Answer
$0$
Work Step by Step
We have: $F(r(t))=\cos^2 t \sin (0) i+\sin^2 t j+\cos t \sin t k=0 i i+\sin^2 t j+\cos t \sin t k$
Since, the surface is the part of the circle $x^2+y^2=1 $ and therefore, the parameterization of the boundary can be written as: $C: r(t)=\cos t i+\sin t j+0 k \implies dr=(-\sin t i+\cos t j+0k) dt$
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.
Now, $$\iint_{S} curl F \cdot dS=\iint_{C} F \cdot dr=\int_0^{2 \pi} (0 i+\sin^2 t j+\cos t \sin t k) \cdot (-\sin t i+ \cos t j+0 k) dt \\=\int_{0}^{2 \pi} \sin^2 t \cos t dt$$
Plug $u=\sin t $ and $du =\cos t dt$
So, $\int_{0}^{2 \pi} \sin^2 t \ \cos t dt=\int_0^{2 \pi} u^2 du= 0$
