Uncertainty of polar Coordinates

You appear bogged down in symbols. With

$$ rho = (x^2 + y^2)^{frac 1 2} $$

you get, w.r.t to $x$:

$$ deltarho = frac 1 2 (x^2 + y^2)^{-frac 1 2} 2 x delta x=frac x {rho} delta x$$

with $y$ giving a similar result. Adding them in quadrature:

$$ (delta rho)^2 = frac 1 {rho^2}[x^2(delta x)^2 + y^2(delta y)^2 ]$$

You can also write that as:

$$ (delta rho)^2 = big(frac x {rho}delta xbig)^2 + big(frac y {rho}delta ybig)^2 $$

so that the error's coordinate dependence is exactly proportional to the variables coordinate dependence.

If the uncertainties don't depend on coordinate:

$$ delta x = delta y equiv Delta $$

then

$$ (delta rho)^2 = frac 1 {rho^2}[x^2 + y^2 ]Delta^2 =frac{rho^2}{rho^2}Delta^2 = Delta^2$$

The other half of the total variance (also of magnitude $Delta^2$) is orthogonal, and goes into the $deltaphi$ part with a scaling of $rho$.

Note that if $x=0$, then:

$$ rho = pm y $$

and an error in $x$, $dx$, leads to:

$$ rho' = (y + dx)^{frac 1 2}= rho(1+frac{dx^2} {y^2})^{frac 1 2} approx rho(1 + frac 1 2 frac{dx^2} {y^2}) approx rho$$

to 1st order. If the uncertainty is the same magnitude as the coordinate, then this is not a good estimator...of course, if this is the case, the 1-sigma circle is going to include or be close to the origin, in which case all the polar coordinates are going to be rightfully uncertain.

If that is the case, then it is best to plot your result on a $rho-theta$ plot and show the 1 sigma circle. If that's not an option, then us a Monte Carlo to take your final $(x, y)$ and add gaussian distributions to each (based on $sigma_x$ and $sigma_y$), and compute a $rho$ and $phi$ histograms, and from there make decisions about errors.

If you are near the origin, you should find that $rho$ doesn't look too gaussian (it will be skewed), and $phi$ will be all over the place.

If you must have your results in $barrho$ and $barphi$, it's not going to look good because it was inherently a poor measurement.