Answer
The spherical coordinates is $\left( {\rho ,\theta ,\phi } \right) = \left( {2,\frac{\pi }{4},\frac{\pi }{6}} \right)$.
Work Step by Step
We have $\left( {x,y,z} \right) = \left( {\frac{{\sqrt 2 }}{2},\frac{{\sqrt 2 }}{2},\sqrt 3 } \right)$.
In spherical coordinates:
1. the radial coordinate is
$\rho = \sqrt {{{\left( {\frac{{\sqrt 2 }}{2}} \right)}^2} + {{\left( {\frac{{\sqrt 2 }}{2}} \right)}^2} + {{\left( {\sqrt 3 } \right)}^2}} = \sqrt {\frac{1}{2} + \frac{1}{2} + 3} = 2$
2. the angular coordinate $\theta$ satisfies
$\tan \theta = \frac{y}{x} = \frac{{\sqrt 2 /2}}{{\sqrt 2 /2}} = 1$, ${\ \ }$ $\theta = \frac{\pi }{4}$
3. the angular coordinate $\phi $ satisfies
$\cos \phi = \frac{z}{\rho } = \frac{1}{2}\sqrt 3 $, ${\ \ }$ $\phi = \frac{\pi }{6}$
Therefore, the spherical coordinates is $\left( {\rho ,\theta ,\phi } \right) = \left( {2,\frac{\pi }{4},\frac{\pi }{6}} \right)$.
