Answer
The point in cylindrical coordinates is
$\left( {r,\theta ,z} \right) = \left( {5,0.927, - 1} \right)$.
The point in spherical coordinates is
$\left( {\rho ,\theta ,\phi } \right) = \left( {\sqrt {26} ,0.927,1.768} \right)$.
Work Step by Step
We have $\left( {x,y,z} \right) = \left( {3,4, - 1} \right)$.
1. Convert to cylindrical coordinates.
The radial coordinate is
$r = \sqrt {{x^2} + {y^2}} = \sqrt {{3^2} + {4^2}} = 5$.
The angular coordinate $\theta$ satisfies $\tan \theta = \frac{y}{x}$. So,
$\tan \theta = \frac{4}{3}$, ${\ \ }$ $\theta \simeq 0.927$
The $z$-coordinate is the same, so $z=-1$.
Thus, the point in cylindrical coordinates is
$\left( {r,\theta ,z} \right) = \left( {5,0.927, - 1} \right)$.
2. Convert to spherical coordinates.
The radial coordinate is
$\rho = \sqrt {{x^2} + {y^2} + {z^2}} = \sqrt {{3^2} + {4^2} + {{\left( { - 1} \right)}^2}} = \sqrt {26} $
The angular coordinate $\theta$ satisfies $\tan \theta = \frac{y}{x}$. So,
$\tan \theta = \frac{4}{3}$, ${\ \ }$ $\theta \simeq 0.927$
The angular coordinate $\phi$ satisfies
$\cos \phi = \frac{z}{\rho } = \frac{{ - 1}}{{\sqrt {26} }}$, ${\ \ }$ $\phi = 1.768$
Thus, the point in spherical coordinates is
$\left( {\rho ,\theta ,\phi } \right) = \left( {\sqrt {26} ,0.927,1.768} \right)$.
