Uniformization theorem for Riemannian 2-manifolds with boundary? Specifically the closed disc. Suppose that $D = \{ x \in \mathbb{R}^2  | ||x||_2 \leq 1 \}$ is the closed disc, and let $g$ be any Riemannian metric on $D$.
I'm wondering about the following version of uniformization: Is $(D,g)$ always conformally equivalent to the usual flat metric? If not, what if $g$ is flat?
I'm trying to understand whether regions bounded by a smooth closed curve in the plane are conformally equivalent. 
I'm asking as a follow up to my question here: https://mathoverflow.net/questions/304274/what-is-e-zeta-delta-0-for-a-delta-the-laplacian-of-a-manifold/304286?noredirect=1#comment758887_304286
(The motivation for all of this comes from some questions in graph theory... so even some pointers for basic references in conformal geometry would be super helpful to me.)
 A: Answering my question in the case that $g$ is flat:
If $(U, g_U)$ and $(V,g_V)$ are Riemannian 2 manifolds, and $\phi$ is the conformal mapping between them, then we have that  $\phi^*g_V$ is conformal to $g_U$. (This is the completely obvious fact that I wasn't thinking of.)
Now $(U, \phi^*(g_V)$ is isometric to $(V,g_V)$, and $(U, g_U)$ is in the same conformal class.  
In particular, for any bounded open regions in the plane with smooth boundary, $U$, we can use the Riemann mapping theorem to find a conformal map $\phi : U \to D$, where $D$ is the closed disc. Thus the metric $\phi^* d_{eucl}$ is conformal to $d_{eucl}$ on $U$. Thus $(U,d_{eucl})$ is in the same conformal class as a metric which makes $(U,g)$ isometric to the usual closed disc. 
In the context of the Osgood/Phillips/Sarnak result (discussed over on my linked mathoverflow question), this is saying that the conformal equivalence class of all metrics on $D$ with fixed perimeter and non-positive average Gaussian curvature includes all of the metrics that come from open regions in the plane bounded by a smooth closed curve. (Though it is presumably much larger.)
Answering my question in the comment of the linked math-overflow page: yes, the Sarnak/Osgood/Phillips theorem implies that extremizing the zeta regularized laplacian determinant over all regions bounded by a fixed perimeter closed curve results in a disc.
(Though I'm not 100% confident in my deductions about this right now.)
