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 $v$.

One technique that I found useful when I was teaching calc III this past semester is to parametrize a surface of the form $z=f(x,y)$ as $\mathbf{r}(x,y)=(x,y,f(x,y))$, get a double integral over some region, and then change coordinates in that integral. This is an alternative to directly choosing your parametrization in the "convenient" coordinates (in this case polar coordinates). One nice thing about this is that $\mathbf{r}_x \times \mathbf{r}_y = (-f_x,-f_y,1)$, so if you remember that, you can avoid doing a cross product. By contrast if you actually parametrize by $r,\theta$, taking $\mathbf{r}(r,\theta)=(r\cos(\theta),r\sin(\theta),f(r\cos(\theta),r\sin(\theta)))$, computing $\mathbf{r}_r \times \mathbf{r}_\theta$ is slightly annoying.
One additional point to pay attention to: when you change coordinates from this Cartesian integral I mentioned to polar, you need the polar Jacobian (the extra factor of $r$). When you parametrize directly in polar coordinates, you do not need the polar Jacobian; this factor of $r$ appears inside $\mathbf{r}_r \times \mathbf{r}_\theta$.
