Your second surface is a paraboloid, not a cylinder. Regardless, the trick to Stokes' theorem is that you have the freedom to choose any surface with that curve as a boundary (since there are infinitely many) and you are guaranteed that the integral over all of them will be equal to each other.
In this case, notice that your curve of intersection is contained entirely within the plane $z=\frac{\sqrt{5}-1}{2}a$. So choose to parametrize that plane:
$$\mathbf{r}(x,y) = \left(x,y,\frac{\sqrt{5}-1}{2}a\right), \hspace{20 pt} x^2+y^2\leq \left(\frac{\sqrt{5}-1}{2}\right)^2a^2$$
The normal vector for a plane is easy, it's just $(0,0,1)$ in this case. Which is a bonus because if we're smart about this, we don't have to compute the whole curl, only the $z$ component which is
$$\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} = -2y$$
And if we're even smarter, we'll recognize that the $z$ component of the curl is an odd function of $y$, and we are integrating over a surface (a disk) with $y$ symmetry, so we can conclude the integral is simply $0$ without parametrizing further.