Faster way to calculate the area of this surface I have to solve this exercise:

Find the area of the portion of the surface $z=xy$ included between the two cylinders $x^2+y^2=1$ and $x^2+y^2=4$

what i did so far: 
I parameterized the surface using cylindircal polar coordinates
$$\
\Phi(\rho,\theta)=\
\begin{pmatrix}\
   \rho\cos(\theta) \\
   \rho\sin(\theta) \\
   \rho^2\cos(\theta)\sin(\theta)
\end{pmatrix},\
\ \ \rho\in[1,2], \ \ \theta\in[0,2\pi]
$$
then i computed the partial derivates of $\Phi$ 
$$\Phi_\rho=\
\begin{pmatrix}\
   \cos(\theta) \\
   \sin(\theta) \\
   2\rho\cos(\theta)\sin(\theta)
\end{pmatrix}\
\text{ and }\
\Phi_\theta=\
\begin{pmatrix}\
   -\rho\sin(\theta) \\
   \rho\cos(\theta) \\
   \rho^2(\cos^2(\theta)-\sin^2(\theta))
\end{pmatrix}\
$$
At this point i knew the answer should be
$$\int_0^{2\pi}d\theta\int_1^2|\Phi_\rho\times\Phi_\theta|d\rho$$
the problem is that the expression $|\Phi_\rho\times\Phi_\theta|$ is massive and it would take me forever to compute its integral. Since this question is supposed to be answered within 3 minutes there must be some kind of trick i can use to do it.
EDIT: 
The correct answer is $2\pi(5\sqrt{5}-2\sqrt{2})/3$ but i don't know how to get there
 A: It is actually pretty simple. The trick is compute the area element
before changing variable to cylindrical polar coordinate $(x,y,z) = (\rho\cos\theta,\rho\sin\theta,z)$.
$$\begin{align}\verb/Area/ 
&= \int_{1 \le \sqrt{x^2+y^2} \le 2}\sqrt{1+\left(\frac{\partial z}{\partial x}\right)^2 + \left(\frac{\partial z}{\partial y}\right)^2} dxdy\\
&= \int_{1 \le \sqrt{x^2+y^2} \le 2}\sqrt{1+y^2 + x^2} dxdy\\
&=  \int_0^{2\pi} \int_1^2\sqrt{1+\rho^2} \rho d\rho d\theta\\
&= \frac{2\pi}{3} \left[ \sqrt{1+\rho^2}^3\right]_{1}^2\\
&= \frac{2\pi}{3}\left(5\sqrt{5}-2\sqrt{2}\right)
\end{align}
$$
A: 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 $2\pi[\frac{1}{3}*(\rho^2+1)^{3/2}]^2_1 = \frac{2\pi}{3}(5^{3/2}-2^{3/2})$.
