Calculate Integral of intersection between two surfaces 
Calculate
$$\int_{\partial F} 2xy\, dx + x^2 \, dy + (1+x-z) \, dz$$
for the intersection of $z=x^2+y^2$ and $2x+2y+z=7$. Go clockwise with respect to the origin.

My attempt:
I first calculate the intersection, which is pretty easy
$$z=x^2+y^2 \wedge 2x+2y+z=7 \Rightarrow \, (x+1)^2+(y+1)^2=7$$
which is a circle.
Then I find the curl of the vector field
\begin{align*}
F = \begin{pmatrix} 2xy \\ x^2 \\ 1+x-z\end{pmatrix} \quad \quad \text{curl }F =\begin{pmatrix} 0 \\ -1 \\ 0\end{pmatrix}  
\end{align*}
By Stokes theorem I have...
\begin{align*}
\int_{\partial F} F \, ds = \iint_F \text{curl } F \,dA
\end{align*}
I was told that I have to look at the circle $(x+1)^2+(y+1)^2=7$ on the plane $z=7$. We can easily find a normal vector on this plane, $n=(0,0,1)$. So we see that only the $z$-component of the curl/rotation is relevant.
\begin{align*}
\iint_{(x+1)^2+(y+1)^2\leq 7} 0 \, dxdy = \int^{\sqrt{7}}_{r=0} \int^{2\pi}_{\varphi=0} 0 \, d\varphi dr = 0
\end{align*}
which is very likely wrong. What is my mistake?

You should consider the surface $S$ given by the intersection of the cylinder $(x+1)^2+(y+1)^2\leq 7+2=9$ with the plane $2x+2y+z=7$. Then $\mathbf{n}=-(2,2,1)/3$ (the ellipse $\partial S$ is clockwise oriented) and by Stokes' theorem:
$$
\begin{align*}\int_{\partial S} \mathbf{F} \, d\mathbf{s} &= \iint_S \text{curl }(\mathbf{F}) \,d\mathbf{S}\\
&=\iint_{(x+1)^2+(y+1)^2\leq 9} (0,-1,0)\cdot(-(2,2,1)) \, dxdy\\
&=\iint_{(x+1)^2+(y+1)^2\leq 9} 2 \, dxdy\\
&=2\, \text{Area}(\{(x+1)^2+(y+1)^2\leq 9\})=2\cdot 9\pi=18\pi.
\end{align*}$$
