Answer
$$V=\frac{14}{3} \pi$$
Work Step by Step
The solid lies below the cone $z= \sqrt {x^{2}+y^{2}}$ and above the ring $1\leq x^{2}+y^{2} \leq4$.
In polar coordinates, we have $x^{2}+y^{2} $ and $x=r \cos \theta$.
Thus, the disk D is given by: $$ D=\left\{(r, \theta) | 1 \leq r \leq 2, \quad 0 \leq \theta \leq 2\pi \right\} $$
and, the volume of the given solid is given by:
$$ \begin{aligned} V&=\iint_{x^{2}+y^{2} \leq 4} \sqrt{x^{2}+y^{2}} d A \\ &=\int_{0}^{2 \pi} \int_{1}^{2} \sqrt{r^{2}} r d r d \theta \\ &=\int_{0}^{2 \pi} d \theta \int_{1}^{2} r^{2} d r \\ &=[\theta]_{0}^{2 \pi}\left[\frac{1}{3} r^{3}\right]_{1}^{2} \\ &=2 \pi\left(\frac{8}{3}-\frac{1}{3}\right) \\ &=\frac{14}{3} \pi. \end{aligned} $$
