Answer
The statement is false.
Work Step by Step
The volume is enclosed by the cone $z=\sqrt {x^{2}+y^{2}} $ and the plane $z=2$.
If we put $z=2$ in the equation of the cone $z = \sqrt {x^{2}+y^{2}}$, we get $ x^{2}+y^{2} = 4 $. This means that the the cone intersects the plane $z=2$ in the circle $x^{2}+y^{2} = 4 $, so the volume is
$$
V=\int_{0}^{2\pi} \int_{0}^{2} \int_{r}^{2} r dz dr d \theta \neq \int_{0}^{2\pi} \int_{0}^{2} \int_{r}^{2} dz dr d \theta .
$$
Therefore, the statement is false.
