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I'm aware of all that transformations from one coordinate system to another, but they don't really help me parametrising a surface. Rather the implicit functions are useful to me.

For example the unit sphere can be described simply by $x^2+y^2+z^2 < 1$, what easily brings the parametrisation:

$$\text{unit sphere:}\:\psi(x,y)=\left(\begin{array}{cc}x\\y\\\pm\sqrt{1-x^2-y^2}\end{array}\right)\quad x\in[0,1] \quad y\in[0,1]$$

From there u can receive the parametrisation in cylindrical or spherical coordinates merely by plugging in $\texttt{cylindrical:}\: x = r\,\cos(\varphi), y = r\,\sin(\varphi)$ or $\texttt{spherical:}\: x = \cos(\varphi)\,\sin(\theta), y = \sin(\varphi)\,\sin(\theta)$, leaving us with:

$$\text{unit sphere:}\:\psi(r,\varphi)=\left(\begin{array}{cc}r\,\cos(\varphi)\\r\,\sin(\varphi)\\\pm\sqrt{1-r^2}\end{array}\right)\quad r\in[0,1] \quad \varphi\in[0,2\,\pi]$$

$$\text{unit sphere:}\:\psi(\varphi,\theta)=\left(\begin{array}{cc}\cos(\varphi)\,\sin(\theta)\\\sin(\varphi)\,\sin(\theta)\\\cos(\theta)\end{array}\right)\quad \varphi\in[0,2\,\pi] \quad \theta\in[0,\pi]$$

Now to my actual question trying this with a cylinder. Since it's implicit function is independent of $z$, I already had a little trouble finding the cartesian parametrisation: $x^2+y^2<1$, so I just came up with:

$$\begin{array}{cc} &\text{unit cylinder:}&\:\psi(y,z)=\left(\begin{array}{cc}\pm\sqrt{1-y^2}\\y\\z\end{array}\right)\quad y\in[0,1] \quad z\in[0,1] \\\\ &&&&&\llap{\text{Now using $\texttt{cylindrical:}\: y = r\,\sin(\varphi), z = z$ or $\texttt{spherical:}\: y = \sin(\varphi)\,\sin(\theta), z = \cos(\theta)$, getting:}} \\\\ &\text{unit cylinder:}&\:\psi(r,z)=\left(\begin{array}{cc}r\,\sin(\varphi)\\r\,\cos(\varphi)\\z\end{array}\right)\quad r\in[0,1] \quad \varphi\in[0,2\,\pi] \\\\ &\text{unit cylinder:}&\:\psi(\varphi,\theta)=\left(\begin{array}{cc}\pm\sqrt{1-\sin^2(\varphi)\,sin^2(\theta)}\\\sin(\varphi)\sin(\theta)\\\cos(\theta)\end{array}\right)\quad \varphi\in[0,2\,\pi] \quad \varphi\in[0,\pi]\end{array}$$

the last parametrisation is exactly what I am looking for: a cylinder in spheric coordinates, but it looks a tad awkward.

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    $\begingroup$ In spherical coordinates $r$ increases as you move up the cylinder (z axis). $r$ is not $1$. $\endgroup$ Commented Jun 23, 2021 at 12:20
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    $\begingroup$ $r \sin{\theta} = 1$, the radius of the circle. $r = \displaystyle \frac{1}{\sin{\theta}}$ $\endgroup$ Commented Jun 23, 2021 at 12:27

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From https://en.wikipedia.org/wiki/Spherical_coordinate_system

r ∈ [0, ∞), θ ∈ [0, π], φ ∈ [0, 2π)

For the unit cylinder:

$r \sin{\theta} = 1$, the radius of the cylinder.

$r = \displaystyle \frac{1}{\sin{\theta}}$

\begin{align} \psi(\theta,\varphi) &= \begin{bmatrix} \cos{\varphi} \\ \sin{\varphi} \\ \displaystyle\frac{\cos{\theta}}{\sin{\theta}} \end{bmatrix} \end{align}

At any given $z$ the radius in the $(x,y)$ plane is $1$. $x$ and $y$ do not depend on $\theta$ only $\varphi$.

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  • $\begingroup$ really interesting parametrisation. didn't thought it'll work, but it brings a cylinder with infinit hight. In fact you could just write $\psi(\varphi,\theta) = \displaystyle{\left[\begin{array}{cc}\cos(\varphi)\\\sin(\varphi)\\\theta\end{array}\right]}$ receiving a cylinder, because it's just renamed. $\endgroup$ Commented Jun 24, 2021 at 16:13

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