Stokes' theorem says that the surface integral of the curl of the vector field F across the surface S is equal to the line integral of F along the boundary of S.
In this case, the boundary is a circle in the x,y-plane with radius 2. (To see this, set z = 0 and rearrange the equation for the paraboloid as x² + y² = 4.)
Then
[tex]\displaystyle \iint_S (\nabla\times\vec F)\cdot d\vec s = \int_{\partial S} \vec F \cdot d\vec r[/tex]
where
[tex]\vec r = 2\cos(t) \,\vec\imath + 2\sin(t) \,\vec\jmath[/tex]
parameterizes the circular boundary of S with 0 ≤ t ≤ 2π. Then
[tex]d\vec r = \dfrac{d\vec r}{dt}\,\dt = (-2\sin(t)\,\vec\imath + 2\cos(t)\,\vec\jmath)\,dt[/tex]
and the integral becomes
[tex]\displaystyle \iint_S (\nabla\times\vec F)\cdot d\vec s = \int_{\partial S} \vec F \cdot d\vec r[/tex]
[tex]\displaystyle \iint_S (\nabla\times\vec F)\cdot d\vec s = \int_0^{2\pi} \vec F(\vec r(t)) \cdot \frac{d\vec r}{dt}\,dt[/tex]
[tex]\displaystyle \iint_S (\nabla\times\vec F)\cdot d\vec s \\= \int_0^{2\pi} ((2\cos(t))^2 \sin(0)\,\vec\imath + (2\sin(t))^2\,\vec\jmath + (2\cos(t))(2\sin(t))\,\vec k) \cdot(-2\sin(t)\,\vec\imath + 2\cos(t)\,\vec\jmath)\,dt[/tex]
[tex]\displaystyle \iint_S (\nabla\times\vec F)\cdot d\vec s = \int_0^{2\pi} 8\sin^2(t)\cos(t) \, dt[/tex]
[tex]\displaystyle \iint_S (\nabla\times\vec F)\cdot d\vec s = \int_0^{2\pi} 8\sin^2(t) \, d(\sin(t))[/tex]
[tex]\displaystyle \iint_S (\nabla\times\vec F)\cdot d\vec s = \frac83 (\sin^3(2\pi)-\sin^3(0)) = \boxed{0}[/tex]
