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!
 A: 1) is the way to go. in this case,
$$\mathrm{e}^x\mathbf{i}+z^2\mathbf{k}=\boldsymbol{\nabla}(\mathrm{e}^x+\tfrac{z^3}{3})\text{,}$$
so the restriction of the vector field to the surface is exact and consequently line integrals over closed curves must vanish.
A: It can be done directly by Stokes:
$$
\iint_S (\nabla \times {\bf V}) \cdot d{\bf S}=\oint_\Gamma{\bf V}\cdot\,d{\bf r}.
$$
You are told that the curve $\Gamma$ is in the $xz$-plane, so its normal will be of the form $\gamma(t)\,{\bf j}$. Then 
$$
\oint_\Gamma{\bf V}\cdot\,d{\bf r}=\int_0^1{\bf V(0,\gamma(t),0)}\cdot \gamma'(t)\,{\bf j}\,dt=-3\int_0^1\gamma(t)\,\gamma'(t)\,dt=-3(\gamma^2(1)-\gamma^2(0))=0,
$$
since $\gamma$ describes is a closed curve. Note that ${\bf V}(0,y,0)={\bf i}-3{\bf j}+y^2{\bf k}$.
A: 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.
