Answer
$\iint_S curl F\cdot dS=\int_C F \cdot dr=0$
Work Step by Step
Apply Stoke's Theorem as: $\iint_S curl F\cdot dS=\int_C F \cdot dr$
We need to write these in the parametric form.
we have $x=\cos t, y=\sin t; z=0$; $0 \leq t \leq 2 \pi$
and $r=\lt \cos t , \sin t, 0 \gt $
Then $dr=\lt -\sin t, \cos t ,0 \gt$
$\int_C F \cdot dr=\int_0^{2 \pi} \lt \cos^2 t ,\sin^2 t,0 \gt \cdot \lt -\sin t, \cos t ,0 \gt dt$
or, $\int_C F \cdot dr=\int_0^{2 \pi}- \cos^2 t \sin t dt+\int_0^{2 \pi}\sin^2 t \cos t dt$
Suppose $\cos t=r \implies -\sin t dt =dr$
and $\sin t=s \implies \cos t dt =ds$
$\int_C F \cdot dr=\int_1^{1} r^2 dr+\int_0^{0} s^2 ds=0$
Consider $F=pi+qj+rk$
We know that curl $F=\nabla \times F$
$ curl F=(\dfrac{\partial r}{\partial x}-\dfrac{\partial q}{\partial z}){i}+(\dfrac{\partial p}{\partial z}-\dfrac{\partial r}{\partial x}){j}+(\dfrac{\partial q}{\partial x}-\dfrac{\partial p}{\partial y}){k}$
This implies that
curl$F=0$
Thus, $\iint_S curl F\cdot dS=0$
Hence, Stoke's Theorem has been verified.
$\iint_S curl F\cdot dS=\int_C F \cdot dr=0$
