Chemistry - Cylindrical coordinate system

The cylindrical coordinate system is a three-dimensional system which essentially extends circular polar coordinates by adding a third coordinate (usually denoted $h$ ) which measures the height of a point above the plane.

A point P is given as $(r,θ, h)$ . In terms of the Cartesian coordinate system:

$r$ is the distance from O to P', the orthogonal projection of the point P onto the XY plane. This is the same as the distance of P to the z-axis.
$θ$ is the angle between the positive x-axis and the line OP', measured anti-clockwise.
$h$ is the same as $z$ .

Some mathematicians indeed use $(r,θ, z)$ .

Cylindrical coordinates are useful in analyzing surfaces that are symmetrical about an axis, with the z-axis chosen as the axis of symmetry. For example, the infinitely long circular cylinder that has the Cartesian equation x² + y² = c² has the very simple equation r = c in cylindrical coordinates. Hence the name "cylindrical" coordinates.

Contents

1 Conversion from cylindrical to Cartesian coordinates

2 Conversion from Cartesian to cylindrical coordinates

3 Conversion from cylindrical to spherical coordinates

4 Conversion from spherical to cylindrical coordinates

5 See also

Conversion from cylindrical to Cartesian coordinates

$x = r cosθ$
$y = r sinθ$
$z = h$

$\begin{vmatrix}dx\\dy\\dz\end{vmatrix} = \begin{vmatrix} \cos\theta&-r\sin\theta&0\\ \sin\theta&r\cos\theta&0\\ 0&0&1 \end{vmatrix} \cdot \begin{vmatrix}dr\\d\theta\\dh\end{vmatrix}$

Conversion from Cartesian to cylindrical coordinates

$r = \sqrt{x^2 + y^2}$
$\theta = \arctan\frac{y}{x}$
$h = z\,$

$\begin{vmatrix}dr\\d\theta\\dh\end{vmatrix} = \begin{vmatrix} \frac{x}{\sqrt{x^2+y^2}}&\frac{y}{\sqrt{x^2+y^2}}&0\\ \frac{-y}{x^2+y^2}&\frac{x}{x^2+y^2}&0\\ 0&0&1 \end{vmatrix} \cdot \begin{vmatrix}dx\\dy\\dz\end{vmatrix}$

Conversion from cylindrical to spherical coordinates

${\rho} = \sqrt{r^2+h^2}$
${\phi} = \theta \qquad$
${\theta'} = \arctan\frac{h}{r} \qquad$

$\begin{vmatrix}d\rho\\d\phi\\d\theta' \end{vmatrix} = \begin{vmatrix} \frac{r}{\sqrt{r^2+h^2}} & 0 & \frac{h}{\sqrt{r^2+h^2}} \\ 0 & 1 & 0 \\ \frac{-h}{r^2+h^2} & 0 & \frac{r}{r^2+h^2} \end{vmatrix} \cdot \begin{vmatrix}dr\\d\theta\\dh\end{vmatrix}$

where φ is the azimuth and θ' is the latitude.

Conversion from spherical to cylindrical coordinates

$r = ρcosθ$
$θ' = φ$
$h = ρsinθ$

$\begin{vmatrix}dr\\d\theta'\\dh\end{vmatrix} = \begin{vmatrix} \cos \theta & 0 & - \rho \sin \theta \\ 0 & 1 & 0 \\ \sin \theta & 0 & \rho \cos \theta \end{vmatrix} \cdot \begin{vmatrix}d\rho\\d\phi\\d\theta\end{vmatrix}$

where φ is azimuth and θ is latitude.