Answer
a) $z^2=1+r\cos \theta-r^2$
b) $z=r^2\cos 2 \theta$
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
$(r\cos \theta)^2-r\cos \theta +(r\sin \theta)^2+z^2=1$
This can be written as:
$r^2(\cos^2 \theta+\sin^2 \theta)-r\cos \theta +z^2=1 $
Hence, we have $z^2=1+r\cos \theta-r^2$
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
$z=(r\cos \theta)^2-(r\sin \theta)^2$
This can be written as:
$z=r^2(\cos^2 \theta-\sin^2 \theta)$
Hence, we have $z=r^2\cos 2 \theta$
