$$ \int_{0}^4 \int_{0}^{\sqrt{4x-x^2}}\int_{0}^a z^2\cdot\sqrt{x^2+y^2}\,\mathrm dz\,\mathrm dy\,\mathrm dx$$
My attempt: It seems that as it is, it would be rather difficult to compute this. After drawing this a set it seems that I'm trying to integrate over the 'positive' half of a ball $(x-2)^2+y^2 \leq 4$ starting from the plane $z=0$ all up to $z=a$.
So I tried to switch to polar coordinates, using $ x = 2 + r\cos\phi,\, y = r\sin\phi$.
$$\int_{0}^{2\pi}\int_{0}^{1}\frac{a^3}{3}\sqrt{4+4r\cos\phi+r^2}r\,\mathrm dr\,\mathrm d\phi$$
Now I'm not sure what to do really, any hint would be great!
I would do the switch to standard polar coordinates, $x = r \cos \phi$ and $y = r \sin \phi$.
Then, your integral should become:
$$\frac{a^3}{3}\int_0^{\frac{\pi}{2}} \int_0^{4\cos \theta} r^2 dr d\theta$$