Answer
a) $z^2=2r^2-4$
b) $z=1-2 r\cos \theta+r \sin \theta$
Work Step by Step
a) We know that in the cylindrical co-ordinates $r^2=x^2+y^2 \implies r=\sqrt{x^2+y^2}$ and $x=r \cos \theta \\ y=r \sin \theta$
Thus, we have
$2(x^2+y^2)-z^2=4$
Thus, it can be written as:
$2r^2-z^2=4$ or, $z^2=2r^2-4$
b) We know that in the cylindrical co-ordinates $r^2=x^2+y^2 \implies r=\sqrt{x^2+y^2}$ and $x=r \cos \theta \\ y=r \sin \theta$
Thus, we have
$2r\cos \theta-r\sin \theta+z=1$
Thus, it can be written as:
$z=1-2 r\cos \theta+r \sin \theta$
