# Center of mass for 2d surface

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

• Would be nice if you could share your working? – glowstonetrees May 27 '18 at 21:06

$$x=r\cos\theta$$

$$y=r\sin\theta$$

$$\bar{x}=\frac{1}{m}\iint x\rho(x,y)dA=\frac{1}{m}\int_0^{\pi/2}\int_0^ar^3\cos\theta\sin\theta drd\theta$$

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.

• I think my i forgot to multiply by $r$ (the Jacobian) when calculating the center coordinates. Thanks. – juleand May 27 '18 at 21:37
• That'll do it! All good, you're welcome – AEngineer May 27 '18 at 21:39