Stokes’ Theorem for arbitrary surface and boundary curve in $xz$-plane? I am tasked to find $\iint (\nabla \times {\bf V}) \cdot d{\bf S}$ for any surface whose bounding curve is in the $xz$-plane, where ${\bf V} = (xy + e^x) {\bf i}+ (x^2 -3y){\bf j} + (y^2 + z^2) {\bf k}$. I have attempted this via two methods and am stuck on both:
1) I’ve tried to employ Stokes’ Theorem directly. In the $xz$-plane, $y=0$, so {\bf V} becomes $e^x {\bf i}+ x^2 {\bf j}+ z^2 {\bf k}$, and $dy=0$ which makes the dot product $e^x dx + z^2 dz$. The problem is parametrising afterward. I’m unsure of how to approach this for an arbitrary curve. My intuition tells me that because this is a closed curve, the integral will sum up to zero, but I don’t know how to mathematically express this.
2) I also attempted to directly integrate the curl. The only interesting point to note is that the $y$-component is 0. Apart from that, I am unable to figure out how to obtain the normal vector to the surface to perform the integral.
Any insight is appreciated, thank you!

You are correct. The answer is zero and you're almost there. Essentially you get zero because the differential form you're left with has a potential function.
$e^x dx + z^2 dz$ can be integrated and becomes $f(x,z) = e^x + z^3/3$. 
Take an arbitrary parameterization, say $r (t) =(x (t), 0, z (t) )$, then plugging that into your line integral you get: $(e^{x (t)}x'(t)+z (t)^3z'(t))dt $ integrating with respect to  $t $ yields $e^{x (t)}+z (t)^3/3 $. Now the final thing to note is that since your curve $r (t) $ begins and ends at the same point, evaluating this integral will result in XXX - XXX = 0.
