calculate center of mass of a circle

Question

Find center of mass of $D=\{(x,y)|x^2+y^2\le4, x^2+y^2\ge2x\}$ if density is given by $$\rho(x,y)={1\over \sqrt{x^2+y^2}}$$ I seem to not understand why the given solution is what it is. in the solution they do the following:

Let $M_1, M_2$ be the masses of the left and right halves of the circle, thus: $$M_1=2\int_0^2 dr\int_{\pi\over 2}^\pi d\theta=2\pi$$ $$M_2=2\int_0^{\pi\over 2} d\theta\int_{2\cos\theta}^2 dr=2\pi-4$$ $$M_{tot}=4\pi-4$$

(I'm saving you the parametric transformation...) but I wanted to integrate at once, using the inequalities as integration boundary indicators:

$x^2+y^2\le4\bigcup x^2+y^2 \ge2x\Rightarrow 2cos\theta\le r \le2$

$M_{tot}=\int_0^{2\pi}\int_{2cos\theta}^2drd\theta=4\pi$

where is my mistake?

Thank you

Answer 1

The mistake is that in polar coordinates, we have $r\geq 0$. So the constraint actually is

$\max(0,2\cos\theta)\leq r\leq 2$

which is why they split up the integral into $M_1$ (where $\cos\theta\leq 0$) and $M_2$ (where $\cos\theta\geq 0$) in the given solution.