$\iiint_W\sqrt{x^2+z^2}\,\mathrm{d}V$, $W$ is limited by the plane $y=4$ and the paraboloid $y=\sqrt{x^2+z^2}$.
I'm trying to solve with spherical coordinates, however I got stuck in the following part:
$$\int_0^{\pi}\int_0^{\pi}\int_0^{\sqrt{20}}\sqrt{r^2\sin^2(\phi)\cos^2(\theta)+r^2\cos^2(\phi)}\,\,r^2\sin(\phi)\,\mathrm{d}r\,\mathrm{d}\theta\,\mathrm{d}\phi$$
I know I can take $r$ out and solve the first integral, however I can't continue after that.
Edit: The answer given in the list of exercises is $\frac{128\pi}{15}$ and still don't know how to get it.
It's more convenient to use cylindical coordinates; set $x=r\cos\phi$ and $z=r\sin\phi$. Then $\phi\in[0,2\pi)$ and $y\in[0,4]$. To find the limits of $r$, note that $r^2=x^2+z^2$, so $0\leq r\leq\sqrt{y}$. Also, the Jacobian is equal to $r$, therefore the integral is equal to $$\int_0^{2\pi}\int_0^4\int_0^{\sqrt{y}}r\cdot r\,drdyd\phi=2\pi\int_0^4\frac{y^{3/2}}{3}\,dy=\frac{128\pi}{15}.$$