Answer
\[\left( 50\sqrt{5},-\frac{\pi }{6},\arccos \left( \frac{1}{\sqrt{5}} \right) \right)\]
Work Step by Step
\[\begin{align}
& \left( 100,-\frac{\pi }{6},50 \right) \\
& \left( r,\theta ,z \right):\left( 100,-\frac{\pi }{6},50 \right)\to r=100,\text{ }\theta =-\frac{\pi }{6},\text{ }z=50 \\
& \text{Cylindrical to spherical }\left( \rho ,\theta ,\phi \right),\text{ }\left( r\ge 0 \right)\text{ See page 807} \\
& \rho =\sqrt{{{r}^{2}}+{{z}^{2}}},\text{ }\theta =\theta ,\text{ }\phi =\arccos \left( \frac{z}{\sqrt{{{r}^{2}}+{{z}^{2}}}} \right) \\
& \rho =\sqrt{{{\left( 100 \right)}^{2}}+{{\left( 50 \right)}^{2}}}=50\sqrt{5} \\
& \theta =-\frac{\pi }{6} \\
& \phi =\arccos \left( \frac{50}{50\sqrt{5}} \right)=\arccos \left( \frac{1}{\sqrt{5}} \right) \\
& \text{The spherical }\left( \rho ,\theta ,\phi \right)\text{ coordinates are:} \\
& \left( 50\sqrt{5},-\frac{\pi }{6},\arccos \left( \frac{1}{\sqrt{5}} \right) \right) \\
\end{align}\]
