Answer
$$\left( {4,\frac{{7\pi }}{6},4\sqrt 3 } \right)$$
Work Step by Step
$$\eqalign{
& \left( {8,\frac{{7\pi }}{6},\frac{\pi }{6}} \right) \cr
& {\text{spherical}}\left( {\rho ,\theta ,\phi } \right):\left( {8,\frac{{7\pi }}{6},\frac{\pi }{6}} \right) \to \rho = 8,{\text{ }}\theta = \frac{{7\pi }}{6},{\text{ }}\phi = \frac{\pi }{6} \cr
& {\text{Spherical to cylindrical }}\left( {r,\theta ,z} \right),{\text{ }}\left( {r \geqslant 0} \right){\text{ See page 807}} \cr
& {r^2} = {\rho ^2}{\sin ^2}\phi ,{\text{ }}\theta = \theta ,{\text{ }}z = \rho \cos \phi \cr
& {r^2} = {\left( 8 \right)^2}{\sin ^2}\left( {\frac{\pi }{6}} \right) = 16 \to r = 4 \cr
& \theta = \frac{{7\pi }}{6} \cr
& z = 8cos\left( {\frac{\pi }{6}} \right) = 4\sqrt 3 \cr
& {\text{The cylindrical }}\left( {r,\theta ,z} \right){\text{ coordinates are:}} \cr
& \left( {4,\frac{{7\pi }}{6},4\sqrt 3 } \right) \cr} $$
