Continuous uniform distribution over a circle with radius R

I started to do this problem with the standard integration techniques, but I cant help but think that there has got to be something I am not seeing. Since it is a uniform distribution, even though x and y are not independent, it seems like there should be some shortcut. Here is the problem:

Take a random point (x, y) which is uniformly distributed over the circle with radius R

$f(x,y) = \begin{cases} \frac{1}{\pi R^2} & x^2 + y^2 \le R^2, \\ 0 & \text{otherwise} \end{cases}$

calculate $g(x)$, $h(y)$, the mean of $x$, $y$, and $xy$.

Let the pair $(X,Y)$ of random variables have uniform distribution on the disk with centre the origin and radius $R$. Then the joint density function $f_{X,Y}$ is $\frac{1}{\pi R^2}$ on the disk, and $0$ elsewhere.

For the density function of $x$, "integrate out" $y$. So we are integrating a constant from $-\sqrt{R^2-x^2}$ to $\sqrt{R^2-x^2}$. The result is $\frac{2\sqrt{R^2-x^2}}{\pi R^2}$ (for $-R\le x\le R$, and $0$ elsewhere).

The density function of $Y$ can be read off by symmetry.

For the mean of $X$, we don't need to calculate at all. By symmetry it is $0$.

The mean of $XY$ is also $0$. The second and fourth quadrant parts cancel the first and third quadrant parts.