Answer
\[
r=z, \quad y=r \sin \theta, \quad x=r \cos \theta, \quad \text { where }\left\{\begin{array}{l}
0 \leq \theta \leq 2 \pi \\
0 \leq r \leq 3
\end{array}\right.
\]
Work Step by Step
If we use cylindrical coordenates
\[
z=z, \quad r \sin \theta=y, \quad r \cos \theta=x
\]
and then
\[
z=\sqrt{x^{2}+y^{2}}=\sqrt{r^{2} \cos ^{2} \theta+r^{2} \sin ^{2} \theta}=\sqrt{r^{2}}=r
\]
Since $z=\sqrt{x^{2}+y^{2}} \geq 0$ and $r \leq 3$, then $0 \leq r \leq 3 .$ Using this, we can parametrize the surface:
\[
r=z, \quad y=r \sin \theta, \quad x=r \cos \theta, \quad \text { where }\left\{\begin{array}{l}
0 \leq \theta \leq 2 \pi \\
0 \leq r \leq 3
\end{array}\right.
\]
