At it for over a day now, must concede.
Lets have a vector field $\vec{F}(x,y,z)=(xz,yz+x^2yz+y^3z+yz^5,2z^4)$ with the surface $\Sigma$ given as $(x^2+y^2+z^4)e^{y^2}=1, x\geq 0$ oriented so its normal at the point $(1,0,0)$ is $\vec{N} = (1,0,0)$.
Calculate the work of this vector field along the edge of $\Sigma$, which is oriented with respect to $\Sigma$'s orientation.
I obviously think this is right, but I'm a little bit nervous that I might be doing something stupid. Still, nothing ventured... Let me know what you think.
With the notation as in the question, $$ \partial \Sigma = \left\{ (x,y,z) \in {\mathbb R^3}\mid (x^2 + y^2 + z^4) e^{y^2} = 1, x = 0\right\}.$$
Therefore we want to evaluate the integral
$$ I= \int _{\partial\Sigma} (yz + y^3 z + yz^5 ) \,dy + 2 z^4 \,dz. $$
There are no $x$ terms, because $x=0$ identically on $\partial \Sigma$. That being the case, the $x$ value is irrelevant, and we can consider this as a problem of integration in $\mathbb R^2$, rather than in $\mathbb R^3$.
Namely, the 'path' $\partial \Sigma$ is also the boundary of $$S = \left\{ (x,y,z) \mid (y^2 + z^4) e^{y^2} \le 1, x = 0\right\},$$ and by Green/Stokes, one has that
$$ I = - \int_S (y + y^3 + 5yz^4)\, dy\wedge dz, $$ where one can view the latter as an integral in the $x=0$ a.k.a. $y/z$ plane.
So consider the double integral (no $x$'s) as an iterated integral, and do the $y$ integral first (i.e., hold $z$ constant): we want to evaluate ($z$ fixed) $$J_z = \int_{y^2 - e^{-y^2} \le -z^4} f(y,z) \, dy, $$ where $f(y,z) = (y +y^3 + 5yz^4)$.
Now, the domain of integration in $J_z$ is symmetric about $y=0$. On the other hand $y\mapsto f(y,z)$ is an odd function. Therefore $J_z = 0$, and $I=0$ also.