## What is the Cylindrical Coordinate System?

The distance from a selected reference axis, the direction from the axis relative to a selected reference direction, and also the distance from a selected reference plane perpendicular to the axis area unit, all accustom outline purpose positions in a cylindrical reference frame. Counting that the aspect of the reference plane faces the purpose, the latter distance is expressed as a positive or negative variety.

## Contributions of Mathematicians

The polar coordinate system is a refinement of René Descartes' (1596–1650) two-dimensional coordinate system, which he invented in 1637.

Sir Isaac Newton (1640–1727) established ten separate coordinate systems many decades after Descartes published his two-dimensional coordinate system. The cylindrical coordinate system is one of them.

## Representation for Cylindrical Coordinates

There is no standard notation for cylindrical coordinates.

The ISO norm 31-11 recommends $\left(\rho ,\phi ,z\right)$, where z is the height and $\rho$is the radial coordinate.

The radius is often abbreviated as r or s, the azimuth $\phi$ as t or $\theta$ , and the third coordinate as h.

A point P's three coordinates $\left(r,\theta ,z\right)$ are defined as follows:

The Euclidean distance from the z-axis to point P is known as the axial distance r or radial distance.

The theta $\theta$ is the angle formed by the reference direction on the chosen plane and the line connecting the origin to P's plane projection.

The signed distance from the chosen plane to the point P is the axial coordinate or height z.

## What is a Coordinate System?

• A coordinate system is needed to represent points in the space. It is possible for the coordinate system to be orthogonal or non-orthogonal.
• Right-handed and left-handed coordinate schemes are also possible. A set of coordinate systems determines it which is a set of planes and/or surfaces.
• A coordinate system is a series of coordinates. At each and every point in space, there are three reference directions.
• The reference point for the coordinate system is the origin. Any other point in space is located using this method. The location of a point in space in relation to the origin is defined. Coordinate directions apply to three reference directions.
• The origin (denoted by O) and a basis composed of three mutually perpendicular vectors make up a three-dimensional Cartesian coordinate system. The three coordinate axes are defined by these vectors: x, y, and z. Axis is a word that can be used to describe each of the axes.
• They are often referred to as the abscissa, ordinate, and applicate axes. Three real numbers specify the coordinates of every point in space.

### Insight into Cylindrical Coordinate System

• The origin of the system is for the purpose of representing the points wherever all 3 coordinates may be given as zero. This can be the intersection between the reference plane and also the axis. The axis is diversely referred to as the cylindrical or longitudinal axis, to differentiate it from the polar axis, that is the ray that lies within the reference plane, beginning at the origin and pointing in the reference direction. The other directions that are perpendicular to the longitudinal axis are referred to as radial lines.
• The distance from the axis is also known as the radial distance or radius, whereas the angular coordinate is usually cited because of the spatial relation or because of the angle. The radius and also the angle area unit are known as the polar coordinates, as they correspond to a two-dimensional coordinate system within the plane through the purpose, parallel to the reference plane. The third coordinate is also known as the peak or altitude (if the reference plane is taken into account horizontal), longitudinal position, or axial position.
• They are also known as "cylindrical polar coordinates" and "polar cylindrical coordinates", and they are used to describe the locations of stars in galaxies. So they are termed as "galactocentric cylindrical polar coordinates".

### An Extension of Polar Coordinate System

• Cylindrical coordinates are a basic three-dimensional extension of two-dimensional polar coordinates. Remember that polar coordinates $\left(r,\theta \right)$ can be used to define the location of a point in the plane. The polar coordinate r represents the point's distance from the origin. The angle between the x-axis and the line segment from the origin to the point is known as the polar coordinate.
• The polar coordinates in the xy-plane are simply combined with the normal z coordinate of Cartesian coordinates to form cylindrical coordinates.
• Simply project a point P in the xy-plane down to a point Q to get its cylindrical coordinates.
• Then measure the polar coordinates $\left(r,\theta \right)$of the point Q, where r is the distance from the origin to Q and is the angle between the positive x-axis and the line segment connecting the origin and Q.
• The z-coordinate is the same as the third cylindrical coordinate. It is the xy-signed plane's distance from point P.
• The cylindrical coordinates are $\left(r,\theta ,z\right)$ of the point P.

## How to Change from One Coordinate System to Another?

For Converting Cartesian $\left(x,y,z\right)$ to Cylindrical $\left(r,\theta ,z\right)$ Coordinate System:

$\begin{array}{c}r=\sqrt{{x}^{2}+{y}^{2}}\\ \theta ={\mathrm{tan}}^{-1}\left(\frac{y}{x}\right)\\ z=z\end{array}$

Example: Convert the Cartesian coordinate $\left(-4,4,6\right)$ to a cylindrical coordinate system.

Solution: Apply the conversion formulas mentioned above to obtain the result.

Substitute x as $-4$, y as 4 and z as 6.

The radius r is expressed as,

$\begin{array}{c}r=\sqrt{{\left(-4\right)}^{2}+{4}^{2}}\\ =\sqrt{16+16}\\ =4\sqrt{2}\end{array}$

The theta is expressed as,

$\begin{array}{c}\theta ={\mathrm{tan}}^{-1}\left(\frac{4}{-4}\right)\\ ={\mathrm{tan}}^{-1}\left(-1\right)\\ =\frac{3\pi }{4}\end{array}$

The z-coordinate remains the same.

$z=6$

Therefore, the cylindrical coordinates are $\left(4\sqrt{2},\frac{3\pi }{4},6\right)$.

For Converting Cylindrical $\left(r,\theta ,z\right)$ to Cartesian $\left(x,y,z\right)$ Coordinate System:

 $\begin{array}{c}x=r\mathrm{cos}\theta \\ y=r\mathrm{sin}\theta \\ z=z\end{array}$

Example: Convert the equation in the Cartesian system $2{x}^{2}-2{y}^{2}=36$ to a cylindrical system.

Solution: Apply the conversion formulas mentioned above to obtain the result.

Substitute x as r times cos of theta, y as r times sine of theta in the equation $2{x}^{2}-2{y}^{2}=36$.

$\begin{array}{c}2{\left(r\mathrm{cos}\theta \right)}^{2}-2{\left(r\mathrm{sin}\theta \right)}^{2}=36\\ 2{r}^{2}{\mathrm{cos}}^{2}\theta -2{r}^{2}{\mathrm{sin}}^{2}\theta =36\\ 2{r}^{2}\left({\mathrm{cos}}^{2}\theta -{\mathrm{sin}}^{2}\theta \right)=36\end{array}$

Apply the double angle formula for cos trigonometric function ${\mathrm{cos}}^{2}x-{\mathrm{sin}}^{2}x=\mathrm{cos}2x$:

$\begin{array}{c}{r}^{2}\mathrm{cos}2\theta =\frac{36}{2}\\ {r}^{2}\mathrm{cos}2\theta =18\end{array}$

The cos function is reciprocal of secant trigonometric function,

$\begin{array}{c}\frac{{r}^{2}}{\mathrm{sec}2\theta }=18\\ {r}^{2}=18\mathrm{sec}2\theta \end{array}$

Thus, the equation in the cylindrical system is ${r}^{2}=18\mathrm{sec}2\theta$.

For Converting Cylindrical $\left(r,\theta ,z\right)$ to Spherical $\left(\rho ,\theta ,\phi \right)$Coordinate System:

$\begin{array}{l}\rho =\sqrt{{r}^{2}+{z}^{2}}\\ \theta =\theta \\ \phi ={\mathrm{cos}}^{-1}\left(\frac{z}{\sqrt{{r}^{2}+{z}^{2}}}\right)\end{array}$

Example: Convert the cylindrical coordinate $\left(2,\frac{\pi }{2},1\right)$ to a spherical coordinate system.

Solution: Apply the conversion formulas mentioned above to obtain the result.

The cylindrical system is written as $\left(r,\theta ,z\right)$ which is equal to $\left(2,\frac{\pi }{2},1\right)$.

The rho is calculated as,

$\begin{array}{c}\rho =\sqrt{{2}^{2}+{1}^{2}}\\ =\sqrt{4+1}\\ =\sqrt{5}\end{array}$

The theta value remains the same.

$\theta =\frac{\pi }{2}$

The azimuth is the inverse cos trigonometric function expressed as, $\phi ={\mathrm{cos}}^{-1}\left(\frac{z}{\sqrt{{r}^{2}+{z}^{2}}}\right)$

$\begin{array}{c}\phi ={\mathrm{cos}}^{-1}\left(\frac{1}{\sqrt{{2}^{2}+{1}^{2}}}\right)\\ ={\mathrm{cos}}^{-1}\left(\frac{1}{\sqrt{5}}\right)\end{array}$

Thus, the spherical coordinates are expressed as, $\left(\sqrt{5},\frac{\pi }{2},{\mathrm{cos}}^{-1}\left(\frac{1}{\sqrt{5}}\right)\right)$.

For Converting Spherical $\left(\rho ,\theta ,\phi \right)$ to Cylindrical $\left(r,\theta ,z\right)$ Coordinate System:

$\begin{array}{l}r=\rho \mathrm{sin}\phi \\ \theta =\theta \\ z=\rho \mathrm{cos}\phi \end{array}$

Example: Convert the equation in the spherical system $\rho =4\mathrm{cos}\phi$ to a cylindrical system.

Solution: Apply the conversion formulas mentioned above to obtain the result.

Multiply both sides of the equation $\rho =4\mathrm{cos}\phi$ by $\rho$,

${\rho }^{2}=4\rho \mathrm{cos}\phi$ ….(1)

We know that r is rho times sin of $\phi$ and z is rho times cos of $\phi$.

Squaring the equations we obtain,

Next, we add both the equations together,

$\begin{array}{c}{r}^{2}+{z}^{2}={\rho }^{2}{\mathrm{sin}}^{2}\phi +{\rho }^{2}{\mathrm{cos}}^{2}\phi \\ {r}^{2}+{z}^{2}={\rho }^{2}\left({\mathrm{sin}}^{2}\phi +{\mathrm{cos}}^{2}\phi \right)\\ {r}^{2}+{z}^{2}={\rho }^{2}\end{array}$

So, we have a relation as ${r}^{2}+{z}^{2}={\rho }^{2}$ ….(2)

Substituting equation (2) in equation (1) and using the rule that $z=\rho \mathrm{cos}\phi$.

${r}^{2}+{z}^{2}=4z$

Thus, the equation in the cylindrical coordinate system is ${r}^{2}+{z}^{2}=4z$.

## Common Mistakes

Consider an example: Convert the Cartesian coordinate $\left(-5,5,10\right)$ to a cylindrical coordinate system.

Solution: Apply the conversion formulas mentioned above to obtain the result.

Substitute x as $-5$, y as 5 and z as 10.

The radius r is expressed as,

$\begin{array}{c}r=\sqrt{{\left(-5\right)}^{2}+{5}^{2}}\\ =\sqrt{25+25}\\ =5\sqrt{2}\end{array}$

The theta is expressed as,

$\begin{array}{c}\theta ={\mathrm{tan}}^{-1}\left(\frac{5}{-5}\right)\\ ={\mathrm{tan}}^{-1}\left(-1\right)\\ =-{\mathrm{tan}}^{-1}\left(1\right)\\ =-\frac{\pi }{4}\end{array}$

We would have solved it like this, but the angle theta is always positive and taken in the anticlockwise direction. The point lies in the second quadrant so, to make it positive we add $\pi$ to it.

$\begin{array}{c}\theta =\pi -\frac{\pi }{4}\\ =\frac{3\pi }{4}\end{array}$

This is the correct way of doing it.

If the value of theta lied in the fourth quadrant, then we need to add $2\pi$ to it.

The z-coordinate remains the same.

$z=10$

Therefore, the cylindrical coordinates are $\left(5\sqrt{2},\frac{3\pi }{4},10\right)$.

## Context and Application

• Cylindrical coordinates are helpful in reference to objects and phenomena that have some motion symmetry regarding the longitudinal axis, like water flow during a straight pipe with spherical crosswise, heat distribution during a metal cylinder, magnetic attraction fields made by an electrical current during a long, straight wire, accretion disks in natural philosophy, and so on.
• The best audio pickup patterns for cardioid microphones can also be determined using such coordinates.
• When dealing with a curved trajectory of a moving object (dynamics) and when the movement is repeated back and forth (oscillation) or round and round (rotation), physicists and engineers use cylindrical coordinates (rotation).

## Formulas

• For converting Cartesian $\left(x,y,z\right)$ to the cylindrical $\left(r,\theta ,z\right)$ coordinate system:

$\begin{array}{c}r=\sqrt{{x}^{2}+{y}^{2}}\\ \theta ={\mathrm{tan}}^{-1}\left(\frac{y}{x}\right)\\ z=z\end{array}$

• For converting cylindrical $\left(r,\theta ,z\right)$ to the Cartesian $\left(x,y,z\right)$ coordinate system:

$\begin{array}{c}x=r\mathrm{cos}\theta \\ y=r\mathrm{sin}\theta \\ z=z\end{array}$

• For converting cylindrical $\left(r,\theta ,z\right)$ to the spherical $\left(\rho ,\theta ,\phi \right)$ coordinate system:

$\begin{array}{l}\rho =\sqrt{{r}^{2}+{z}^{2}}\\ \theta =\theta \\ \phi ={\mathrm{cos}}^{-1}\left(\frac{z}{\sqrt{{r}^{2}+{z}^{2}}}\right)\end{array}$

• From spherical $\left(\rho ,\theta ,\phi \right)$ to cylindrical $\left(r,\theta ,z\right)$ coordinate:

$\begin{array}{l}r=\rho \mathrm{sin}\phi \\ \theta =\theta \\ z=\rho \mathrm{cos}\phi \end{array}$

## Practice Problem

Question: Convert the equation in the Cartesian form $4{x}^{2}-4{y}^{2}=24$ to a cylindrical system.

Solution: Apply the conversion formulas mentioned below to obtain the result.

$\begin{array}{c}x=r\mathrm{cos}\theta \\ y=r\mathrm{sin}\theta \\ z=z\end{array}$

Substitute x as r times cos of theta, y as r times sine of theta in the equation, $4{x}^{2}-4{y}^{2}=24$

$\begin{array}{c}4{\left(r\mathrm{cos}\theta \right)}^{2}-4{\left(r\mathrm{sin}\theta \right)}^{2}=24\\ 4{r}^{2}{\mathrm{cos}}^{2}\theta -4{r}^{2}{\mathrm{sin}}^{2}\theta =24\\ 4{r}^{2}\left({\mathrm{cos}}^{2}\theta -{\mathrm{sin}}^{2}\theta \right)=24\end{array}$

Apply the double angle formula for cos trigonometric function ${\mathrm{cos}}^{2}x-{\mathrm{sin}}^{2}x=\mathrm{cos}2x$

$\begin{array}{c}{r}^{2}\mathrm{cos}2\theta =\frac{24}{4}\\ {r}^{2}\mathrm{cos}2\theta =6\end{array}$

The cos function is reciprocal of secant trigonometric function,

$\begin{array}{c}\frac{{r}^{2}}{\mathrm{sec}2\theta }=6\\ {r}^{2}=6\mathrm{sec}2\theta \end{array}$

Thus, the equation in the cylindrical system is ${r}^{2}=6\mathrm{sec}2\theta$.

### Want more help with your geometry homework?

We've got you covered with step-by-step solutions to millions of textbook problems, subject matter experts on standby 24/7 when you're stumped, and more.
Check out a sample geometry Q&A solution here!

*Response times may vary by subject and question complexity. Median response time is 34 minutes for paid subscribers and may be longer for promotional offers.

### Search. Solve. Succeed!

Study smarter access to millions of step-by step textbook solutions, our Q&A library, and AI powered Math Solver. Plus, you get 30 questions to ask an expert each month.

Tagged in
MathGeometry

### Cylindrical Coordinates

• Copyright, Community Guidelines, DSA & other Legal Resources: Learneo Legal Center
• bartleby, a Learneo, Inc. business