Answer
a) $r(u,v)=\lt u \cos v, u \sin v, 1-u \cos v-u \sin v\gt$ where $0 \le u \le 3$; $0 \le v \le 2\pi$
b) $r(u,v)=\lt 1-u \cos v-u \sin v, u \cos v, u \sin v\gt$ where $0 \le u \le 3$; $0 \le v \le 2\pi$
Work Step by Step
a) Here, we have $x^2+z^2=4$ and $z=1-(x+y)$
Thus, $x=r \cos \theta, y=r \sin \theta , z=z$
Then, we have $r(u,v)=\lt u \cos v, u \sin v, 1-u \cos v-u \sin v\gt$ where $0 \le u \le 3$; $0 \le v \le 2\pi$
b) Here, we have $y^2+z^2=9$ and $x=1-(y+z)$
Thus, $x=r \cos \theta, y=r \sin \theta , z=z$
Then, we have $r(u,v)=\lt 1-u \cos v-u \sin v, u \cos v, u \sin v\gt$ where $0 \le u \le 3$; $0 \le v \le 2\pi$
Hence, our answers are:
a) $r(u,v)=\lt u \cos v, u \sin v, 1-u \cos v-u \sin v\gt$ where $0 \le u \le 3$; $0 \le v \le 2\pi$
b) $r(u,v)=\lt 1-u \cos v-u \sin v, u \cos v, u \sin v\gt$ where $0 \le u \le 3$; $0 \le v \le 2\pi$
