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