# Gaussian curvature of a surface of revolution

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.

• For those who want a reference for this, it appears e.g. in Kristopher Tapp, Differential Geometry of Curves and Surfaces, example 4.19. Commented Jun 7, 2021 at 23:50

We have $${\bf f}_u = (f'\cos\ v,f'\sin\ v,g'),\ {\bf f}_v = f(-\sin\ v, \cos\ v,0).$$ 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 $$\alpha$$ has unit speed, $$$$\tag{A} (f')^2+(g')^2=1.$$$$ So $$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$$
Hence the product of the eigenvalues is $$f''/g' (-g'/f)=-f''/f$$.
• For this to work, we need e.g. $g' \ne 0$. Can this be deduced from the assumptions? Commented Jun 17, 2022 at 23:04
The curves $$\gamma_c(s)=f(s,c)$$ ($$c$$ a constant) are geodesics for symmetry reasons (each of them lies in a plane such that the symmetry with respect to this plane preserves the surface). The vector field $$w:=\partial/\partial v$$ (in your parametrization of the surface) thus satisfies the Jacobi (=geodesic deviation) equation $$w''+Rw=0$$. Notice that $$w=f(u)e$$ where $$e$$ is of length 1 and orthogonal to the geodesic, hence $$e'=0$$. We thus have $$w''+Rw=(f''+Rf)e=0$$, from where $$R=-f''/f$$.