Curvature of a metric defined on an open disc in $\mathbb{R}^2$ Let $D$ be an open disc centred at the origin in $\mathbb{R}^2$. Give $D$ a Riemannian metric of the the form $(dx^2+dy^2)/f(r)^2$, where $r=\sqrt{x^2+y^2}$ and $f(r)>0$. Show that the curvature of this metric is $K = ff'' - (f')^2 + ff'/r$. 
First of all, I am confused as to what the question means by "curvature of the metric". I have previously only studies how to calculate the Gauss curvature of embedded surfaces, and am not sure how to proceed in this case. Many thanks for you help.
 A: You are given in the euclidean plane a Riemannian metric that is conformally equivalent to the ordinary euclidean metric:
$$ds^2= E(x,y)(dx^2+dy^2)\ ,\tag{1}$$
and the local scaling factor is  dependent only on $r:=\sqrt{x^2+y^2}$:
$$E(x,y)={1\over \bigl(f(r)\bigr)^2}$$
for some given function $r\mapsto f(r)>0$. This will make the computations simpler; see below.
It is a central fact of differential geometry that the datum $(1)$ defines in the plane a metric regime different from the usual one that one can study without reference to an embedding of some surface $S$ into ${\mathbb R}^3$. In particular it makes sense to talk about geodesics, Gaussian curvature, and other concepts of surface theory. The so-called Theorema egregium  allows us to compute the Gauss curvature at any given point $(x,y)$ of our "abstract" surface $S$. Look here
http://en.wikipedia.org/wiki/Gaussian_curvature
under the heading "Alternative formulas". In our case the following alternative formula is relevant: When $F=0$ and $E=G=e^{\sigma(x,y)}$ then
$$K(x,y)=-{1\over2 e^\sigma}\ \Delta\sigma\ .$$
Now express $\sigma$ by your $f$ and use repeatedly the rule
$${\partial u\over\partial x}={x\over r}{du\over dr},\qquad {\partial u\over\partial y}={y\over r}{du\over dr}$$
when computing $\Delta\sigma$.
