TITLE: Gaussian curvaure of a surface revolution
QUESTION [6 upvotes]: Let $\alpha(s)=(f(s), g(s))$ be a plane curve parametrized by arc length on the $yz$-plane and assume that $f(s)>0.$ The surface revolution attained by rotating the curve parameterized by $\alpha$ about the z-axis which is given by $f(u,v) = (f(u)\cos v, f(u)\sin v, g(u)).$ 
How do I show that it has Gaussian curvature $K(u,v) = -\frac{f''(u)}{f(u)}.$
I am wondering how it is derived? I computed the normal vector, the first and second fundamental forms but I still can't derive it.

REPLY [5 votes]: $$ {\bf f}_u = (f'\cos\ v,f'\sin\ v,g'),\ {\bf f}_v = f(-\sin\ v, \cos\ v,0)$$
Note that $$N =\frac{ {\bf f}_u\times {\bf f}_v }{|{\bf f}_u\times {\bf f}_v|} =\frac{f(-g'\cos\ v, -g'\sin\ v, f')}{|{\bf f}_u\times {\bf f}_v |} $$
Since $\alpha(s)$ has unit speed, so $$(f')^2+(g')^2=1\ (A)$$ So $$N =  
(-g'\cos\ v, -g'\sin\ v, f').$$
Here $N_u= -(g''\cos\ v,g''\sin\ v, f''),\ N_v = (g'\sin\ v, -g'\cos\ v,0)$
Since $f'f''+g'g''=0$ from $(A)$ $(i.e.,\ g''=-\frac{f'f''}{g'})$, $$N_u = f''(f'/g'\cos\ v,f'/g'\sin\ v,1)=f''/g' {\bf f}_u$$ and $$
N_v=-g'/f {\bf f}_v$$
Hence product of eigencvalues is $f''/g' (-g'/f)=-f''/f$