Answer
$$ r^2\sin^2\theta +z^2\leq 9
, \quad 0\leq\theta\leq \pi/4 \ or \ 5\pi/4\leq\theta\leq 2\pi.$$
Work Step by Step
Since
$$ x=r\cos \theta,\quad y=r\cos \theta, \quad z=z,$$
then $$ y^2+z^2\leq 9, \quad x\geq y $$
takes the form
$$ r^2\sin^2\theta +z^2\leq 9 .$$
Moreover, $$ x\geq y\Longrightarrow \cos\theta \geq \sin\theta\\
\Longrightarrow 0\leq\theta\leq \pi/4 \ or \ 5\pi/4\leq\theta\leq 2\pi.$$
Hence, $$ r^2\sin^2\theta +z^2\leq 9
, \quad 0\leq\theta\leq \pi/4 \ or \ 5\pi/4\leq\theta\leq 2\pi.$$
