Answer
$u=x, y=e^{u} \cos v \quad z=e^{u} \sin v$ is a parametrization for the surface.
Work Step by Step
Since we revolved this function about the $x$ axis, $x$ is a free parameter, ie. we can take $x=u,$ and then
\[
e^{u} \sin v =z \text { and } e^{u} \cos v=y
\]
$u=x, y=e^{u} \cos v \quad z=e^{u} \sin v$ is a parametrization for the surface.
