I'll start with some critique. First of all, your normal vector isn't quite correct: from the equation of the plane $-2x+z=0$, we get the normal vector $\mathbf{n}=\langle-2,0,1\rangle$ (or it could be its opposite, but this one gives the upward orientation, consistent with the counterclockwise orientation of the curve $C$). Fortunately, it doesn't affect your solution because the first component of curl is zero.

Second, it is a really bad habit to drop differentials, representing the variables of integration, from integral notation! For example, the last line of your computation should be written as
$$\iint_S (2x-xz)\,dx\,dy=\iint_S (2x-2x^2)\,dx\,dy=\int_{-\pi/2}^{\pi/2}\int_0^{2\cos\theta} (2r\cos\theta-2r^2\cos^2\theta)r\,dr\,d\theta.$$

Third, you must be much more clear regarding domains of integration. The "equality"
$$\iint_S \operatorname{curl}F\cdot\mathbf{n}\,dS=\iint_S (2x-2x^2)\,dx\,dy$$
is wrong because the domains of integration in these two integrals are **NOT** the same and thus cannot be denoted by the same letter $\color{red}{S}$. If $S$ stands for the portion of the plane cut out by the paraboloid (or cylinder), then it's rightfully used in the first integral, but not in the second. The second one represents a double integral over a region $D$ in the $xy$-plane after you effectively parameterized the surface $S$. And this region $D$ is the disk $(x-1)^2+y^2=1$, that you correctly found. And to integrate over this $D$, it certainly makes sense to switch to polar coordinates.

In the end of the day, you did get a correct double integral in polar coordinates (also see above), so you can finish solving this problem by evaluating that integral. (I presume you can do that, and you don't need us to give you the answer.)

Now, a very short **main answer** to your **main question**: *YES*, we are allowed to choose any such surface. :-)

The surface associated to your triangle is defined by: $\lambda_1\in[0,1],\ \lambda_2\in[0,1]$ such that $\lambda_1+\lambda_2\le 1$
$$
t(\lambda_1,\lambda_2)=\lambda_1(v_2-v_1)+\lambda_2(v_3-v_1)+v1=(4\lambda_2,4\lambda_1,1)
$$
the Curvilinear abscissa are:
$$
\gamma_1(\lambda_1,\lambda_2)=\frac{\partial t}{\partial_{\lambda_1}}=(0,4,0)
$$
$$
\gamma_2(\lambda_1,\lambda_2)=\frac{\partial t}{\partial_{\lambda_2}}=(4,0,0)
$$
Thus surface element $dS$ is
$$
dS=\gamma_1\wedge \gamma_2 d\lambda_1d\lambda_2=(0,0,-16) d\lambda_1d\lambda_2
$$
So
\begin{align}
\int\int curl F.dS&=\int_{\lambda_1=0}^1\int_{\lambda_2=0}^{1-\lambda_1} \underbrace{(-3z^2,-z^2,-x)\circ t(\lambda_1,\lambda_2)}_{(-3,-1,-4\lambda_2)} \cdot \underbrace{dS(\lambda_1,\lambda_2)}_{(0,0,-16) d\lambda_1d\lambda_2} \\
&=\int_{\lambda_1=0}^1\int_{\lambda_2=0}^{1-\lambda_1}64\lambda_2d\lambda_1d\lambda_2 \\ &=64\int_{\lambda_1=0}^1\frac{1}{2}(1-\lambda_1)^2d\lambda_1\\&=\frac{32}{3}
\end{align}

## Best Answer

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: $$ \begin{align*}\int_{\partial S} \mathbf{F} \, d\mathbf{s} &= \iint_S \text{curl }(\mathbf{F}) \,d\mathbf{S}\\ &=\iint_{(x+1)^2+(y+1)^2\leq 9} (0,-1,0)\cdot(-(2,2,1)) \, dxdy\\ &=\iint_{(x+1)^2+(y+1)^2\leq 9} 2 \, dxdy\\ &=2\, \text{Area}(\{(x+1)^2+(y+1)^2\leq 9\})=2\cdot 9\pi=18\pi. \end{align*}$$