Riemannian metric given in polar coordinates the Riemannian metric of the euclidean plane is given in polar coordinates as
\begin{align*}
ds^2=dr^2+r^2d\theta^2.
\end{align*}
Consider more generally,
\begin{align*}
ds^2=dr^2+\psi(r)^2d\theta^2,
\end{align*}
where $\psi>0$ is a smooth function. 
It seems to me a very natural question to ask what happens if the function $\psi$ changes. Is it possible to 'tell' what kind of Riemannian manifold correspond to such a metric if we choose different functions $\psi$ ( e.g. $\psi^2(r)=r,r^3,r^4,.. etc.)$ 
For example, if we choose $\psi(x)^2=\sinh(x)^2$ we 'obtain' the hyperbolic plane, i.e. the hyperbolic metric in polar coordinates is equal to the corresponding metric. Do other choices of $\psi$ have some significant meaning, or are there maybe well known examples which can be found in some literature? 
Best regards
 A: The exponential mapping creates a metric of the form 
\begin{align*}
ds^2=dr^2+\psi(r,\theta)^2d\theta^2,
\end{align*}
locally in a sufficiently small neighbourhood of any point $p$ on a two-dimensional manifold. Your condition expresses a local symmetry: on a sufficiently small neighbourhood of $p$ the metric is invariant with respect to a translation in the $\theta$ coordinate (we could call it a 'local rotation').
It makes a lot of difference if you want that kind of metric in one point or in every point, and in the latter case if the function $\psi$ has to be the same for every point (it isn't in your examples with $\psi^2(r)=r,r^3\ldots$).
If the same function $\psi(r)$ is valid around every point $p$ then we are in the realm of surfaces of constant curvature; to see this, note that the curvature for a general $\psi(r)$ is given by
$$K=-\frac1{\sqrt G}\frac{\partial^2\sqrt G}{\partial r^2}$$
where $G=1.\psi^2(r)$ is the determinant of the matrix expressing the components of the metric tensor. This shows that the function $\psi(r)$ uniquely determines the curvature at the origin of the coordinate system. If all origins have the same function $\psi(r)$ then all origins have the same scalar curvature.
Thus constant curvature implies $\psi''/\psi$ must be constant. Depending on the sign of the constant (the opposite of the sign of the curvature), the solutions of that differential equation are: first-degree polynomials (your example of a Euclidean plane), a scaled, shifted version of the hyperbolic sine (your example of a hyperbolic sphere) and a scaled, shifted version of the ordinary sine (a sphere). The shift can be eliminated by requiring $\psi(0)=0.$
The sphere of radius $R$ has $\psi(r)=\sin(r/R).$
