Center of Mass of a Circle How would one find the center of mass of a circle? The center of mass of a rod is given by:
$$\frac{1}{M}\int^{L}_{0}\rho x dx$$
So, for a sphere, it would be an area integral, such as:
$$\frac{1}{M}\int^{?}_{?} \rho ? dA$$ 
The "?" mean I don't know what to put there. 
This is an area integral, which means it needs to be taken over the two axises, correct? 

Your first integral is correct. However, to find the center of mass of a sphere requires three integrals, one for each coordinate. Each of those integrals will look like the one you wrote down.
$$
\bar x = \frac{1}{M}\int_{S} \rho xdA
$$
$$
\bar y = \frac{1}{M}\int_{S} \rho ydA
$$
$$
\bar z = \frac{1}{M}\int_{S} \rho zdA
$$
The limits of integration will have to cover the whole sphere so if you use spherical coordinates, which you should, the vertical angle will go from $0$ to $\pi$ and the azimuth will go from $0$ to $2\pi$. Of course, you'll have to convert $x,y$ and $z$ to whatever coordinate system you are using.
Let's set up $\bar x$ in spherical coordinates as an example. In spherical coordinates
$$
dA = (Rd\theta)(R\sin \theta d\phi) = R^2\sin \theta d\theta d\phi
$$
where I am using $\theta$ for the vertical angle and $\phi$ for the azimuth.
$$
\bar x = \frac{1}{M}\int_{S} \rho xdA = \frac{1}{M}\int_{0}^{2\pi}\int_{0}^{\pi}\rho xR^2\sin \theta d\theta d\phi = \frac{R^2}{M}\int_{0}^{2\pi}\int_{0}^{\pi}\rho x\sin \theta d\theta d\phi
$$
Substituting $x = R\sin \theta \cos \phi$ gives us 
$$
\bar x = \frac{R^2}{M}\int_{0}^{2\pi}\int_{0}^{\pi}\rho(R\sin \theta \cos \phi)\sin \theta d\theta d\phi = \frac{R^3}{M}\int_{0}^{2\pi}\int_{0}^{\pi}\rho \sin^2 \theta \cos \phi d\theta d\phi
$$
