Faster way to calculate the area of this surface

It is actually pretty simple. The trick is compute the area element before changing variable to cylindrical polar coordinate $(x,y,z) = (\rho\cos\theta,\rho\sin\theta,z)$.

$$\begin{align}\verb/Area/ &= \int_{1 \le \sqrt{x^2+y^2} \le 2}\sqrt{1+\left(\frac{\partial z}{\partial x}\right)^2 + \left(\frac{\partial z}{\partial y}\right)^2} dxdy\\ &= \int_{1 \le \sqrt{x^2+y^2} \le 2}\sqrt{1+y^2 + x^2} dxdy\\ &= \int_0^{2\pi} \int_1^2\sqrt{1+\rho^2} \rho d\rho d\theta\\ &= \frac{2\pi}{3} \left[ \sqrt{1+\rho^2}^3\right]_{1}^2\\ &= \frac{2\pi}{3}\left(5\sqrt{5}-2\sqrt{2}\right) \end{align} $$

It's actually not that hard to work out your cross product, using $Cos(\theta)^2 + Sin(\theta)^2 = 1$:

$|\Phi_\rho \times \Phi_\theta| = \begin{vmatrix}\rho^2*(Cos(\theta)^2-Sin(\theta)^2)Sin(\theta)-2\rho^2*Cos(\theta)^2Sin(\theta)\\-2\rho*Cos(\theta)Sin(\theta)^2 -\rho^2*(Cos(\theta)^2-Sin(\theta)^2)Cos(\theta)\\ \rho*Cos(\theta)^2 + \rho*Sin(\theta)^2\end{vmatrix} = \begin{vmatrix}-\rho^2*Sin(\theta)^2\\-\rho^2*Cos(\theta)^2\\\rho\end{vmatrix} = \sqrt{\rho^4 + \rho^2} = \rho*(\rho^2+1)^{1/2}$. The integration gives $2\pi[\frac{1}{3}*(\rho^2+1)^{3/2}]^2_1 = \frac{2\pi}{3}(5^{3/2}-2^{3/2})$.