-\pi/4}_{\pi/4} \cos \theta \cdot\frac{R^3}{3} d\theta$$
$$=  \frac{R^3}{3} \int^{-\pi/4}_{\pi/4} \cos \theta \cdot d\theta$$
$$=  \sqrt{2}\frac{R^3}{3}$$
Therefore
$$\bar x = \sqrt{2}\frac{R^3}{3} \times \frac{\sigma}{M} $$
$$= \sqrt{2}\frac{R^3}{3} \times \frac{\sigma}{\frac{\sigma \pi R^2}{4}}$$
$$= \frac{4 \sqrt{2} R}{3\pi}$$
Geometrically, then center of mass lies on the bisector of the quarter circle, and
about 60% of the way out from the vertex to the rim.
