computing the $y_{cm}$

Question

Suppose I have a half disc and the coordinates axes at the centre of base of the disc. For the given system, I have surface mass density $S$ as $$S=S_0 sin\theta$$($S_0$ being positive constant). I need to get to the center of mass coordinates. Since the half disc is in $x-y$ plane, so $$z_{cm}=0$$Also since $sin(\frac{\pi}{2}+\theta)=sin(\frac{\pi}{2}-\theta)$, so it turns out that the surface mass density of two points symmetrical about the $y$ axis are equal. This means the mass of left and right part of half disc are equal, so $$x_{cm}=0$$Now for the $y$ centre of mass, I have $$y_{cm}=\frac{\int ydm}{\int dm}=\frac{\int y S_0sin\theta dA}{S_0sin\theta dA}$$How do I compute this integral? Do I use polar coodinates or cartesian ones? I do not see a way of expressing $dA$ in terms of $dx$ and $dy$. Does this turn out to be a surface integral?(I've never evaluated one before)

Thanks in advance. I'm sorry if this looks more of a Maths SE question than a Physics SE question. This is indeed, a physics question.

Answer 1

I'd recommend performing the integrals in polar coordinates $(r, \theta)$ on the plane. In these coordinates, the area element is $$ dA = r\,dr\,d\theta $$ If you're having trouble even after this suggestion, comment with your confusion.

Answer 2

$$y_{cm}=\frac{\int ydm}{\int dm}=\frac{\int_0^\pi\int_0^R(rsin\theta)\times (Ssin\theta)\times rdrd\theta}{\int_0^\pi\int_0^R(Ssin\theta)\times rdrd\theta}=\frac{\int_0^\pi\int_0^Rr^2sin^2\theta drd\theta}{\int_0^\pi\int_0^Rrsin\theta drd\theta}$$ $$y_{cm}=\frac{R}{3}\times\frac{\int_0^\pi(1-cos2\theta)d\theta}{\int_0^\pi sin\theta d\theta}=\frac{\pi R}{6}$$ Whoa, I computed a double integral for the first time!