= Solution
By <Stokes theorem>, it suffices to calculate the boundary <line integral>. On the $xz$ plane, $y=0$ and $dy=0$, so
$$
\mathbf V\cdot d\mathbf r=e^x\,dx+z^2\,dz=d\left(e^x+\frac{z^3}{3}\right).
$$
This <exact differential> has zero integral around a closed curve. Therefore
$$
\boxed{\iint_S(\nabla\times\mathbf V)\cdot\hat{\mathbf n}\,dS=0.}
$$
The argument applies to any admissible spanning surface, including a curved one; the surface itself need not lie in the $xz$ plane.
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