## Calculus 8th Edition

The equation of a sphere whose center is at $(0,0,\dfrac{1}{2})$ with radius $\dfrac{1}{2}$.
The conversion of rectangular coordinates to spherical coordinates is given as: $x=\rho \sin \phi \cos \theta; y=\rho \sin \phi \sin \theta;z=\rho \cos \phi$ Here, $\rho=\sqrt {x^2+y^2+z^2}$; $\phi =\cos^{-1} [\dfrac{z}{\rho}]; \theta=\cos^{-1}[\dfrac{x}{\rho \sin \phi}]$ Here, we have $\rho =\cos \phi$ This can be written as: $\rho^2=\rho \cos \phi$ But $x=\rho \sin \phi \cos \theta; y=\rho \sin \phi \sin \theta;z=\rho \cos \phi$ Here, $\rho=\sqrt {x^2+y^2+z^2}$; $\phi =\cos^{-1} [\dfrac{z}{\rho}]; \theta=\cos^{-1}[\dfrac{x}{\rho \sin \phi}]$ So, we have $x^2+y^2+z^2=z$ and $x^2+y^2+(z-\dfrac{1}{2})^2=\dfrac{1}{4}$ Also, $x^2+y^2+(z-\dfrac{1}{2})^2=(\dfrac{1}{2})^2$ This is the equation of a sphere whose center is at $(0,0,\dfrac{1}{2})$ with radius $\dfrac{1}{2}$.