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 $v$.

When you use the cross product method, the Jacobian takes care of itself. Along the surface the $\vec r=\langle x,y,z\rangle=\langle r\cos\theta,r\sin\theta,2\ln r\rangle$. Then the total differential along the surface is
$$d\vec r=\langle\cos\theta,\sin\theta,\frac2r\rangle\,dr+\langle-r\sin\theta,r\cos\theta,0\rangle\,d\theta$$
Then we can get the vector areal element via the cross product
$$\begin{align}d^2\vec A&=\pm\langle\cos\theta,\sin\theta,\frac2r\rangle\,dr\times\langle-r\sin\theta,r\cos\theta,0\rangle\,d\theta\\
&=\pm\langle-2\cos\theta,-2\sin\theta,r\rangle\,dr\,d\theta\end{align}$$
Then the scalar areal element is
$$d^2A=\left|\left|d^2\vec A\right|\right|=\sqrt{4+r^2}\,dr\,d\theta$$
Then
$$A=\int d^2A=\int_0^{2\pi}\int_1^\sqrt5\sqrt{4+r^2}\,dr\,d\theta$$
In the inner integral, let $r=2\sinh u$. Then
$$\begin{align}\int\sqrt{4+4^2}\,dr&=\int2^2\cosh^2u\,du=2\int(\cosh2u+1)du=2\left(\frac12\sinh2u+u\right)+C_1\\
&=2\left(\sinh u\cosh u+u\right)+C_1\\
&=2\left(\frac r2\sqrt{\frac{r^2}{4}+1}+\ln\left(\frac r2+\sqrt{\frac{r^2}4+1}\right)\right)+C_1\\
&=\frac12r\sqrt{4+r^2}+2\ln\left(r+\sqrt{4+r^2}\right)+C\end{align}$$
So
$$\begin{align}A&=2\pi\left[\frac12r\sqrt{4+r^2}+2\ln\left(r+\sqrt{4+r^4}\right)\right]_1^{\sqrt5}\\
&=2\pi\left[\frac123\sqrt5+2\ln\left(\sqrt5+3\right)-\frac12\sqrt5-2\ln2\right]=2\pi\left[\sqrt5+4\ln\left(\frac{\sqrt5+1}2\right)\right]\end{align}$$
You just knew the golden ratio was coming when you saw the $\sqrt5$ in the limits!
