Evaluating the surface integral where S is part of a plane and inside a cylinder I'm working on a problem that states:

Evaluate $\iint_S (x^2z+y^2z)\,dS$ where $S$ is part of the plane $z=4+x+y$ that lies inside the cylinder $x^2+y^2=4$.

I want to use the following surface integral formula:

$$\iint_S f(x,y,z)\,dS = \iint_D f\left(\vec r\left(u, v\right)\right) \left\lVert\vec r_u \times \vec r_v \right\rVert \,dA$$

It requires the surface to be parametrized, however. How can I parameterize the surface that was given in this problem to start?
EDIT: I was thinking of going polar.
$$ r^2 = 4 $$
$$ r = 2 $$
$$ z = 4 + 2cos(\theta)+2sin(\theta) $$
 A: The surface $S$ is the part of the plane $z=4+x+y$ that lies inside the cylinder $x^2+y^2=4$. We want to describe $S$ as a parametric surface $\vec r=\vec r(u,v)$. This is most easily accomplished in cylindrical coordinates, Where the equation of the plane takes the form $z=4+\rho\cos{\phi}+\rho\sin{\phi}=4+\rho(\cos{\phi}+\sin{\phi})$, and the equation of the cylinder takes the form $\rho=2$.
We have the following parametric representation of $S$: 
$$\vec{r}(\rho,\phi)=\langle\rho\cos{\phi},\rho\sin{\phi},4+\rho(\cos{\phi}+\sin{\phi})\rangle,~~~0\leq\rho\leq2,0\leq\phi\leq2\pi.$$
Computing the partial derivatives and their cross product, we get
$$\vec{r}_\rho=\langle\cos{\phi},\sin{\phi},\cos{\phi}+\sin{\phi}\rangle$$
$$\vec{r}_\phi=\langle-\rho\sin{\phi},\rho\cos{\phi},\rho(\cos{\phi}-\sin{\phi})\rangle$$
$$\vec{r}_\rho\times\vec{r}_\phi=\langle-\rho,-\rho,\rho\rangle.$$
Thus, $\|\vec{r}_\rho\times\vec{r}_\phi\|=\sqrt{3}\rho$.

Rewriting the integrand in cylindrical coordinates, we have
$$f(x,y,z)=x^2z+y^2z=(x^2+y^2)z=\rho^2z.$$
Recall that on the surface $S$, we have $z=4+\rho(\cos{\phi}+\sin{\phi})$, so then
$$f(\vec{r}(\rho,\phi))=\rho^2(4+\rho(\cos{\phi}+\sin{\phi})).$$
Putting it all together, the surface integral evaluates to
$$I=\iint_{S}f(\vec{r})\,dS=\iint_{D}f(\vec{r}(\rho,\phi))\|\vec{r}_\rho\times\vec{r}_\phi\|\,dA\\
=\sqrt{3}\int_{0}^{2\pi}\int_{0}^{2}\rho^3(4+\rho(\cos{\phi}+\sin{\phi}))\,d\rho\,d\phi$$
We can make the evaluation of the above integral even easier by switching the order of integration and noting that $\cos\phi+\sin\phi$ is periodic in $\phi$ with period $2\pi$ and that the integral over one period is automatically zero. Hence, the value of the surface integral is:
$$I=\sqrt{3}\int_{0}^{2\pi}\int_{0}^{2}\rho^3(4+\rho(\cos{\phi}+\sin{\phi}))\,d\rho\,d\phi=\sqrt{3}\int_{0}^{2}\int_{0}^{2\pi}(4\rho^3+\rho^4(\cos{\phi}+\sin{\phi}))\,d\phi\,d\rho\\
=\sqrt{3}\int_{0}^{2}\int_{0}^{2\pi}4\rho^3\,d\phi\,d\rho\\
=2\sqrt{3}\pi\int_{0}^{2}4\rho^3\,d\rho\\
=2\sqrt{3}\pi(2^4)\\
=2^5\sqrt{3}\pi.$$
