Use calculus!
For any region $R$ with density function $\rho(x,y)$ we can compute a number of quantities from it.
First off, mass:
$$m=\iint_R\rho(x,y)\ dA$$
Now we need the first moments, or the moments of mass:
$$M_x = \iint_Ry\ \rho(x,y)\ dA\\M_y=\iint_Rx\ \rho(x,y)\ dA$$
The center of mass is then
$$(\bar x, \bar y) = \left(\frac{M_y}{m}, \frac{M_x}{m}\right)$$
Now applying this to the problem in question, we can first of all set $\rho(x,y)=1$. Doing this, several terms drop out or cancel, and we have
$$(\bar x, \bar y)=\left(\frac{\iint_Rx\ dA}{\iint_RdA},\frac{\iint_Ry\ dA}{\iint_RdA}\right)$$
$\iint_RdA$ is really just the area of $R$, which in this case is exactly $\frac{\pi r^2}{2}$, so substituting into the previous equation yields
$$(\bar x, \bar y)=\left(\frac{2\iint_Rx\ dA}{\pi r^2},\frac{2\iint_Ry\ dA}{\pi r^2}\right)$$
All that remains is to evaluate those double integrals, which is accomplished by choosing suitable bounds in order to replace them with one or more iterated integral.