Answer
a) $z^2=2r^2-4$ and
b) $z=1-2 r\cos \theta+r \sin \theta$
Work Step by Step
a) In the cylindrical coordinate system, we have $x=r \cos \theta \\ y=r \sin \theta \\z=z$
Conversion of rectangular to cylindrical coordinate system, we have $r^2=x^2+y^2 \\ \tan \theta=\dfrac{y}{x} \\z=z$
$2(x^2+y^2)-z^2=4$
Re-arrange it as: $2r^2-z^2=4$ or, $z^2=2r^2-4$
b) In the cylindrical coordinate system, we have $x=r \cos \theta \\ y=r \sin \theta \\z=z$
Conversion of rectangular to cylindrical coordinate system, we have $r^2=x^2+y^2 \\ \tan \theta=\dfrac{y}{x} \\z=z$
Now, $2r\cos \theta-r\sin \theta+z=1$
Re-arrange the above equations as:
$z=1-2 r\cos \theta+r \sin \theta$
