Answer
The volume is $\frac{16\pi}{3}$.
Work Step by Step
In polar coordinates, the disk $x^2+y^2\leq 4$ is written as $0\leq r\leq 2$ when $0\leq \theta\leq 2\pi$.
Then, the volume of the solid under the cone $z=\sqrt{x^2+y^2}$ or $z=\sqrt{r^2}$ and above the disk $x^2+y^2\leq 4$ is given by $V=\int_0^2\int_0^{2\pi}\sqrt{r^2}\cdot rd\theta dr$.
Evaluate $V$:
$V=\int_0^2\int_0^{2\pi}r^2d\theta dr$
$=\int_0^2[r^2\theta]_0^{2\pi}dr$
$=\int_0^22\pi r^2 dr$
$=\frac{2\pi r^3}{3}]_0^2$
$=\frac{16\pi}{3}$
