Finding the Center of Mass of a disk when a part of it is cut out.

The center of mass can probably be computed by using difficult integrals, but it can also be computed in a very simple way.

Let's call $\vec{OG}$ the center of mass, $m_{big}$ and $m_{small}$ the mass of the big and small disk respectively. By using the center of mass definition, we have,

$$ \vec{OG} = \frac{1}{m_{big} + m_{small}}(m_{big} \vec{OO} + m_{small} \vec{OO'}) $$

Since the small disk is cut out and not added we should not use a positive mass, but a negative mass for $m_{small}$, therefore, since the surface of the hole is 4 times smaller than the total surface,

\begin{align} m_{small} &= -\frac{1}{4} m_{big} \end{align}

By substituting this value into the first equation, we obtain,

$$\vec{OG} = - \frac{1}{3}\vec{OO'}$$

Given the symmetry you noted, it seems not worth it to do the integral in polar coordinates when you know the center of mass will only have an $x$. Let's rotate your picture upside down. Notice that computing the integral will give us

$$\iint\limits_{\text{shaded region}}^{} x\:dA = \iint\limits_{\text{small circle on right side}}^{} x\:dA$$

by symmetry because $x$ is an odd function. Next, you want integrate w.r.t. $x$ first because the square roots will cancel

$$\bar{x} =\frac{4}{3\pi R^2}\int_{-\frac{R}{2}}^\frac{R}{2} \int_{\frac{R}{2}-\sqrt{\frac{R^2}{4}-y^2}}^{\frac{R}{2}+\sqrt{\frac{R^2}{4}-y^2}} x\:dx \:dy = \frac{4}{3\pi R}\int_{-\frac{R}{2}}^\frac{R}{2} \sqrt{\frac{R^2}{4}-y^2}\:dy = \frac{R}{6}$$

I am assuming the mass per unit area is one for simplicity.

Using symmetry there is no need to actually compute an integral other than to determine masses.

In general, if you have two essentially disjoint sets $A,B$ then $\int_{A \cup B} f = \int_A f + \int_B f$.

If we let $\bar{x}_A = {1 \over \int_A dm } \int_A x dm$, then $\bar{x}_{A \cup B} = { \int_A dm \over \int_{A \cup B} dm } \bar{x}_A + { \int_B dm \over \int_{A \cup B} dm } \bar{x}_B$.

Furthermore, if we translate an object by $d$ we have $\bar{x}_{A +\{d\}} = \bar{x}_A + d$.

Let $A$ be the big disc with the little disk cut out and let $B$ be the little disk.

If we take the centre of the big disk at the origin, we have $\bar{x}_{A \cup B} = 0$, $\bar{x}_B = ({R \over 2},0)$, $\int_{A \cup B} dm = \pi R^2$, $\int_{B} dm = {1 \over 4} \pi R^2$ (and $\int_{A} dm = {3 \over 4} \pi R^2$, of course).

Solving the above for $\bar{x}_A$ gives $\bar{x}_A = -({R \over 6},0)$.