Chapter 13 - Vector Geometry - 13.7 Cylindrical and Spherical Coordinates - Exercises - Page 699: 13

$$ r^2+z^2\leq 4.$$

Since $$ x=r\cos \theta,\quad y=r\cos \theta, \quad z=z,$$ then $$ y^2+z^2\leq4, \quad x=0$$ $$ x=0 \Longrightarrow r\cos \theta =0\Longrightarrow \theta =\frac{\pi}{2} \ or \ \frac{3\pi}{2} (r\neq 0).$$ and hence we have $$ r^2\sin^2\theta+z^2=r^2+z^2\leq 4, \quad (\theta =\frac{\pi}{2} \ or \ \frac{3\pi}{2} ).$$