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.
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 $\Delta\sigma$.