I have calculated with Mathematica the integral:
$$\int_{-h/2}^{h/2}dz\int_{-R}^{R}dx\int_{-\sqrt{R^2-x^2}}^{\sqrt{R^2-x^2}}dy(x^2+z^2)$$
the result is: $$\dfrac{\pi}{12}h^3R^2+\dfrac{\pi}{4}hR^4$$
I am surprised about the $\pi$: where does it comes from?
At some point you are integrating something like:
$$ \int_{-1}^{1}\sqrt{1-x^2}dx $$
The graph of the function $f(x)=\sqrt{1-x^2}$ looks like this:
So you are trying to integrate (find the area) of half the circle. Of course, you would expect $\pi$.
