Answer
$$\left( {2\sqrt {13} , - \frac{\pi }{6},\arccos \left( {\frac{3}{{\sqrt {13} }}} \right)} \right)$$
Work Step by Step
$$\eqalign{
& \left( {4, - \frac{\pi }{6},6} \right) \cr
& \left( {r,\theta ,z} \right):\left( {4, - \frac{\pi }{6},6} \right) \to r = 4,{\text{ }}\theta = - \frac{\pi }{6},{\text{ }}z = 6 \cr
& {\text{Cylindrical to spherical }}\left( {\rho ,\theta ,\phi } \right),{\text{ }}\left( {r \geqslant 0} \right){\text{ See page 807}} \cr
& \rho = \sqrt {{r^2} + {z^2}} ,{\text{ }}\theta = \theta ,{\text{ }}\phi = \arccos \left( {\frac{z}{{\sqrt {{r^2} + {z^2}} }}} \right) \cr
& \rho = \sqrt {{{\left( 4 \right)}^2} + {{\left( 6 \right)}^2}} = \sqrt {52} = \sqrt {4 \cdot 13} = 2\sqrt {13} \cr
& \theta = - \frac{\pi }{6} \cr
& \phi = \arccos \left( {\frac{6}{{2\sqrt {13} }}} \right) = \arccos \left( {\frac{3}{{\sqrt {13} }}} \right) \cr
& {\text{The spherical }}\left( {\rho ,\theta ,\phi } \right){\text{ coordinates are:}} \cr
& \left( {2\sqrt {13} , - \frac{\pi }{6},\arccos \left( {\frac{3}{{\sqrt {13} }}} \right)} \right) \cr} $$
