Can a region always be parametrized by a single function? Can some connected region in $\Bbb R^n$, possibly with some other nice conditions on the region, always be parametrized by a single function $\vec r(x_1, x_2, \dots, x_k)$ (even if it may be easier to not restrict yourself to this)?  Or do you sometimes have to use different parametrizations for different parts of the region?

Edit:
I was being intentionally vague with "nice" because I don't know what, if any, other conditions might be necessary. By parametrization I mean an injective, differentiable function that maps some parameters $x_1,\dots,x_k$, where $k \le n$, to the connected region in $R^n$ that we're concerned with.
 A: A coordinate mapping (such as polar coordinates in the plane) is only unique if the implicit function theorem is valid on the region. The implicit function theorem in n-dimensional Euclidean space states that if the Jacobian matrix, i.e. the matrix of partial derivatives, of the coordinatized mappings defining the graph of the function on an open subset of $\Bbb R^n$ is invertible, then there's a unique continuously differentiable map from the open set $U$ into the open set $V$ where the graph of the function $g$ is equal to a level set of $f$.  That is $(\mathbf x, g(\mathbf x)) = \{(\mathbf x,\mathbf y) | f(\mathbf x,\mathbf y) =c \}$ where $c$ is a real number.
If the maps under consideration are coordinate mappings and $c = 0$, then $f = g^{-1}$ and the inverse function theorem then gives the "change of coordinate" by "equivalently" in terms of the components of either $f$ or $g$. Otherwise, multiple coordinate mappings are applicable. 
Here's a standard example, shamelessly stolen from a Wikipedia page (lol) : Take a region in $R^3$ parametrised by polar coordinates $(R, \theta)$. We can go to a new coordinate system (Cartesian coordinates) by defining functions $x(R, \theta) = R \cos(\theta)$ and $y(R, \theta) = R \sin(\theta)$. This makes it possible given any point $(R, \theta)$ to find corresponding Cartesian coordinates $(x, y)$.
When can we go back and convert Cartesian into polar coordinates? By the previous example, it is sufficient to have $\det J \ne 0$, with $$J  =\begin{bmatrix}
 \frac{\partial x(R,\theta)}{\partial R} & \frac{\partial x(R,\theta)}{\partial \theta} \\
 \frac{\partial y(R,\theta)}{\partial R} & \frac{\partial y(R,\theta)}{\partial \theta} \\
\end{bmatrix}=
 \begin{bmatrix}
 \cos \theta & -R \sin \theta \\
 \sin \theta & R \cos \theta
\end{bmatrix}$$.
Since $\det J = R$, conversion back to polar coordinates is possible if $R \ne 0$. So it remains to check the case $R = 0$. It is easy to see that in case $R = 0$, our coordinate transformation is not invertible: at the origin, the value of $θ$ is not well-defined.

I know, it's really confusing sometimes, but it's really important to understand under which conditions this is possible. Differential geometry really depends entirely on this framework.For good discussions of this question and lots of good examples in low dimensions, take a look at Vector Calculus by Peter Baxandall and Hans Liebeck. More abstract discussions can be found in either Analysis on Manifolds by James Munkres or Analysis in Euclidean Space by Kenneth Hoffman. 
A: There are topological constraints, even in the plane.
For instance, an annulus cannot be continuously parametrized by the unit square or unit disk because an annulus is not simply connected.
On the other hand, the Riemann mapping theorem says that every non-empty simply connected open subset of the plane can be nicely parametrized by the open disk. Here nicely means biholomorphically.
