When given Cartesian coordinates of the form \(\displaystyle (x,y,z)\) to cylindrical coordinates of the form \(\displaystyle (r,\theta,z)\), the first and third terms are the most straightforward.

\(\displaystyle r=\sqrt{x^2+y^2}\)

\(\displaystyle z=z\)

Care should be taken, however, when calculating \(\displaystyle \theta\). 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 \(\displaystyle |\theta|>90^{\circ};\frac{\pi}{2}\)

For our coordinates \(\displaystyle (12,-5,31)\)

 \(\displaystyle \begin{align*}&r=\sqrt{(12)^2+(-5)^2}=13\\&\theta=arctan(\frac{-5}{12})=-22.62^{\circ}\\&z=31\end{align*}\)

When given Cartesian coordinates of the form \(\displaystyle (x,y,z)\) to cylindrical coordinates of the form \(\displaystyle (r,\theta,z)\), the first and third terms are the most straightforward.

\(\displaystyle r=\sqrt{x^2+y^2}\)

\(\displaystyle z=z\)

Care should be taken, however, when calculating \(\displaystyle \theta\). 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 \(\displaystyle |\theta|>90^{\circ};\frac{\pi}{2}\)

For our coordinates \(\displaystyle (-24,10,32)\)

 \(\displaystyle \begin{align*}&r=\sqrt{(-24)^2+(10)^2}=26\\&\theta=arctan(\frac{10}{-24})=157.38^{\circ}\\&z=32\end{align*}\)

When given Cartesian coordinates of the form \(\displaystyle (x,y,z)\) to cylindrical coordinates of the form \(\displaystyle (r,\theta,z)\), the first and third terms are the most straightforward.

\(\displaystyle r=\sqrt{x^2+y^2}\)

\(\displaystyle z=z\)

Care should be taken, however, when calculating \(\displaystyle \theta\). 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 \(\displaystyle |\theta|>90^{\circ};\frac{\pi}{2}\)

For our coordinates \(\displaystyle (20,14,25)\)

 \(\displaystyle \begin{align*}&r=\sqrt{(20)^2+(14)^2}=24.41\\&\theta=arctan(\frac{14}{20})=34.99^{\circ}\\&z=25\end{align*}\)

When given Cartesian coordinates of the form \(\displaystyle (x,y,z)\) to cylindrical coordinates of the form \(\displaystyle (r,\theta,z)\), the first and third terms are the most straightforward.

\(\displaystyle r=\sqrt{x^2+y^2}\)

\(\displaystyle z=z\)

Care should be taken, however, when calculating \(\displaystyle \theta\). 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 \(\displaystyle |\theta|>90^{\circ};\frac{\pi}{2}\)

For our coordinates \(\displaystyle (-9,12,15)\)

 \(\displaystyle \begin{align*}&r=\sqrt{(-9)^2+(12)^2}=15\\&\theta=arctan(\frac{12}{-9})=126.87^{\circ}\\&z=15\end{align*}\)

When given Cartesian coordinates of the form \(\displaystyle (x,y,z)\) to cylindrical coordinates of the form \(\displaystyle (r,\theta,z)\), the first and third terms are the most straightforward.

\(\displaystyle r=\sqrt{x^2+y^2}\)

\(\displaystyle z=z\)

Care should be taken, however, when calculating \(\displaystyle \theta\). 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 \(\displaystyle |\theta|>90^{\circ};\frac{\pi}{2}\)

For our coordinates \(\displaystyle (36,-160,18)\)

 \(\displaystyle \begin{align*}&r=\sqrt{(36)^2+(-160)^2}=164\\&\theta=arctan(\frac{-160}{36})=-77.32^{\circ}\\&z=18\end{align*}\)

When given Cartesian coordinates of the form \(\displaystyle (x,y,z)\) to cylindrical coordinates of the form \(\displaystyle (r,\theta,z)\), the first and third terms are the most straightforward.

\(\displaystyle r=\sqrt{x^2+y^2}\)

\(\displaystyle z=z\)

Care should be taken, however, when calculating \(\displaystyle \theta\). 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 \(\displaystyle |\theta|>90^{\circ};\frac{\pi}{2}\)

For our coordinates \(\displaystyle (-1.5,-2,10)\)

 \(\displaystyle \begin{align*}&r=\sqrt{(-1.5)^2+(-2)^2}=2.5\\&\theta=arctan(\frac{-2}{-1.5})=-126.87^{\circ}\\&z=10\end{align*}\)

When given Cartesian coordinates of the form \(\displaystyle (x,y,z)\) to cylindrical coordinates of the form \(\displaystyle (r,\theta,z)\), the first and third terms are the most straightforward.

\(\displaystyle r=\sqrt{x^2+y^2}\)

\(\displaystyle z=z\)

Care should be taken, however, when calculating \(\displaystyle \theta\). 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 \(\displaystyle |\theta|>90^{\circ};\frac{\pi}{2}\)

For our coordinates \(\displaystyle (-50,120,100)\)

 \(\displaystyle \begin{align*}&r=\sqrt{(-50)^2+(120)^2}=130\\&\theta=arctan(\frac{120}{-50})=112.62^{\circ}\\&z=100\end{align*}\)

When given Cartesian coordinates of the form \(\displaystyle (x,y,z)\) to cylindrical coordinates of the form \(\displaystyle (r,\theta,z)\), the first and third terms are the most straightforward.

\(\displaystyle r=\sqrt{x^2+y^2}\)

\(\displaystyle z=z\)

Care should be taken, however, when calculating \(\displaystyle \theta\). 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 \(\displaystyle |\theta|>90^{\circ};\frac{\pi}{2}\)

For our coordinates \(\displaystyle (5,-25,-6)\)

 \(\displaystyle \begin{align*}&r=\sqrt{(5)^2+(-25)^2}=25.5\\&\theta=arctan(\frac{-25}{5})=-78.69^{\circ}\\&z=-6\end{align*}\)

When given Cartesian coordinates of the form \(\displaystyle (x,y,z)\) to cylindrical coordinates of the form \(\displaystyle (r,\theta,z)\), the first and third terms are the most straightforward.

\(\displaystyle r=\sqrt{x^2+y^2}\)

\(\displaystyle z=z\)

Care should be taken, however, when calculating \(\displaystyle \theta\). 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 \(\displaystyle |\theta|>90^{\circ};\frac{\pi}{2}\)

For our coordinates \(\displaystyle (1,-1.707,6)\)

 \(\displaystyle \begin{align*}&r=\sqrt{(1)^2+(-1.71)^2}=1.98\\&\theta=arctan(\frac{-1.71}{1})=-59.64^{\circ}\\&z=6\end{align*}\)

When given Cartesian coordinates of the form \(\displaystyle (x,y,z)\) to cylindrical coordinates of the form \(\displaystyle (r,\theta,z)\), the first and third terms are the most straightforward.

\(\displaystyle r=\sqrt{x^2+y^2}\)

\(\displaystyle z=z\)

Care should be taken, however, when calculating \(\displaystyle \theta\). 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 \(\displaystyle |\theta|>90^{\circ};\frac{\pi}{2}\)

For our coordinates \(\displaystyle (7,21,22)\)

 \(\displaystyle \begin{align*}&r=\sqrt{(7)^2+(21)^2}=22.14\\&\theta=arctan(\frac{21}{7})=71.57^{\circ}\\&z=22\end{align*}\)