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 

 \(\displaystyle \begin{align*}&(0,19,2)\\&\rho=\sqrt{(0)^2+(19)^2+(2)^2}=19.1\\&\theta=arctan(\frac{19}{0})=90^{\circ}\\&\phi=arccos(\frac{2}{19.1})=83.99^{\circ}\end{align*}\)

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 

 \(\displaystyle \begin{align*}&(-15,-32,-44)\\&\rho=\sqrt{(-15)^2+(-32)^2+(-44)^2}=56.44\\&\theta=arctan(\frac{-32}{-15})=-115.11^{\circ}\\&\phi=arccos(\frac{-44}{56.44})=141.23^{\circ}\end{align*}\)

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 

 \(\displaystyle \begin{align*}&(12,-26,-14)\\&\rho=\sqrt{(12)^2+(-26)^2+(-14)^2}=31.87\\&\theta=arctan(\frac{-26}{12})=-65.22^{\circ}\\&\phi=arccos(\frac{-14}{31.87})=116.05^{\circ}\end{align*}\)

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 

 \(\displaystyle \begin{align*}&(36,-33,-14)\\&\rho=\sqrt{(36)^2+(-33)^2+(-14)^2}=50.8\\&\theta=arctan(\frac{-33}{36})=-42.51^{\circ}\\&\phi=arccos(\frac{-14}{50.8})=106^{\circ}\end{align*}\)

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 

 \(\displaystyle \begin{align*}&(-30,-33,26)\\&\rho=\sqrt{(-30)^2+(-33)^2+(26)^2}=51.62\\&\theta=arctan(\frac{-33}{-30})=-132.27^{\circ}\\&\phi=arccos(\frac{26}{51.62})=59.76^{\circ}\end{align*}\)

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 

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 

 \(\displaystyle \begin{align*}&(-15,-34,13)\\&\rho=\sqrt{(-15)^2+(-34)^2+(13)^2}=39.37\\&\theta=arctan(\frac{-34}{-15})=-113.81^{\circ}\\&\phi=arccos(\frac{13}{39.37})=70.72^{\circ}\end{align*}\)

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 

 \(\displaystyle \begin{align*}&(-23,-33,-25)\\&\rho=\sqrt{(-23)^2+(-33)^2+(-25)^2}=47.36\\&\theta=arctan(\frac{-33}{-23})=-124.88^{\circ}\\&\phi=arccos(\frac{-25}{47.36})=121.86^{\circ}\end{align*}\)

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 

 \(\displaystyle \begin{align*}&(21,34,-47)\\&\rho=\sqrt{(21)^2+(34)^2+(-47)^2}=61.69\\&\theta=arctan(\frac{34}{21})=58.3^{\circ}\\&\phi=arccos(\frac{-47}{61.69})=139.63^{\circ}\end{align*}\)

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 

 \(\displaystyle \begin{align*}&(-26,-40,26)\\&\rho=\sqrt{(-26)^2+(-40)^2+(26)^2}=54.33\\&\theta=arctan(\frac{-40}{-26})=-123.02^{\circ}\\&\phi=arccos(\frac{26}{54.33})=61.41^{\circ}\end{align*}\)