Answer
$\iint_S curl F\cdot dS=\int_C F \cdot dr=0$
Work Step by Step
Apply Stoke's Theorem : $\iint_S curl F\cdot dS=\int_C F \cdot dr$
In our problem, we are given that the paraboloid and the xy plane intersect on a circle, that is, $z=0; x^2+y^2=1$
Need to write these in the parametric form such as: $x=\cos t, y=\sin t; z=0$; $0 \leq t \leq 2 \pi$
Thus, $r=\lt \cos t , \sin t, 0 \gt$ and $dr=\lt -\sin t, \cos t ,0 \gt$
Now, $F=x^2 i+y^2 j+z^2 k $ and $F=\lt \cos^2 t ,\sin^2 t,0 \gt$
Then, we have $\int_C F \cdot dr=\int_0^{2 \pi}- \cos^2 t \sin t dt+\int_0^{2 \pi}\sin^2 t \cos t dt$
Plug $m=\cos t$ and $-\sin t dt =dm$
and $n=\sin t$ and $\cos t dt =dn$
$\int_C F \cdot dr=\int_1^{1} p^2 dp+\int_0^{0} k^2 dk=0$
when $F=ai+bj+ck$
Then, $curl F=(\dfrac{\partial c}{\partial x}-\dfrac{\partial b}{\partial z})\hat{i}+(\dfrac{\partial a}{\partial z}-\dfrac{\partial c}{\partial x})\hat{j}+(\dfrac{\partial b}{\partial x}-\dfrac{\partial a}{\partial y})\hat{k}$
This gives: $curl(F)=0$
so, $\iint_S curl (F) \cdot dS=0$
Thus, Stoke's Theorem is verified.
$\iint_S curl F\cdot dS=\int_C F \cdot dr=0$
