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
 A: We may analyze the moments with respect to the $y$-axis. The full large disc has moment $\pi R^2\cdot0$, but we have to subtract the moment of the left out smaller disc, which is $\pi\left(R\over2\right)^2\cdot{R\over2}$. If $(\rho,0)$ is the center of mass we obtain the equation
$$\pi R^2\cdot0-\pi\left(R\over2\right)^2\cdot{R\over2}={3\pi\over4} R^2\cdot\rho\ ,$$
since the resulting shape has area ${3\pi\over4}R^2$. Solving for $\rho$ leads to
$$\rho=-{R\over6}\ .$$
A: 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}
A: Examining your integral,
$$x = \frac{8}{\pi R^2} \times\left(\int_{0}^{R} x\left(\sqrt{R^2 - x^2} - \sqrt{\frac{R^2}{4} - \left(x - \frac{R}{2}\right)^2}\right)dx\right), $$
you can write
\begin{multline}
2\int_{0}^{R} x\left(\sqrt{R^2 - x^2} - \sqrt{\frac{R^2}{4} - \left(x - \frac{R}{2}\right)^2}\right)dx
= \\
2\int_{0}^{R} x \sqrt{R^2 - x^2} \,dx - 2\int_{0}^{R}x\sqrt{\frac{R^2}{4} - \left(x - \frac{R}{2}\right)^2}\,dx
\end{multline}
where the first term on the right-hand side should come out to the mass moment of a half disk of radius $R$ around its diameter, that is, the integral multiplied by $2$ should come out to
$\frac{4R}{3\pi} \cdot \frac{\pi R^2}{2} = \frac23 R^3$,
and the second term should come out to the mass moment of a disk of radius $\frac R2$ around a tangent line, that is,
$\frac R2 \cdot \frac{\pi R^2}{4} = \frac\pi8 R^3$.
Note that the $\pi$ is not canceled in the second term.
Adding the two terms and then
dividing by the mass in the two quadrants (which is $\frac14{\pi R^2}$),
I get
$$ \frac{8R}{3\pi} - \frac R2$$
for the $x$-coordinate of the COM of the first and fourth quadrants.
Combining this with the other two quadrants, keeping in mind that there is twice as much mass in the second and third quadrants as in the first and fourth,
the term $\frac{8R}{3\pi}$  (multplied by the mass in the first and fourth quadrants) cancels the term 
$-\frac{4R}{3\pi}$ (multplied by the mass in the second and third quadrants),
leaving only the term $- \frac R2$ multiplied by the mass in the first and fourth quadrants; then dividing by the total mass we get $- \frac R3.$
I think a better approach is to resist computing a COM until the very end;
before that, the only thing you should compute is the mass moments around the $y$ axis. You could write your integral as 
$$2 \int_{0}^{R} x\left(\sqrt{R^2 - x^2} - \sqrt{\frac{R^2}{4} - \left(x - \frac{R}{2}\right)^2}\right)dx, $$
giving you the mass moment on the right side of the axis, then add the
(negative) mass moment on the left side.
The result is 
$$2 \int_{0}^{R} x\left(\sqrt{R^2 - x^2} - \sqrt{\frac{R^2}{4} - \left(x - \frac{R}{2}\right)^2}\right)dx + 2 \int_{-R}^{0} x\sqrt{R^2 - x^2}\,dx, $$
which you can simplify so that the $\sqrt{R^2 - x^2}$ terms cancel and all you are left to compute is
$$2 \int_{0}^{R} x\left(- \sqrt{\frac{R^2}{4} - \left(x - \frac{R}{2}\right)^2}\right)dx. $$
At the very end you divide by the mass, which you should have computed to be
$\frac34 \pi R^2.$
