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Let $\vec{F}$ be the vector field $\vec{F}\left(x,y,z\right)=\left(z,x,y\right)$. Let $\rm S$ be portion of the surface $x^{2}+y^{2}+z=1$ lying above the $\rm XY$-plane, oriented upward. Evaluate the surface integral

$$\int_{\rm S}{\rm Curl}\vec{F}\cdot\vec{{\rm d}S}$$

I know how to evaluate the surface integral directly, but I don't know how to apply Stokes' theorem to solve it. Can someone please walk me through this? Thank you for your help!

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    $\begingroup$ What have you tried? The theorem is very explicit $$\int_{\rm S}\left(\vec{\nabla}\times\vec{F}\right)\cdot\vec{{\rm d}S}=\int_{\partial {\rm S}}\vec{F}\cdot\vec{{\rm d}\ell}$$ Do you see what $\partial{\rm S}$ is? $\endgroup$ – eranreches Dec 9 '17 at 19:39
  • $\begingroup$ Not really, I tried doing it but the computation was very similar to direct evaluation, so I wasn't sure if I was doing it right. $\endgroup$ – Jack Harold Dec 9 '17 at 19:43
  • $\begingroup$ The computation is pretty simple, so don't think you are doing wrong if your solution seems to be too simple. It will be easier for us to help you if you show us your way of thinking. So add your trial! $\endgroup$ – eranreches Dec 9 '17 at 19:48
  • $\begingroup$ Can you show the computation so I can check it please? I have it on paper and it's really messy. $\endgroup$ – Jack Harold Dec 9 '17 at 19:58
  • $\begingroup$ It is far better correcting your own solution rather than reading the solutions of others. So take a couple of minutes and write it down. Then I can point out your mistake, and post an answer. $\endgroup$ – eranreches Dec 9 '17 at 20:10
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It's actually pretty straightforward. Let the parametrization of the circle $x^2+y^2=1$ laying on the $xy$ plane be $\gamma(t)=(\cos t,\sin t,0)$, where $t\in[0,2\pi]$. Then $\gamma'(t)=(-\sin t, \cos t,0)$ and

\begin{align} \int_{\rm S}{\rm Curl}\vec{F}\cdot\vec{{\rm d}S} =\int_{0}^{2\pi} f(\gamma(t))\cdot\gamma'(t){\rm d}t =\int_{0}^{2\pi} (0,\cos t,\sin t)\cdot(-\sin t,\cos t,0){\rm d}t =\int_{0}^{2\pi} (\cos t)^2{\rm d}t \end{align}

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