I'm trying to solve this question in preparation for my upcoming uni subject, i believe i understand it properly but im not quite sure, would like some assurance or guidance.
(a) Use Stokes Theorem to Evaluate $$\int\int_s\triangledown\times F\cdot dS$$ Where $$F=yz\hat i-xy \hat j+\hat k$$ and S is the hemisphere $x^2+y^2+(z-1)^2=4$ with $z\geq 1$ oriented upward
My answer is $-4\pi$ (working below) can someone confirm if this is correct, if not why. Thanks.
$$x^2+y^2+(1-1)^2=4$$ $$x^{2}+y^{2}=4$$ $$r(t)=<2cos(t),2sin(t),0>$$ $$F=[2sin(t)(1)\hat i-2cos(t)\cdot 2sin(t)+1]$$ giving $$\int^{2\pi}_{0}\left[((2\sin t)(-2\sin t))-(2\cos t\cdot 2\sin t)(2\cos t)\right]dt$$ and then solving this to get $$-4\pi$$
Somehow you end up calculating the right integral but for the life of me I don't understand your expression $F=[2sin(t)(1)\hat i-2cos(t)\cdot 2sin(t)+1]\,.$
A correct way of naming the theorem is "the Stokes theorem" or "Stokes' theorem".
A reader friendly solution would tell us that you are calculating the line integral of $F$ over the boundary of the hemisphere. Have you thought about orientation? Why is the result not $+4\pi\,?$
It would not hurt to say a few words why the correct equation for that boundary is $x^2+y^2=4\,.$
With the parametrization $t\mapsto \gamma(t):=(r\cos t,r\sin t,1)$ of that boundary $(r=2)$ the line integral is then
\begin{align} \int_0^{2\pi}F(\gamma(t))\cdot \dot \gamma(t)\,dt&=\int_0^{2\pi}\underbrace{\begin{pmatrix}r\sin t\\-r^2\cos t\sin t\\1\end{pmatrix}}_{\begin{pmatrix}yz\\-xy\\1\end{pmatrix}}\cdot\begin{pmatrix}-r\sin t\\r\cos t\\0\end{pmatrix}\,dt\\ &=\int_0^{2\pi}-r^2\sin^2 t-r^3\cos^2 t\sin t\,dt=-r^2\int_0^{2\pi}\sin^2t\,dt\\ &=-4\pi\,. \end{align}