Answer
\[\left( 54\sqrt{3},-\frac{5\pi }{6},\frac{\pi }{3} \right)\]
Work Step by Step
\[\begin{align}
& \left( 81,-\frac{5\pi }{6},27\sqrt{3} \right) \\
& \left( r,\theta ,z \right):\left( 81,-\frac{5\pi }{6},27\sqrt{3} \right)\to r=81,\text{ }\theta =-\frac{5\pi }{6},\text{ }z=27\sqrt{3} \\
& \text{Cylindrical to spherical }\left( \rho ,\theta ,\phi \right),\text{ }\left( r\ge 0 \right)\text{ See page 807} \\
& \rho =\sqrt{{{r}^{2}}+{{z}^{2}}},\text{ }\theta =\theta ,\text{ }\phi =\arccos \left( \frac{z}{\sqrt{{{r}^{2}}+{{z}^{2}}}} \right) \\
& \rho =\sqrt{{{\left( 81 \right)}^{2}}+{{\left( 27\sqrt{3} \right)}^{2}}}=54\sqrt{3} \\
& \theta =-\frac{5\pi }{6} \\
& \phi =\arccos \left( \frac{27\sqrt{3}}{54\sqrt{3}} \right)=\arccos \left( \frac{1}{2} \right)=\frac{\pi }{3} \\
& \text{The spherical }\left( \rho ,\theta ,\phi \right)\text{ coordinates are:} \\
& \left( 54\sqrt{3},-\frac{5\pi }{6},\frac{\pi }{3} \right) \\
\end{align}\]
