Answer
Stoke's Theorem has been verified.
$\iint_S curl F\cdot dS=\int_C F \cdot dr=0$
Work Step by Step
Stoke's Theorem states that $\iint_S curl F\cdot dS=\int_C F \cdot dr$
We are given that the paraboloid and the xy plane intersect on a circle, that is, $z=0; x^2+y^2=1$
Write these in the parametric form as: $x=\cos t, y=\sin t; z=0$; $0 \leq t \leq 2 \pi$
Thus, $r=\lt \cos t , \sin t, 0 \gt \implies dr=\lt -\sin t, \cos t ,0 \gt$
and $F=x^2 i+y^2 j+z^2 k \implies F=\lt \cos^2 t ,\sin^2 t,0 \gt$
Now, $\int_C F \cdot dr=\int_0^{2 \pi} \lt \cos^2 t ,\sin^2 t,0 \gt \cdot \lt -\sin t, \cos t ,0 \gt dt$
or, $\int_C F \cdot dr=\int_0^{2 \pi}- \cos^2 t \sin t dt+\int_0^{2 \pi}\sin^2 t \cos t dt$
Plug $\cos t=p \implies -\sin t dt =dp$
and $\sin t=k \implies \cos t dt =dk$
$\int_C F \cdot dr=\int_1^{1} p^2 dp+\int_0^{0} k^2 dk=0$
Use definition of curl $F=\nabla \times F$
Consider $F=ai+bj+ck$
That is, curl $F=(\dfrac{\partial c}{\partial x}-\dfrac{\partial b}{\partial z})\hat{i}+(\dfrac{\partial a}{\partial z}-\dfrac{\partial c}{\partial x})\hat{j}+(\dfrac{\partial b}{\partial x}-\dfrac{\partial a}{\partial y})\hat{k}$
So, we have
curl$F=0$
Thus, $\iint_S curl F\cdot dS=0$
Hence, Stoke's Theorem has been verified.
$\iint_S curl F\cdot dS=\int_C F \cdot dr=0$
