Answer
$$\left( {2\sqrt 2 ,\frac{{2\pi }}{3},\frac{{3\pi }}{4}} \right)$$
Work Step by Step
$$\eqalign{
& \left( {2,\frac{{2\pi }}{3}, - 2} \right) \cr
& \left( {r,\theta ,z} \right):\left( {2,\frac{{2\pi }}{3}, - 2} \right) \to r = 2,{\text{ }}\theta = \frac{{2\pi }}{3},{\text{ }}z = - 2 \cr
& {\text{Cylindrical to spherical }}\left( {\rho ,\theta ,\phi } \right),{\text{ }}\left( {r \geqslant 0} \right){\text{ See page 807}} \cr
& \rho = \sqrt {{r^2} + {z^2}} ,{\text{ }}\theta = \theta ,{\text{ }}\phi = \arccos \left( {\frac{z}{{\sqrt {{r^2} + {z^2}} }}} \right) \cr
& \rho = \sqrt {{{\left( 2 \right)}^2} + {{\left( { - 2} \right)}^2}} = \sqrt 8 = \sqrt {4 \cdot 2} = 2\sqrt 2 \cr
& \theta = \frac{{2\pi }}{3} \cr
& \phi = \arccos \left( {\frac{{ - 2}}{{2\sqrt 2 }}} \right) = \frac{{3\pi }}{4} \cr
& {\text{The spherical }}\left( {\rho ,\theta ,\phi } \right){\text{ coordinates are:}} \cr
& \left( {2\sqrt 2 ,\frac{{2\pi }}{3},\frac{{3\pi }}{4}} \right) \cr} $$
