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 $\displaystyle (6,6,-6)$

$\displaystyle \begin{align*}&r=6^2+6^2=8.49\\&\theta=arctan(\frac{6}{6})=45^{\circ}\\&z=-6\end{align*}$

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 $\displaystyle (-9,7,-4)$

$\displaystyle \begin{align*}&r=\sqrt{(-9)^2+(7)^2}=11.4\\&\theta=arctan(\frac{7}{-9})=142.13^{\circ}\\&z=-4\end{align*}$

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 $\displaystyle (8,1,7)$

$\displaystyle \begin{align*}&r=\sqrt{(8)^2+(1)^2}=8.06\\&\theta=arctan(\frac{1}{8})=7.13^{\circ}\\&z=7\end{align*}$

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 $\displaystyle (3,4,7)$

$\displaystyle \begin{align*}&r=\sqrt{(3)^2+(4)^2}=5\\&\theta=arctan(\frac{4}{3})=53.13^{\circ}\\&z=7\end{align*}$

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 $\displaystyle (-5,-7,5)$

$\displaystyle \begin{align*}&r=\sqrt{(-5)^2+(-7)^2}=8.6\\&\theta=arctan(\frac{-7}{-5})=-125.54^{\circ}\\&z=5\end{align*}$

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 $\displaystyle (-8,4,1)$

$\displaystyle \begin{align*}&r=\sqrt{(-8)^2+(4)^2}=8.94\\&\theta=arctan(\frac{4}{-8})=153.43^{\circ}\\&z=1\end{align*}$

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 $\displaystyle (-4,-1,-4)$

$\displaystyle \begin{align*}&r=\sqrt{(-4)^2+(-1)^2}=4.12\\&\theta=arctan(\frac{-1}{-4})=-165.96^{\circ}\\&z=-4\end{align*}$

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 $\displaystyle (3,6,-5)$

$\displaystyle \begin{align*}&r=\sqrt{(3)^2+(6)^2}=6.71\\&\theta=arctan(\frac{6}{3})=63.43^{\circ}\\&z=-5\end{align*}$

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 $\displaystyle (-6,-6,-4)$

$\displaystyle \begin{align*}&r=\sqrt{(-6)^2+(-6)^2}=8.49\\&\theta=arctan(\frac{-6}{-6})=-135^{\circ}\\&z=-4\end{align*}$

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 $\displaystyle (9,-1,4)$

$\displaystyle \begin{align*}&r=\sqrt{(9)^2+(-1)^2}=9.06\\&\theta=arctan(\frac{-1}{9})=-6.34^{\circ}\\&z=4\end{align*}$