Gauss curvature of points that are covered by a certain parametrization How do I compute the Gauss curvature of the points of 
$T^2$ = {$(x, y, z) \in \Bbb{R}^3$ | $x$ = (cos$u$ + 2)cos$v$, $y$ = (cos$u$ + 2)sin$v$, $z$ = sin$u$, $u, v \in \Bbb{R}$}
that are covered by the parametrization 
$F(u, v)$ = ((cos$u$ + 2)cos$v$, (cos$u$ + 2)sin$v$, sin$u$) where $0 < u < \pi$, $0 < v < 2\pi$?
I'm not really sure where to start. How do I approach this problem? 
 A: Given a smooth regular and injective curve $\alpha(u)=(f(u),0,g(u))$, with $f(u)>0$ always, the revolution parametrization $$X(u,v) =(f(u)\cos v, f(u)\sin v,g(u)) $$is also regular, injective, and covers the revolution surface, excluding a meridian. If $\alpha$ has unit speed, the induced metric (first fundamental form) is then ${\rm d}s^2 = {\rm d}u^2+f(u)^2 {\rm d}v^2$. In other words, $E=1$, $F=0$ and $G(u,v)=f(u)^2$. With a little more patience one can compute the coefficients of the second fundamental form of $X$ ($(e,f,g)$, $(L,M,N)$, $(h_{ij})$, or whatever notation you've been taught) as $L= \langle X_{uu},N\rangle$, etc., where $$N(u,v) = \frac{X_u \times X_v}{\|X_u \times X_v\|}.$$ Using that $$K = \frac{LM-N^2}{EG-F^2}, $$ one can show that we'll have $$K = -\frac{f''}{f} $$ for this particular $X$. For $f(u)=(\cos u +2)$, we conclude that the Gaussian curvature of the torus is given by $$K(u,v)=\frac{\cos u}{2+\cos u}. $$
See, for example, p. $100$ in John Oprea's Differential Geometry and Its Applications (first edition, I think).
