domain of surface of revolution Let $0<b<a,(u,v) \in \mathbb{R} \times \mathbb{R}$. Then the map $g(u,v):=((a+b\cos u)\cos v,(a+b\cos u)\sin v,b\sin u)$ defines a torus.
I wonder for $g$ to be a surface does it really need $(u,v) \in \mathbb{R} \times \mathbb{R}$?  
If let $(u,v) \in U \times V$ where $U,V$ are both open in $R$ , will $g$ still be a surface in the context of differential geometry? (I know it may not be a surface of revolution though)
 A: A continuously differentiable map 
$${\bf f}: \quad\Omega\to{\mathbb R}^3,\quad (u,v)\mapsto{\bf x}(u,v)=\bigl(x_1(u,v),x_2(u,v),x_3(u,v)\bigr)$$
defined on an open set $\Omega\subset{\mathbb R}^2$ is called an immersion, if for all $(u,v)\in\Omega$ the differential $d{\bf f}(u,v)$ has rank $2$, i.e., if ${\bf x}_u\times{\bf x}_v\ne{\bf 0}$ for all $(u,v)$. 
When this "technical condition" is fulfilled then ${\bf f}$ maps sufficiently small pieces of $\Omega$ bijectively onto small pieces of a (large) smooth surface $S:={\bf f}(\Omega)$. But globally the map ${\bf f}$ need not be one-one: There might be self-intersections, or some parts of $S$ are covered several times.
Such is the case in your example: One computes
$${\bf g}_u\times{\bf g}_v=(a+b\cos u)\bigl(-b\cos u\cos v, -b \cos\sin v,-b\sin u\bigr)\ ,$$
which is easily seen to be $\ne{\bf 0}$ for all $(u,v)\in{\mathbb R}^2$. Therefore this ${\bf g}$ is indeed an immersion. But the function ${\bf g}$ is doubly periodic; whence any two points $(u,v)$ differing by $(2k\pi,2\ell\pi)$ are mapped to the same point of $S$. Nevertheless it  makes sense to leave the map ${\bf g}$ as it stands, because restricting $u$, $v$ to a period square introduces artificial seams.
It is another matter when you want to compute the area of $S$. In this case you have to make sure that $S$ is covered exactly once by the parametrization (up to said seams, which form a set of measure zero).
