Answer
a) $z^2=1+r\cos \theta-r^2$
b) $z=r^2\cos 2 \theta$
Work Step by Step
a) In the cylindrical co-ordinates system, we have $r^2=x^2+y^2 \\ x=r \cos \theta \\ y=r \sin \theta$
Now, we have $(r\cos \theta)^2-r\cos \theta +(r\sin \theta)^2+z^2=1$
or, $r^2(\cos^2 \theta+\sin^2 \theta)-r\cos \theta +z^2=1 $
so, $z^2=1+r\cos \theta-r^2$
b) In the cylindrical co-ordinates system, we have $r^2=x^2+y^2 \\ x=r \cos \theta \\ y=r \sin \theta$
Now, we have
$z=(r\cos \theta)^2-(r\sin \theta)^2$
This gives:
$z=r^2(\cos^2 \theta-\sin^2 \theta)$
Thus, $z=r^2\cos 2 \theta$
