## Calculus (3rd Edition)

$$r^2\leq 9 , \quad 0\leq\theta\leq \pi/4 \ or \ 5\pi/4\leq\theta\leq 2\pi.$$
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, $$=r^2\leq 9 , \quad 0\leq\theta\leq \pi/4 \ or \ 5\pi/4\leq\theta\leq 2\pi.$$