This is a rather simple problem, yet i had a hard time understanding the results (if they are correct).
The problem: calculate mass of a quarter of a circle in the positive quadrant in $\mathbb{R}^2$ with radius $a$, where the density is given by $\rho(x,y)=y$, and find the center of mass.
I calculated the required mass using double integral and using polar coordinates ($0\leq r \leq a, 0 \leq \theta \leq \frac{\pi}{4}$), and got:
Mass = $\frac{a^3}{3}$
Center = $(1,\frac{\pi}{4})$ (i.e, constant and independent of the radius)
It didn't made sense to me that the center stays the same.
Thanks
$$x=r\cos\theta$$
$$y=r\sin\theta$$
$$\bar{x}=\frac{1}{m}\iint x\rho(x,y)dA=\frac{1}{m}\int_0^{\pi/2}\int_0^ar^3\cos\theta\sin\theta drd\theta$$
Likewise
$$\bar{y}=\frac{1}{m}\iint y\rho(x,y)dA=\frac{1}{m}\int_0^{\pi/2}\int_0^a r^3\sin^2\theta drd\theta$$
So the $a$ terms do not cancel in the center of mass because the $r$ term is already cubed before integration.