Cylindrical coordinates are an alternate three-dimensional coordinate system to the Cartesian coordinate system. Cylindrical coordinates have the form (*r, θ, z*), where *r* is the distance in the *xy* plane, θ is the angle of *r* with respect to the *x*-axis, and *z* is the component on the *z*-axis. This coordinate system can have advantages over the Cartesian system when graphing cylindrical figures such as tubes or tanks.

Here, we will learn about the formulas that we can use to transform from Cartesian coordinates to cylindrical coordinates. Then, we will look at some practice problems in which we will apply these formulas.

##### TRIGONOMETRY

**Relevant for**…

Learning to transform from Cartesian to cylindrical coordinates.

##### TRIGONOMETRY

**Relevant for**…

Learning to transform from Cartesian to cylindrical coordinates.

## How to transform from Cartesian coordinates to cylindrical coordinates?

Cylindrical coordinates can be more convenient when we want to graph cylinders, tubes, or similar figures. This coordinate system is used in calculus since it allows using an easier reference system for cylindrical figures and finding derivatives or integrals becomes easier.

The cylindrical coordinate system has the form $latex (r, \theta, z)$, where *r* is the distance from the origin to the location of the point in the xy plane and θ is the angle formed by the line and the x-axis.

This coordinate system is considered as an extension to the third dimension of the polar coordinate system. Comparing these coordinates with the Cartesian coordinates, $latex (x, y, z)$, we see that the component of the third dimension, *z*, is the same.

We can use a right triangle and the Pythagorean theorem to find the value of *r* in terms of *x* and *y*. The *x* and *y* coordinates form the legs of the triangle and *r* forms the hypotenuse. Therefore, we have the relationship:

$latex {{r}^2}={{x}^2}+{{y}^2}$

$latex r=\sqrt{{{x}^2}+{{y}^2}}$ |

To find the angle *θ, *we use the inverse tangent function. The tangent function of an angle in a triangle is equal to the opposite side divided by the adjacent side. In this case, the opposite side is equal to the *y *coordinate and the adjacent side is the *x *coordinate. Therefore, we have:

$latex \theta=\tan^{-1}(\frac{y}{x})$ |

A complication in finding the angle *θ *is that many times, the calculator does not return the correct value for the angle. This is because the range of the inverse tangent function goes from $latex – \frac{\pi}{2}$ to $latex \frac{\pi}{2}$ and this does not cover the four quadrants of the Cartesian plane.

We can solve this by using the following table to correct the angles:

Quadrant | Value of $latex {{\tan}^{-1}}$ |

I | The calculator gives the correct value |

II | We have to add 180° to the value of the calculator |

III | We have to add 180° to the value of the calculator |

IV | We have to add 360° to the value of the calculator |

## Cartesian to cylindrical coordinates – Examples with answers

The conversion formulas from Cartesian to cylindrical coordinates are applied to solve the following examples. Try to solve the problems yourself before looking at the answer.

**EXAMPLE 1**

If we have the Cartesian coordinates (2, 2, 5), what is the equivalence in cylindrical coordinates?

##### Solution

We know that we have $latex x = 2, ~ y = 2, ~ z = 5$. Therefore, we have to find *r, θ* and *z* to form the cylindrical coordinates. We find *r*, using the following equation:

$latex r=\sqrt{{{x}^2}+{{y}^2}}$

$latex r=\sqrt{{{2}^2}+{{2}^2}}$

$latex r=\sqrt{4+4}$

$latex r=\sqrt{8}$

We find *θ,* using the following formula:

$latex \theta={{\tan}^{-1}}(\frac{y}{x})$

$latex \theta={{\tan}^{-1}}(\frac{2}{2})$

$latex \theta={{\tan}^{-1}}(1)$

$latex \theta=45$° $latex =\frac{\pi}{4}$

The component in *z* is the same, so the cylindrical coordinates are $latex (\sqrt{8}, \frac{\pi}{4}, 5)$.

**EXAMPLE 2**

What is the point (-3, 6, 3) equal to in cylindrical coordinates?

##### Solution

We have the values $latex x=-3,~y=6,~z=3$. We have to find *r, θ* and *z,* using these values. We start with *r*:

$latex r=\sqrt{{{x}^2}+{{y}^2}}$

$latex r=\sqrt{{{(-3)}^2}+{{6}^2}}$

$latex r=\sqrt{9+36}$

$latex r=\sqrt{45}$

$latex r=3\sqrt{5}$

Now, we find the angle *θ,* using the following formula:

$latex \theta={{\tan}^{-1}}(\frac{y}{x})$

$latex \theta={{\tan}^{-1}}(\frac{6}{-3})$

$latex \theta={{\tan}^{-1}}(-2)$

$latex \theta=-63.4$°

However, we note that the point is located in the second quadrant since the component in *x* is negative and the component in *y* is positive. Therefore, we have to add 180° to the obtained value. That means the angle is $latex \theta=-63.4+180=116.6$°.

The component in *z* is the same. Therefore, the point in cylindrical coordinates is $latex (3\sqrt{5}, 116.6^{\circ}, 3)$.

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**EXAMPLE 3**

If we have the point (-4, -1, -3) in Cartesian coordinates, what is its equivalence in cylindrical coordinates?

##### Solution

We can recognize the values $latex x = -4, ~ y = -1, ~ z = -3$. To find this point in cylindrical coordinates, we have to find *r, θ* and *z*. We can find *r*, using the following equation:

$latex r=\sqrt{{{x}^2}+{{y}^2}}$

$latex r=\sqrt{{{(-4)}^2}+{{(-1)}^2}}$

$latex r=\sqrt{16+1}$

$latex r=\sqrt{17}$

Now, we find the angle *θ,* using the following formula:

$latex \theta={{\tan}^{-1}}(\frac{y}{x})$

$latex \theta={{\tan}^{-1}}(\frac{-1}{-4})$

$latex \theta=14$°

In this case, the point is in the third quadrant since both the *x* component and the *y* component are negative. Therefore, we have to add 180° to the angle obtained to find the correct angle. The correct angle is $latex \theta = 14 + 180 = 194$°.

Taking into account that the *z* component is the same, the cylindrical coordinates are $latex (\sqrt{17}, 194^{\circ}, -3)$.

**EXAMPLE 4**

Determine the equivalence in cylindrical coordinates of the point (2, -6, 4).

##### Solution

We recognize the values $latex x = 2, ~ y = -6, ~ z = 4$. We use these values to find *r, θ* and *z* and form the cylindrical coordinates. We find *r*, using the following equation:

$latex r=\sqrt{{{x}^2}+{{y}^2}}$

$latex r=\sqrt{{{2}^2}+{{(-6)}^2}}$

$latex r=\sqrt{4+36}$

$latex r=\sqrt{40}$

$latex r=2\sqrt{10}$

The angle *θ,* is found as follows:

$latex \theta={{\tan}^{-1}}(\frac{y}{x})$

$latex \theta={{\tan}^{-1}}(\frac{-6}{2})$

$latex \theta={{\tan}^{-1}}(-3)$

$latex \theta=-71.6$°

We note that the point is in the fourth quadrant since the component in *x* is positive and the component in *y* is negative. Therefore, we have to add 360° to the angle given by the calculator to find the correct value. The correct angle is $latex \theta=-71.6+360=288.4$°.

We keep the *z* component, so the cylindrical coordinates are $latex (2\sqrt{10}, 288.4^{\circ}, 4)$.

## Cartesian to Cylindrical coordinates – Practice problems

Put into practice what you have learned about transforming Cartesian to cylindrical coordinates by solving the following problems. Select an answer and check it to make sure you selected the correct one.

## See also

Interested in learning more about cylindrical coordinates? Take a look at these pages:

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