Answer
The equation in rectangular coordinates:
${\left( {x - 1} \right)^2} + {y^2} + {z^2} = 1$
This is the equation of a sphere of radius $1$ centered at $\left( {1,0,0} \right)$.
Work Step by Step
In spherical coordinates we have $x = \rho \sin \phi \cos \theta $ and ${\rho ^2} = {x^2} + {y^2} + {z^2}$.
Convert the spherical equation $\rho = 2\cos \theta \sin \phi $ to rectangular coordinates:
$\rho = 2\cos \theta \sin \phi $
$\rho = 2\frac{x}{\rho }$, ${\ \ }$ $2x = {\rho ^2} = {x^2} + {y^2} + {z^2}$
Thus, the spherical equation $\rho = 2\cos \theta \sin \phi $ in rectangular coordinates is
$2x = {x^2} + {y^2} + {z^2}$
${x^2} + {y^2} + {z^2} - 2x = 0$
${\left( {x - 1} \right)^2} - 1 + {y^2} + {z^2} = 0$
${\left( {x - 1} \right)^2} + {y^2} + {z^2} = 1$
This is the equation of a sphere of radius $1$ centered at $\left( {1,0,0} \right)$.
