Parametrizing the surface $z=\log(x^2+y^2)$ 
Let $S$ be the surface given by $$z = \log(x^2+y^2),$$ with $1\leq x^2+y^2\leq5$. Find the surface area of $S$.

I'm thinking the approach should be $$A(s) = \iint_D \ |\textbf{T}_u\times \textbf{T}_v|  \,du\,dv ,$$   where $T_u, T_v$ are tangent vectors to the surface (and of course their cross product is the normal).
Now my issue is the parametrization bit. I'm having a very hard time grasping the concept of parametrization of surfaces. I'd be really grateful if someone can help with parametrizing the surface, i.e. $x(u,v)$, $y(u,v)$ and $z(u,v)$ and the domains of $u$ and $v$.

If a geometric object has a particular symmetry, generally parameterizations that reflect that symmetry result in easier computations. In our case, $S$ is the surface of a graph whose domain $A := \{1 \leq x^2 + y^2 \leq 5\}$ is an annulus centered at the origin. This suggests letting one of our parameterization variables be the distance $r$ of a point $p \in A$ to the origin, that set, setting $$r := \sqrt{x^2 + y^2}.$$ Then, we can see that our domain $A$ takes on an especially simple form: $$A := \{1 \leq r \leq \sqrt{5}\} .$$ When we use this $r$ for one of our parameters, usually we take an angular parameter $\theta$ for the other (the coordinate system $(r, \theta)$ is called polar coordinates). Since our region sweeps all the way around our origin, we should let $\theta$ vary over some (half-open) interval of length $2 \pi$, e.g., $[0, 2 \pi]$.
Drawing a picture, we see using elementary trigonometry that the transformation that expresses $x, y$ in terms of our polar coordinates $r, \theta$ is
$$\left\{\begin{array}{rcl}x(r, \theta) & = & r \cos \theta \\ y(r, \theta) & = & r \sin \theta\end{array}\right.$$
