Let $S$ be a part of the paraboloid $z=1-x^2-y^2$ such that $z\geq 2|y|$. They ask to calculate $$ \int_C\frac{-y}{x^2+y^2}dx+\frac{x}{x^2+y^2}dy+\frac{1+e^z}{1+z^2}dz $$ where the curve $C$ is traversed once in an anticlockwise direction if it is observed from the point $(0,0,1)$.
It is easy to see that the curl of $F$ is $(0,0,0)$. So my initial idea was to use Stokes' theorem with which the answer would be zero.
But realizing that $F$ is not a continuous field this is not possible, now in reality it would be necessary to look for a surface that has two borders: one of them $C$ and another $C_0$ (which would be easier to calculate). The following occurs to me, taking the same surface but bounded above with $z=15/16$. That would make the new surface no longer go through the z axis.
My attempt
Let $\lambda(t)=(\frac{1}{4}\cos t,\frac{1}{4}\sin t,\frac{15}{16})$ then $\lambda'(t)=(-\frac{1}{4}\sin t,\frac{1}{4}\cos t,0)$. And we have $F(\lambda(t))=(-4\sin t,4\cos t,\frac{1+e^{15/16}}{1+(15/16)^2})$. \begin{align} \int_{C_0} F\cdot dr&=\int_{C} F\cdot dr+\iint_S (\nabla\times F)\cdot dS=\int_C F\cdot dr\\ \int_{C_0} F\cdot dr&=\int_0^{2\pi}\cos^2 t+\sin^2 tdt=2\pi \end{align}
What you've argued is that if $C_0$ is a second curve within your surface that circulates once counterclockwise around the "north pole" of the paraboloid, then $\int_C F\,dx = \int_{C_0} F\,dx$. You can take $C_0$ to be the intersection of the paraboloid with the plane $z=9/10$ if you like, as long as you're sure that the plane does not intersect $C$.
For this transformation to make sense, you need the integral around $C_0$ to be easier to compute than that around $C$; for instance you might notice that in cylindrical coordinates, $F = d\theta$.