I can't find the Centre of Mass I am currently trying to find the centre of mass (COM) with a general coordinate R (radius of big circle) of a circle which is missing another circle, with half of the radius of the big circle (R/2). Half of this smaller circle is in the first quadrant, and half is in the fourth quadrant.
I initially tried with an arbitrary value of R, R=2, and got the COM to be 2/6 (the expected value which I know to be true, as x of COM should be R/6). I then proceeded to try and find a general proof of some sort by using R as the value of the radius. 
I seem to be getting some contradiction. For example, my mass moment around the Y-axis ($\frac{1}{3}R^3$) is different to the calculated value when R=2 (have checked this by putting R=2 in $\frac{1}{3}R^3$, of course). This formula for the mass moment around the y-axis was calculated from the definition of a mass moment about the y-axis:
$M_y = \int_{b}^{a} x[f(x)-g(x)] dx$.
This inevtiably means that my calculation of the x-coordinate of the COM for quadrants one and four is incorrect...
$$x = \frac{8}{\pi R^2} \times\left(\int_{0}^{R} x\left(\sqrt{R^2 - x^2} - \sqrt{\frac{R^2}{4} - \left(x - \frac{R}{2}\right)^2}\right)dx\right) $$
Which comes out to be:
$x = \frac{8R}{3\pi}$
This gives the general formula for the x-coordinate of the centre of mass of this shape within the first quadrant, which, as quadrant one and four are symetrical, will be the x-coordinate of the centre of mass for both quadrant one and four (with the corresponding y value being 0); $(\frac{8R}{3\pi},0)$. This x-coordinate must still undergo a weighted average, of sorts, against the normal semi-circle in quadrants two and three. 
The expected output of this general formula for the x-coordinate of the COM for the value R=2 is 0.6977... (an earlier calculated value). However, I seem to get 1.6977..., +1 to the correct value.
Quadrant's two and three will have the COM at position:
$(\frac{4R}{3\pi},0)$. Using this, we can later perform the weighted average aforementioned.
My guess is that the answer to the integral for the mass-moment is incorrect, but as to how to fix that I am unsure.
Could someone possibly help me out please. I'm spending way too much time on this and have gotten myself very confused.
Thanks,
Aidanaidan12

Your integral should be
$$\frac{8}{\pi R^2} \left(\int_{0}^{R} x\left(\sqrt{R^2 - x^2} - \sqrt{\frac{R^2}{4} - \left(x - \frac{R}{2}\right)^2}\right)dx\right) = \frac{8R}{3\pi} - \frac R2$$
and then you get
$$x_{\text{cm}} = \frac{\left(-\frac{4R}{3\pi}\right)\cdot \frac{R^2\pi}2 + \left(\frac{8R}{3\pi} - \frac R2\right)\cdot \frac{R^2\pi}4}{\frac{3R^2}4} =-\frac{R}6$$
which is the correct result.
The integral is much simpler in polar coordinates. The area of the shape is $A = \frac{3R^2}4$ so
\begin{align}
x_{\text{cm}}&= \frac4{3R^2}\int_{\text{shape}} x\,dxdy \\
&= \frac4{3R^2}\left[\int_{\phi= -\frac\pi2}^{\frac\pi2}\int_{r=R\cos\phi}^R r^2\cos\phi \,drd\phi + \int_{\phi= \frac\pi2}^{\frac{3\pi}2}\int_{r=0}^R r^2\cos\phi \,drd\phi\right]\\
&= \frac4{3R^2}\cdot \frac{R^3}3\left[\int_{\phi= -\frac\pi2}^{\frac\pi2} (1-\cos^3\phi)\cos\phi \,d\phi + \int_{\phi= \frac\pi2}^{\frac{3\pi}2}\cos\phi \,d\phi\right]\\
&= \frac4{3R^2}\cdot \frac{R^3}3\left[\left(2-\frac{3\pi}8\right) -2\right]\\
&= -\frac{R}6
\end{align}
