When given Cartesian coordinates of the form $\displaystyle (x,y,z)$ to cylindrical coordinates of the form $\displaystyle (\rho,\theta,\phi)$, it would be useful to calculate the $\displaystyle \rho$ term first, as we'll derive $\displaystyle \phi$ from it.

$\displaystyle \rho=\sqrt{x^2+y^2+z^2}$

Next, begin calculating our angles. Care should be taken, however, when calculating them. Beginning with $\displaystyle \theta$, the formula for it is as follows: 

$\displaystyle \theta=arctan(\frac{y}{x})$

However, it is important to be mindful of the signs of both $\displaystyle y$ and $\displaystyle x$, bearing in mind which quadrant the point lies; this will determine the value of $\displaystyle \theta$:

It is something to bear in mind when making a calculation using a calculator; negative $\displaystyle y$ values by convention create a negative $\displaystyle \theta$, while negative $\displaystyle x$ values lead to $\displaystyle |\theta|>90^{\circ};\frac{\pi}{2}$

To calculate $\displaystyle \phi$, we can make use of our $\displaystyle \rho$ value. It is found as

$\displaystyle \phi=arccos(\frac{z}{\rho})$, where $\displaystyle 0\leq\phi\leq180^{\circ}$

For our coordinates 

 $\displaystyle \begin{align*}&(17,11,37)\\&\rho=\sqrt{(17)^2+(11)^2+(37)^2}=42.18\\&\theta=arctan(\frac{11}{17})=32.91^{\circ}\\&\phi=arccos(\frac{37}{42.18})=28.69^{\circ}\end{align*}$

When given Cartesian coordinates of the form $\displaystyle (x,y,z)$ to cylindrical coordinates of the form $\displaystyle (\rho,\theta,\phi)$, it would be useful to calculate the $\displaystyle \rho$ term first, as we'll derive $\displaystyle \phi$ from it.

$\displaystyle \rho=\sqrt{x^2+y^2+z^2}$

Next, begin calculating our angles. Care should be taken, however, when calculating them. Beginning with $\displaystyle \theta$, the formula for it is as follows: 

$\displaystyle \theta=arctan(\frac{y}{x})$

However, it is important to be mindful of the signs of both $\displaystyle y$ and $\displaystyle x$, bearing in mind which quadrant the point lies; this will determine the value of $\displaystyle \theta$:

It is something to bear in mind when making a calculation using a calculator; negative $\displaystyle y$ values by convention create a negative $\displaystyle \theta$, while negative $\displaystyle x$ values lead to $\displaystyle |\theta|>90^{\circ};\frac{\pi}{2}$

To calculate $\displaystyle \phi$, we can make use of our $\displaystyle \rho$ value. It is found as

$\displaystyle \phi=arccos(\frac{z}{\rho})$, where $\displaystyle 0\leq\phi\leq180^{\circ}$

For our coordinates 

 $\displaystyle \begin{align*}&(-18,-14,35)\\&\rho=\sqrt{(-18)^2+(-14)^2+(35)^2}=41.77\\&\theta=arctan(\frac{-14}{-18})=-142.13^{\circ}\\&\phi=arccos(\frac{35}{41.77})=33.09^{\circ}\end{align*}$

$\displaystyle \begin{align*}&\text{When converting Cartesian coordinates:} (x,y,z)\\&\text{to spherical coordinates: }(\rho,\theta,\phi)\text{, it is useful to}\\& \text{calculate the }\rho\text{ term first:}\\& \rho=\sqrt{x^2+y^2+z^2}\\&\text{Next, calculate angles, being prudent to be careful when calculating them}.\\&\text{The formula for }\theta\text{ is as follows:}\\& \theta=arctan(\frac{y}{x})\\&\text{Keep in mind the signs of }x\text{ and }y\text{, and}\\& \text{which quadrant the point lies in.}\\& \text{Negative }y\text{ values lead to a negative }\theta\text{; negative }x\\&\text{values lead to }|\theta|>90^{\circ}\\&\text{To calculate }\phi\text{, we can use our }\rho\text{ value:}\\&\phi=arccos(\frac{z}{\rho})\text{ where }0\leq\phi\leq180^{\circ}\\&\text{For our coordinates: }(25,12,111)\\&\rho=\sqrt{(25)^2+(12)^2+(111)^2}=114.41\\&\theta=arctan(\frac{12}{25})=25.64^{\circ}\\&\phi=arccos(\frac{111}{114.41})=14.03^{\circ}\end{align*}$

$\displaystyle \begin{align*}&\text{When converting Cartesian coordinates:} (x,y,z)\\&\text{to spherical coordinates: }(\rho,\theta,\phi)\text{, it is useful to}\\& \text{calculate the }\rho\text{ term first:}\\& \rho=\sqrt{x^2+y^2+z^2}\\&\text{Next, calculate angles, being prudent when calculating them}.\\&\text{The formula for }\theta\text{ is as follows:}\\& \theta=arctan(\frac{y}{x})\\&\text{Keep in mind the signs of }x\text{ and }y\text{, and}\\& \text{which quadrant the point lies in.}\\& \text{Negative }y\text{ values lead to a negative }\theta\text{; negative }x\\&\text{values lead to }|\theta|>90^{\circ}\\&\text{To calculate }\phi\text{, we can use our }\rho\text{ value:}\\&\phi=arccos(\frac{z}{\rho})\text{ where }0\leq\phi\leq180^{\circ}\\&\text{For our coordinates: }(-71,-55,-115)\\&\rho=\sqrt{(-71)^2+(-55)^2+(-115)^2}=145.91\\&\theta=arctan(\frac{-55}{-71})=-142.24^{\circ}\\&\phi=arccos(\frac{-115}{145.91})=142.01^{\circ}\end{align*}$

$\displaystyle \begin{align*}&\text{When converting Cartesian coordinates:} (x,y,z)\\&\text{to spherical coordinates: }(\rho,\theta,\phi)\text{, it is useful to}\\& \text{calculate the }\rho\text{ term first:}\\& \rho=\sqrt{x^2+y^2+z^2}\\&\text{Next, calculate angles, being prudent when calculating them}.\\&\text{The formula for }\theta\text{ is as follows:}\\& \theta=arctan(\frac{y}{x})\\&\text{Keep in mind the signs of }x\text{ and }y\text{, and}\\& \text{which quadrant the point lies in.}\\& \text{Negative }y\text{ values lead to a negative }\theta\text{; negative }x\\&\text{values lead to }|\theta|>90^{\circ}\\&\text{To calculate }\phi\text{, we can use our }\rho\text{ value:}\\&\phi=arccos(\frac{z}{\rho})\text{ where }0\leq\phi\leq180^{\circ}\\&\text{For our coordinates: }(132,44,-6)\\&\rho=\sqrt{(132)^2+(44)^2+(-6)^2}=139.27\\&\theta=arctan(\frac{44}{132})=18.43^{\circ}\\&\phi=arccos(\frac{-6}{139.27})=92.47^{\circ}\end{align*}$

$\displaystyle \begin{align*}&\text{When converting Cartesian coordinates:} (x,y,z)\\&\text{to spherical coordinates: }(\rho,\theta,\phi)\text{, it is useful to}\\& \text{calculate the }\rho\text{ term first:}\\& \rho=\sqrt{x^2+y^2+z^2}\\&\text{Next, calculate angles, being prudent when calculating them}.\\&\text{The formula for }\theta\text{ is as follows:}\\& \theta=arctan(\frac{y}{x})\\&\text{Keep in mind the signs of }x\text{ and }y\text{, and}\\& \text{which quadrant the point lies in.}\\& \text{Negative }y\text{ values lead to a negative }\theta\text{; negative }x\\&\text{values lead to }|\theta|>90^{\circ}\\&\text{To calculate }\phi\text{, we can use our }\rho\text{ value:}\\&\phi=arccos(\frac{z}{\rho})\text{ where }0\leq\phi\leq180^{\circ}\\&\text{For our coordinates: }(13,67,7)\\&\rho=\sqrt{(13)^2+(67)^2+(7)^2}=68.61\\&\theta=arctan(\frac{67}{13})=79.02^{\circ}\\&\phi=arccos(\frac{7}{68.61})=84.14^{\circ}\end{align*}$

$\displaystyle \begin{align*}&\text{When converting Cartesian coordinates:} (x,y,z)\\&\text{to spherical coordinates: }(\rho,\theta,\phi)\text{, it is useful to}\\& \text{calculate the }\rho\text{ term first:}\\& \rho=\sqrt{x^2+y^2+z^2}\\&\text{Next, calculate angles, being prudent when calculating them}.\\&\text{The formula for }\theta\text{ is as follows:}\\& \theta=arctan(\frac{y}{x})\\&\text{Keep in mind the signs of }x\text{ and }y\text{, and}\\& \text{which quadrant the point lies in.}\\& \text{Negative }y\text{ values lead to a negative }\theta\text{; negative }x\\&\text{values lead to }|\theta|>90^{\circ}\\&\text{To calculate }\phi\text{, we can use our }\rho\text{ value:}\\&\phi=arccos(\frac{z}{\rho})\text{ where }0\leq\phi\leq180^{\circ}\\&\text{For our coordinates: }(149,-85,-119)\\&\rho=\sqrt{(149)^2+(-85)^2+(-119)^2}=208.77\\&\theta=arctan(\frac{-85}{149})=-29.7^{\circ}\\&\phi=arccos(\frac{-119}{208.77})=124.75^{\circ}\end{align*}$

$\displaystyle \begin{align*}&\text{When converting Cartesian coordinates:} (x,y,z)\\&\text{to spherical coordinates: }(\rho,\theta,\phi)\text{, it is useful to}\\& \text{calculate the }\rho\text{ term first:}\\& \rho=\sqrt{x^2+y^2+z^2}\\&\text{Next, calculate angles, being prudent when calculating them}.\\&\text{The formula for }\theta\text{ is as follows:}\\& \theta=arctan(\frac{y}{x})\\&\text{Keep in mind the signs of }x\text{ and }y\text{, and}\\& \text{which quadrant the point lies in.}\\& \text{Negative }y\text{ values lead to a negative }\theta\text{; negative }x\\&\text{values lead to }|\theta|>90^{\circ}\\&\text{To calculate }\phi\text{, we can use our }\rho\text{ value:}\\&\phi=arccos(\frac{z}{\rho})\text{ where }0\leq\phi\leq180^{\circ}\\&\text{For our coordinates: }(-117,-131,-29)\\&\rho=\sqrt{(-117)^2+(-131)^2+(-29)^2}=178.02\\&\theta=arctan(\frac{-131}{-117})=-131.77^{\circ}\\&\phi=arccos(\frac{-29}{178.02})=99.38^{\circ}\end{align*}$

$\displaystyle \begin{align*}&\text{When converting Cartesian coordinates:} (x,y,z)\\&\text{to spherical coordinates: }(\rho,\theta,\phi)\text{, it is useful to}\\& \text{calculate the }\rho\text{ term first:}\\& \rho=\sqrt{x^2+y^2+z^2}\\&\text{Next, calculate angles, being prudent when calculating them}.\\&\text{The formula for }\theta\text{ is as follows:}\\& \theta=arctan(\frac{y}{x})\\&\text{Keep in mind the signs of }x\text{ and }y\text{, and}\\& \text{which quadrant the point lies in.}\\& \text{Negative }y\text{ values lead to a negative }\theta\text{; negative }x\\&\text{values lead to }|\theta|>90^{\circ}\\&\text{To calculate }\phi\text{, we can use our }\rho\text{ value:}\\&\phi=arccos(\frac{z}{\rho})\text{ where }0\leq\phi\leq180^{\circ}\\&\text{For our coordinates: }(-16,-40,79)\\&\rho=\sqrt{(-16)^2+(-40)^2+(79)^2}=89.98\\&\theta=arctan(\frac{-40}{-16})=-111.8^{\circ}\\&\phi=arccos(\frac{79}{89.98})=28.61^{\circ}\end{align*}$

$\displaystyle \begin{align*}&\text{When converting Cartesian coordinates:} (x,y,z)\\&\text{to spherical coordinates: }(\rho,\theta,\phi)\text{, it is useful to}\\& \text{calculate the }\rho\text{ term first:}\\& \rho=\sqrt{x^2+y^2+z^2}\\&\text{Next, calculate angles, being prudent when calculating them}.\\&\text{The formula for }\theta\text{ is as follows:}\\& \theta=arctan(\frac{y}{x})\\&\text{Keep in mind the signs of }x\text{ and }y\text{, and}\\& \text{which quadrant the point lies in.}\\& \text{Negative }y\text{ values lead to a negative }\theta\text{; negative }x\\&\text{values lead to }|\theta|>90^{\circ}\\&\text{To calculate }\phi\text{, we can use our }\rho\text{ value:}\\&\phi=arccos(\frac{z}{\rho})\text{ where }0\leq\phi\leq180^{\circ}\\&\text{For our coordinates: }(38,82,130)\\&\rho=\sqrt{(38)^2+(82)^2+(130)^2}=158.33\\&\theta=arctan(\frac{82}{38})=65.14^{\circ}\\&\phi=arccos(\frac{130}{158.33})=34.81^{\circ}\end{align*}$