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 $z = 0$
$(2): \quad$ The $z$-axis of $\CC$ coincides with the $z$-axis of $\PP$.
$(3): \quad$ 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

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