Answer
$$8 \pi$$
Work Step by Step
Stokes' Theorem states that $\iint_{S} curl F \cdot dS=\iint_{C} F \cdot dr$
where, $C$ corresponds to the boundary of the surface oriented counter-clockwise.
We have the parameterization representation for the given surface as: $r=\lt 2 \cos t, 2 \sin t, 1 \gt$ and $ dr = \lt -2 \sin t , 2 \cos t , 0 \gt$
So, $F[r(t)]=\lt -4 \sin t , 2 \sin t , 6 \cos t \gt$
$$\iint_{S} curl F \cdot dS=\iint_{C} F \cdot dr= \int_{0}^{2 \pi} \lt -4 \sin t , 2 \sin t j, 6 \cos t \gt \cdot \lt -2 \sin t , 2 \cos t j, 0 \gt\\=\int_{0}^{2 \pi} 8 \sin^2 t+4 \cos t \sin t dt \\=\int_{0}^{2 \pi} 8 (\dfrac{1-\cos 2t}{2})+(2) \times (2) \times \cos t \sin t dt \\=\int_{0}^{2 \pi} 4+2 \sin 2t -4 \cos 2t dt \\=\int_{0}^{2 \pi} 2 \sin t -8 \times (\cos 2t) dt+ \int_0^{2 \pi} 4 dt \\=[4t-2 \sin 2t -\cos 2t]_0^{2 \pi} \\=4 [2 \pi -0-1-(0-0-1)] \\=8 \pi$$
