Cylindrical coordinate system
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The cylindrical coordinate system is a three-dimensional system which essentially extends circular polar coordinates by adding a third coordinate (usually denoted <math>h</math>) which measures the height of a point above the plane.
A point P is given as <math>(r, \theta, h)</math>. In terms of the Cartesian coordinate system:
- <math>r</math> 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.
- <math>\theta</math> is the angle between the positive x-axis and the line OP', measured anti-clockwise.
- <math>h</math> is the same as <math>z</math>.
Some mathematicians indeed use <math>(r, \theta, z)</math>. It is also common in physics to use <math>(\rho, \phi, z)</math> to denote these coordinates.
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.
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Conversion from cylindrical to Cartesian coordinates
<math>x = r \cos\theta</math>
<math>y = r \sin\theta</math>
<math>z = h</math><math> \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} </math>
Conversion from Cartesian to cylindrical coordinates
<math>r = \sqrt{x^2 + y^2}</math>
<math>\theta = \arctan\frac{y}{x}</math>
<math>h = z\,</math><math> \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}
</math>
Conversion from cylindrical to spherical coordinates
<math>{\rho} = \sqrt{r^2+h^2}</math>
<math>{\phi} = \theta \qquad </math>
<math>{\theta'} = \arctan\frac{h}{r} \qquad </math><math> \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}
</math>
where φ is the azimuth and θ' is the latitude.
Conversion from spherical to cylindrical coordinates
<math>{r} = \rho \cos \theta </math>
<math>{\theta'} = \phi </math>
<math>{h} = \rho \sin \theta </math><math> \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} </math>
where φ is azimuth and θ is latitude.
Line and volume elements
In many problems involving cylindrical polar coordinates, it is useful to know the line and volume elements; these are used in integration to solve problems involving paths and volumes.
The line element is <math>dl\ = dr\ + r\,d\theta\ + dz</math>.
The volume element is <math>dV\ = r\,dr\,d\theta\,dz</math>.
It is also important in many cases to be able to find the gradient of a vector field in cylindrical polar coordinates. The gradient can be worked out from first principals, if one knows theta, r and z in terms of cartesian coordinates, but the general equation is given below.
<math>\nabla \equiv \mathbf{\hat r}\frac{\partial}{\partial r} + \boldsymbol{\hat \theta}\frac{1}{r}\frac{\partial}{\partial \theta} + \mathbf{\hat z}\frac{\partial}{\partial z}</math>.