Conversion between Cartesian and Cylindrical Coordinates
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Theorem
Let $S$ be a $3$-dimensional space.
Let a Cartesian $3$-space $\CC$ be applied to $S$.
Let a cylindrical coordinate system $\PP$ be superimposed upon $\CC$ such that:
- $(1): \quad$ The origin of $\CC$ and $P$ coincide with the pole of the polar coordinate plane of $\PP$ corresponding to $z = 0$
- $(3): \quad$ The $x$-axis of $\CC$ coincides with the polar axis the polar coordinate plane of $\PP$ corresponding to $z = 0$.
Let $p$ be a point in $S$.
Let $p$ be specified as $p = \tuple {r, \theta, z}$ expressed in the cylindrical coordinates of $\PP$.
Then $p$ is expressed as $\tuple {r \cos \theta, r \sin \theta, z}$ in $\CC$.
Contrariwise, let $p$ be expressed as $\tuple {x, y, z}$ in the cartesian coordinates of $\CC$.
Then $p$ is expressed in $\PP$ as $\tuple {r, \theta, z}$, where:
\(\ds r\) | \(=\) | \(\ds \sqrt {x^2 + y^2}\) | ||||||||||||
\(\ds \theta\) | \(=\) | \(\ds \arctan \dfrac y x + \pi \sqbrk {x < 0 \text{ or } y < 0} + \pi \sqbrk {x > 0 \text{ and } y < 0}\) | ||||||||||||
\(\ds z\) | \(=\) | \(\ds z\) |
where:
- $\sqbrk {\, \cdot \,}$ is Iverson's convention.
- $\arctan$ denotes the arctangent function.
Proof
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Sources
- 1998: David Nelson: The Penguin Dictionary of Mathematics (2nd ed.) ... (previous) ... (next): cylindrical coordinate system
- 2008: David Nelson: The Penguin Dictionary of Mathematics (4th ed.) ... (previous) ... (next): cylindrical coordinate system