Jacobian in coordinate transformations when the new coordinates are multifunction's of old ones. Suppose that you have polar coordinates in terms of the cartesian coordinates:
$$ r^2 = x^2 + y^2 \tag{1}$$
$$ \theta = \tan^{-1} \frac{y}{x}$$
The Jacobian is given as:
$$ \begin{bmatrix} \frac{\partial r}{\partial x} & \frac{\partial r}{\partial y} \\ 
\frac{\partial \theta}{\partial x} & \frac{\partial \theta}{\partial y} \end{bmatrix}$$
Now here is my question, in almost every textbook and video I see we take the positive square root branch in $(1)$:
$$ r = \sqrt{x^2 +y^2}$$
But really there is an ambiguity here because one could have taken the negative square root and calculate the Jacobian. For each definition, we get a jacobian. How do we choose the correct jacobian for the transformation?
 A: What we have for sure, even for negative $r$ and $\theta$ outside of a standard range, is $x=r\cos\theta$ and $y=r\sin\theta$.
Then $r^{2}=x^{2}+y^{2}$ is certainly true. We can hold $y$ constant and implicitly differentiate with respect to $x$, to get $2r\dfrac{\partial r}{\partial x}=2x$. That means that (if $r\ne0$), we have $\dfrac{\partial r}{\partial x}=x/r=\cos\theta$, whether $r$ is positive or negative. Similarly, we can hold $x$ constant to find that $\dfrac{\partial r}{\partial y}=y/r=\sin\theta$.
Analogously, if $x\ne0$, $\dfrac{y}{x}=\tan\theta$ is true, so we can hold $y$ constant and differentiate with respect to $x$ to find $-\dfrac{y}{x^{2}}=\dfrac{\partial\theta}{\partial x}\sec^{2}\theta$ so $\dfrac{\partial\theta}{\partial x}=-\dfrac{y}{r^{2}}=-\dfrac{\sin\theta}{r}$. And similarly we can hold $x$ constant and differentiate with respect to $y$ to find $\dfrac{1}{x}=\dfrac{\partial\theta}{\partial y}\sec^{2}\theta$ and $\dfrac{\partial\theta}{\partial y}=\dfrac{\cos\theta}{r}$.
These answers check out, since $\begin{bmatrix}\cos\theta & \sin\theta\\
-\dfrac{\sin\theta}{r} & \dfrac{\cos\theta}{r}
\end{bmatrix}^{-1}=\begin{bmatrix}\cos\theta & -r\sin\theta\\
\sin\theta & r\cos\theta
\end{bmatrix}=\begin{bmatrix}\dfrac{\partial x}{\partial r} & \dfrac{\partial x}{\partial\theta}\\
\dfrac{\partial y}{\partial r} & \dfrac{\partial y}{\partial\theta}
\end{bmatrix}$.
The derivation of the Jacobian above assumed that $x\ne0$ (which implies $r\ne0$). But we could get the same results by assuming $y\ne0$ and working with $\dfrac{x}{y}=\cot\theta$ instead, so it's still right even if $x=0$, as long as we don't have $\left(x,y\right)=\left(0,0\right)$, which is kind of a bad point for polar coordinates (as alluded to in Hans Lundmark's comment).
A: @Peek-a-boo's comment solved my query. The multifunction ambiguity is removed by the constraints we set on the values which the new coordinates can take. For example in the polar example in question, the ambiguity is removed after defining $r>0$. Discussion on $\theta$ can be found here
