Compute $\oint_L \mathbf{A} \cdot d\mathbf{L}$ where $A=(yz+2z,xy-x+z,xy+5y)$ and $L$ is the line of intersection between the cylinder $x^2+z^2=4$ and the plane $x+y=2$. It is oriented such that the tangential vector at $(2,0,0)$ is $(0,0,1)$
Attempted solution
The rotation of the vector field is found to be $(x+4,2,y-1-z)$. A parametrization for the surface $S$ with $L$ as its boundary is given by $$\mathbf{r}(x,z)=\begin{cases} x=x \\ y=2-x \\z=z \end{cases}$$ for points $x,z$ such that $x^2+z^2 \leq 4$. This gives the normal vector $\hat{n} = \dfrac{\dfrac{\partial \mathbf{r}}{\partial x} \times \dfrac{\partial \mathbf{r}}{\partial z}}{|\dfrac{\partial \mathbf{r}}{\partial x} \times \dfrac{\partial \mathbf{r}}{\partial z}|} = (-1,-1,0)/\sqrt{2}\: $. Using Stokes' theorem we obtain $$\oint_L \mathbf{A} \cdot d\mathbf{L} = \iint_S \nabla \times \mathbf{A} \cdot d\mathbf{S} = \iint_S (x+4,2,y-1-z)\cdot (-1,-1,0)\: dS= \\= \iint_S-x-6 \: dS$$
Here I don't see how to continue. How can I determine $dS$ and the limits?