# Precalculus

## Analytic Geometry # Cylindrical Coordinates

To determine the position of a point in three-dimensional space, cylindrical coordinates are often used. Cylindrical coordinates are introduced as follows. Choose a plane π and define on it a polar coordinate system with a pole O and a polar axis Ox. Cylindrical coordinates extend the polar coordinate system by adding the vertical z-axis passing through the pole O and perpendicular to the plane π.

Let M be any point in space, N is its orthogonal projection onto the plane π, and $$\left({\rho, \varphi}\right)$$ are the polar coordinates of point N on the plane π. The point Mz is the projection of point M on the z-axis so that

$OM_z = z,\;-\infty \lt z \lt \infty.$

The ordered triplet of numbers $$\left({\rho, \varphi, z}\right)$$ is called the cylindrical coordinates of the point M.

If some point M has coordinates $$\rho,$$ $$\varphi,$$ $$z$$ this is denoted as $$M\left({\rho, \varphi, z}\right).$$

The name "cylindrical coordinates" is due to the fact that the coordinate surface $$\rho = const$$ (that is, the surface all the points of which have the same $$\rho-$$coordinate) is a cylinder with the generatrix parallel to the $$z-$$axis.

## Converting Between Cylindrical and Cartesian Coordinates

Let the cylindrical and Cartesian coordinate systems have a common origin at point $$O.$$ If you choose the axes of the Cartesian coordinate system as indicated in the figure, then the Cartesian coordinates $$\left({x, y, z}\right)$$ of the point M will be related to its cylindrical coordinates $$\left({\rho, \varphi, z}\right)$$ by the formulas

$x = \rho\cos\varphi,\;\;y = \rho\sin\varphi,\;\;z = z.$

The inverse conversion from Cartesian to cylindrical coordinates is given by the expressions

$\rho = \sqrt{x^2 + y^2},\;\;\tan\varphi = \frac{y}{x},\;\;z=z.$

## Solved Problems

Click or tap a problem to see the solution.

### Example 1

Find cylindrical coordinates of the points

1. $$A\left({-1,-1,1}\right)$$
2. $$B\left({-3,4,5}\right)$$
3. $$C\left({0,-2,-1}\right)$$

### Example 2

Given the cylindrical coordinates of points find their Cartesian coordinates.

1. $$K\left({1,\frac{\pi}{4},-1}\right)$$
2. $$L\left({\sqrt{3},\frac{2\pi}{3},0}\right)$$
3. $$M\left({4,0,-4}\right)$$

### Example 3

Given two points in cylindrical coordinates $$A\left({3,\frac{\pi}{2},4}\right)$$ and $$B\left({1,\frac{3\pi}{2},7}\right)$$ find the distance between them.

### Example 4

Given the cylindrical coordinates of the point $$M\left({\rho,\varphi, z}\right)$$ find the angle $$\alpha$$ between vector $$\mathbf{OM}$$ and the $$x-$$axis.

### Example 1.

Find cylindrical coordinates of the points

1. $$A\left({-1,-1,1}\right)$$
2. $$B\left({-3,4,5}\right)$$
3. $$C\left({0,-2,-1}\right)$$

Solution.

Calculate the polar radius $$\rho$$ and the polar angle $$\varphi$$ using the formulas

$\rho = \sqrt{x^2 + y^2},\;\;\tan\varphi = \frac{y}{x}.$

The $$z-$$coordinate (applicate) remains unchanged.

1. $$A\left({-1,-1,1}\right)$$
$\rho = \sqrt{\left({-1}\right)^2 + \left({-1}\right)^2} = \sqrt{2},\;\tan\varphi = \frac{-1}{-1} = 1, \Rightarrow \varphi = \frac{\pi}{4};$
$A\left({-1,-1,1}\right) \mapsto A\left({\sqrt{2},\frac{\pi}{4},1}\right);$
2. $$B\left({-3,4,5}\right)$$
$\rho = \sqrt{\left({-3}\right)^2 + 4^2} = 5,\;\tan\varphi = \frac{4}{-3} = -\frac{4}{3}, \Rightarrow \varphi = \arctan{\left({-\frac{4}{3}}\right)} = -\arctan\frac{4}{3};$
$B\left({-3,4,5}\right) \mapsto B\left({5,-\arctan\frac{4}{3},5}\right);$
3. $$C\left({0,-2,-1}\right)$$
$\rho = \sqrt{0^2 + \left({-2}\right)^2} = 2,\;\tan\varphi = \frac{-2}{0} = -\infty, \Rightarrow \varphi = \frac{3\pi}{2};$
$C\left({0,-2,-1}\right) \mapsto C\left({2,\frac{3\pi}{2},-1}\right).$

### Example 2.

Given the cylindrical coordinates of points find their Cartesian coordinates.

1. $$K\left({1,\frac{\pi}{4},-1}\right)$$
2. $$L\left({\sqrt{3},\frac{2\pi}{3},0}\right)$$
3. $$M\left({4,0,-4}\right)$$

Solution.

To convert from cylindrical to Cartesian coordinate system, we use the formulas

$x = \rho\cos\varphi,\;\;y = \rho\sin\varphi,\;\;z = z.$
1. $$K\left({1,\frac{\pi}{4},-1}\right)$$
$x = 1\cdot\cos\frac{\pi}{4} = 1\cdot1 = 1,\;y = 1\cdot\sin\frac{\pi}{4} = 1\cdot1 = 1,\;z = -1;$
$K\left({1,\frac{\pi}{4},-1}\right) \mapsto K\left({1,1,-1}\right);$
2. $$L\left({\sqrt{3},\frac{2\pi}{3},0}\right)$$
$x = \sqrt{3}\cos\frac{2\pi}{3} = \sqrt{3}\left({-\frac{1}{2}}\right)= -\frac{\sqrt{3}}{2},\;y = \sqrt{3}\sin\frac{2\pi}{3} = \sqrt{3}\cdot\frac{\sqrt{3}}{2} = \frac{3}{2},\;z = 0;$
$L\left({\sqrt{3},\frac{2\pi}{3},0}\right) \mapsto L\left({-\frac{\sqrt{3}}{2},\frac{3}{2},0}\right);$
3. $$M\left({4,0,-4}\right)$$
$x = 4\cos0 = 4\cdot1 = 4,\;y = 4\sin0 = 4\cdot0 = 0,\;z = -4;$
$M\left({4,0,-4}\right) \mapsto M\left({4,0,-4}\right).$

### Example 3.

Given two points in cylindrical coordinates $$A\left({3,\frac{\pi}{2},4}\right)$$ and $$B\left({1,\frac{3\pi}{2},7}\right)$$ find the distance between them.

Solution.

Let's convert the cylindrical coordinates of the points to Cartesian ones:

$x_A = \rho\cos\varphi = 3\cos\frac{\pi}{2} = 3\cdot0 = 0,\;\;y_A = \rho\sin\varphi = 3\sin\frac{\pi}{2} = 3\cdot1 = 3,\;\;z_A = 4;$
$x_B = \rho\cos\varphi = 1\cos\frac{3\pi}{2} = 1\cdot0 = 0,\;\;y_B = \rho\sin\varphi = 1\sin\frac{3\pi}{2} = 1\cdot\left({-1}\right) = -1,\;\;z_B = 7.$

Hence points $$A$$ and $$B$$ have the following coordinates $$x,y,z:$$

$A\left({0,3,4}\right) \text{ and } B\left({0,-1,7}\right).$

Now you can easily calculate the distance between these points:

$d\left({AB}\right) = \sqrt{\left({x_B - x_A}\right)^2 + \left({y_B - y_A}\right)^2 + \left({z_B - z_A}\right)^2} = \sqrt{\left({0 - 0}\right)^2 + \left({-1 - 3}\right)^2 + \left({7 - 4}\right)^2} = \sqrt{0 + 16 + 9} = \sqrt{25} = 5.$

### Example 4.

Given the cylindrical coordinates of the point $$M\left({\rho,\varphi, z}\right)$$ find the angle $$\alpha$$ between vector $$\mathbf{OM}$$ and the $$x-$$axis.

Solution.

In the figure below, the height $$MN$$ is perpendicular to the $$xy-$$plane (plane π). The straight line $$NL$$ lies in the $$xy-$$plane and is perpendicular to the $$x-$$axis. Therefore, by the theorem of three perpendiculars, the line $$ML$$ is also perpendicular to the $$x-$$axis, i.e. triangle $$MLO$$ is a right triangle.

The cosine of the angle $$\alpha = \angle LOM$$ in the right triangle $$MLO$$ is given by the formula

$\cos\alpha = \frac{OL}{OM}.$

Express the leg $$OL$$ in terms of known cylindrical coordinates:

$OL = x = \rho\cos\varphi.$

According to the Pythagorean theorem,

$OM = \sqrt{ON^2 + MN^2} = \sqrt{\rho^2 + z^2}.$

As a result we get

$\cos\alpha = \frac{OL}{OM} = \frac{\rho\cos\varphi}{\sqrt{\rho^2 + z^2}}.$