I try to reformulate the commentaries of fgp as an answer.
First consider a probability space $(\Omega, \mathcal{A}, \mathbb{P}')$ and a random variable uniformly distributed on $[0,2\pi]$:
$$X: (\Omega, \mathcal{A}, \mathbb{P}') \rightarrow ([0,2\pi],\mathcal{B}([0,2\pi])).$$
Furthermore consider the parametrization of the unit circle
$$p: [0,2\pi] \rightarrow S^1; \quad x \mapsto e^{i\cdot x}$$
which is continuous.
Now consider the space $(S^1,\mathcal{B}(S^1))$ where $\mathcal{B}(S^1)$ is the $\sigma$-algebra generated by the open sets of $S^1$.
We define $\mathbb{P}$ to be the probability measure induced by the map
$$p\circ X: (\Omega, \mathcal{A}, \mathbb{P}') \rightarrow (S^1,\mathcal{B}(S^1)); \quad \omega \mapsto e^{i X(\omega)}.$$
Then we have
\begin{equation}
P(Z\in A) = \frac{1}{2\pi i}\int_{A}{\frac{1}{z}}dz \qquad \forall A \in \mathcal{B}(S^1)
\end{equation}
or more generally
$$E[f(Z)]=\frac{1}{2\pi i}\int_{S^1}{\frac{f(z)}{z}}dz$$
for any measurable, bounded function $f: S^1 \rightarrow \mathbb{R}$
So we must be careful what the measurable space really is. In the case of the "density" in the question we are considering the probability space $(S^1,\mathcal{B}(S^1), \mathbb{P})$ and on this we get can give the value of $\mathbb{P}$ via the above formula.
Hence the above is not a density with respect to the Lebesgue-measure on $\mathbb{C}$, but a way of formulating the value with the help of the parametrization of (or - to be more precise - a contour integral along) the unit circle $S^1$.
By definition, the marginal density of $X$ is simply $$f_X(x) = \int_{y=-1}^1 f_{X,Y}(x,y) \, dy = \int_{y=-\sqrt{1-x^2}}^{\sqrt{1-x^2}} \frac{1}{\pi} \, dy.$$ The second equality arises from the fact that $$f_{X,Y}(x,y) = \frac{1}{\pi} \mathbb{1}(x^2+y^2 \le 1),$$ from which we see that for a given $X = x$, the support of $Y$ is then $-\sqrt{1-x^2} \le Y \le \sqrt{1-x^2}$.
Best Answer
The area of the disk is $\pi$. We have a uniform distribution on the disk, so the probability of landing in a part of the disk with area $A$ is proportional to $A$, say $kA$. Since the probability of landing in the unit disk is $1$, we have $k\pi=1$ and therefore $k=\dfrac{1}{\pi}$. Thus the probability of landing in a part of the disk with area $A$ is $\dfrac{A}{\pi}$.
Now what function $f(x,y)$ is it that integrated over any part of the disk with area $A$ gives result $\dfrac{A}{\pi}$? The constant function $\dfrac{1}{\pi}$.