Let D be an open disc centred at the origin in $ \Bbb R^2 $. Give D a Riemannian metric of the form $ (dx^2 + dy^2)/f(r)^2 $, where $ r = \sqrt{x^2 + y^2} $ and $ f(r) > 0 $. Show that the Gauss curvature of this metric is $ K = f f'' - (f')^2 + f f' / r $.
This looks like it shouldn't be too bad, but I can't work out how to do it! My supervisor said that it isn't too bad, just pretty messy, and we didn't have time to do it in the supervision. If someone would be able to help me get (substantially) underway, then I'd be most appreciative! (I don't expect a full typset solution since this would probably take quite a long time!)
This is an example sheet question, completely non-examinable!
Thanks! :)
Cartesian coordinates are isothermal for your metric, so writing $\lambda = 1/f(r)$ you have the standard formula $K = -(\Delta \log \lambda)/\lambda^{2}$. (The computation is indeed a bit messy, but the formula comes out as stated.)