Texture mapping and conformal transformation I'm a 3D programmer with background in physics, interested to better know how texture mapping can be made using conformal maps for simple surfaces.
I want to texture map a paraboloid:
$(x,y,z)=(x,c(x^2+z^2),z)$
$c \in \mathbb{R}$ is a parameter. I can equivalently write it as $F(\rho,\theta)$ in polar coordinates:
$$ x(\rho,\theta) = \rho \cos \theta \\ y(\rho,\theta)=c \rho^2\\ z(\rho,\theta) = \rho \sin \theta$$
My target is to find the coordinates $(u,v)$ for which the area element $dS$ is constant and such that they are orthogonal. I'm computing the area element in the usual way:
$$dS = \left\lVert \frac{\partial F}{\partial \rho} \times \frac{\partial F}{\partial \theta} \right\rVert = \det \begin{pmatrix}
 & \hat i & \hat j & \hat k \\ 
 & \cos \theta & 2c\rho & \sin \theta \\
 & -\rho \sin \theta & 0 & \rho \cos \theta \\
\end{pmatrix} = \lvert\rho\rvert \sqrt{4c^2\rho^2+1}$$
The simple polar coordinates are not the right choice I argue. How can I find those $u,v$ coordinates?
I hope to have put the correct tag to this question since I think that being this transformation from the cartesian grid where the coordinates are orthogonal, the map must preserve it.
 A: There are two things you are asking in your question, and it is far from obvious that they will coincide. On the one hand, you ask about a constant area element, while on the other hand you ask about a conformal map. Just because the map maintains the orthogonality of your coordinate lines does not mean that the map as a whole will neccessarily be conformal.
Conformal
Let's start with the conformality requirement, and let's also preserve the rotational symmetry and stick to polar coordinates. If your map was conformal, then the scale factor between texture coordinates and object coordinates would be independent of direction.
To compute the path length for an infinitesimal change in parameters, compute the total derivative of the path length in object space:
\begin{align*}
\mathrm d s^2 &=
\left\lVert\frac{\partial F}{\partial\rho}\mathrm d\rho\right\rVert^2 +
\left\lVert\frac{\partial F}{\partial\theta}\mathrm d\theta\right\rVert^2 =
\left\lVert\begin{pmatrix}\cos\theta\\2c\rho\\\sin\theta
\end{pmatrix}\mathrm d\rho\right\rVert^2 +
\left\lVert\begin{pmatrix}-\rho\sin\theta\\0\\\rho\cos\theta
\end{pmatrix}\mathrm d\theta\right\rVert^2 \\&=
\left(1+4c^2\rho^2\right)(\mathrm d\rho)^2 + \rho^2(\mathrm d\theta)^2
\end{align*}
Now do the same for texture space, where you have $G(r,\theta)$ like this:
\begin{align*}
u &= r\cos\theta \\
v &= r\sin\theta
\end{align*}
\begin{align*}
\mathrm d s'^2 &=
\left\lVert\frac{\partial G}{\partial r}\mathrm d r\right\rVert^2 +
\left\lVert\frac{\partial G}{\partial\theta}\mathrm d\theta\right\rVert^2 =
\left\lVert\begin{pmatrix}\cos\theta\\\sin\theta
\end{pmatrix}\mathrm d r\right\rVert^2 +
\left\lVert\begin{pmatrix}-r\sin\theta\\r\cos\theta
\end{pmatrix}\mathrm d\theta\right\rVert^2 \\&=
(\mathrm dr)^2 + r^2(\mathrm d\theta)^2
\end{align*}
Now you want the ratio between these two directional derivatives to be the same.
\begin{align*}
\frac{\left(1+4c^2\rho^2\right)(\mathrm d\rho)^2}{(\mathrm dr)^2}
&= \frac{\rho^2(\mathrm d\theta)^2}{r^2(\mathrm d\theta)^2} \\
\frac{\mathrm dr}{\mathrm d\rho} &= \frac{r}{\rho}\sqrt{4c^2\rho^2+1}
\end{align*}
This is a differential equation, which you can type into Wolfram Alpha. The result will look somewhat like this:
$$ r(\rho) = k_1\frac{\rho}{1+\sqrt{4c^2\rho^2+1}}e^{\sqrt{4c^2\rho^2+1}} $$
The constant $k_1$ will determine the global scale of your texture, you can choose it to suite your needs.
Constant area element
So now about this constant area element. This would be a parametrization such that equal changes in $r$ and $\theta$ correspond to equal area on the paraboloid, no matter the current position. Let's have a look at this area element, similar to what you already did in your question.
\begin{align*}
\left\lvert\mathrm dS\right\rvert &=
\left\lVert\frac{\partial F}{\partial\rho}\mathrm d\rho\times
\frac{\partial F}{\partial\theta}\mathrm d\theta\right\rVert =
\left\lVert\begin{pmatrix}\cos\theta\\2c\rho\\\sin\theta
\end{pmatrix}\times\begin{pmatrix}-\rho\sin\theta\\0\\\rho\cos\theta
\end{pmatrix}\,\mathrm d\rho\,\mathrm d\theta\right\rVert =
\left\lVert\begin{pmatrix}
2c\rho^2\cos\theta\\
\rho\\
2c\rho^2\sin\theta
\end{pmatrix}\,\mathrm d\rho\,\mathrm d\theta\right\rVert \\&=
\left\lvert\rho\sqrt{4c^2\rho^2+1}\,\mathrm d\rho\,\mathrm d\theta\right\rvert
\end{align*}
Nothing new so far, just confirming your computation and adding a slightly different notation. Now let's introduce a constant $a$ for the constant area element you want for our texture coordinates. This is a unitless scale factor.
\begin{align*}
\mathrm dS &= a\,\mathrm d r\,\mathrm d\theta =
\rho\sqrt{4c^2\rho^2+1}\,\mathrm d\rho\,\mathrm d\theta \\
\frac{\mathrm dr}{\mathrm d\rho} &=
\frac{\rho}{a}\sqrt{4c^2\rho^2+1} \\
r(\rho) &= \int\frac{\rho}{a}\sqrt{4c^2\rho^2+1}\,\mathrm d\rho
= \frac{\left(4c^2\rho^2+1\right)^{\frac32}}{12ac^2} + k_2
\end{align*}
You could also use Wolfram Alpha to compute this integral, or directly type in the differential equation. You'd probably want to choose the constant $k_2$ in such a way that for $\rho=0$ you get $r=0$.
\begin{align*}
k_2 &= -\frac1{12ac^2} \\
r(\rho) &= \frac{\left(4c^2\rho^2+1\right)^{\frac32}-1}{12ac^2}
\end{align*}
As you can see, this form is very different from the one obtained for the conformal map, so constant area element is something else entirely.
