Answer
a) $\rho =2 \sin \phi \cos \theta$
b)$\rho =\dfrac{1}{\sin \phi( \cos \theta+2\sin \theta)+3 \cos \phi}$
Work Step by Step
Conversion of rectangular to spherical coordinates is as follows:
$x=\rho \sin \phi \cos \theta; y=\rho \sin \phi \sin \theta;z=\rho \cos \phi$
and
$\rho=\sqrt {x^2+y^2+z^2}$;
$\cos \phi =\dfrac{z}{\rho}$; $\cos \theta=\dfrac{x}{\rho \sin \phi}$
a) Here, $x^2-2x+y^2+z^2=0$
or, $\rho^2 -2 \rho \sin \phi \cos \theta=0$
Hence,$\rho =2 \sin \phi \cos \theta$
b) Here, $x+2y+3z=1$
or,
$\rho \sin \phi \cos \theta+2\rho \sin \phi \sin \theta+3\rho \cos \phi=1$
or, $\rho[ \sin \phi( \cos \theta+2\sin \theta)+3 \cos \phi] =1$
Hence,$\rho =\dfrac{1}{\sin \phi( \cos \theta+2\sin \theta)+3 \cos \phi}$
