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Let $\vec F(x,y,z)=(x^2+y^2+\sin(xy)$, $e^x$+$2xy, -yz$). Calculate the flow of $\nabla\times\vec F$ through the following surface: $$ S=\{(x,y,z) x^2 + 2y^2 + 3z^2=10, y\ge 0\} $$

I figured that if you added $S_{aux} =\{(x,y,z)\in \Bbb R^3 : x^2 + 3z^2 =10\}$ to S you had a closed surface and therefore could use the divergence theorem. This would mean that: $$ \iint_{S+S_{aux}}\nabla\times\vec F \mathrm{d}s = \iiint_{V} \nabla\circ(\nabla\times\vec F) \mathrm{d}v. $$
Since $\nabla\circ(\nabla\times\vec F)=0$, then: $$ \iiint_{V} \nabla\circ(\nabla\times\vec F)\mathrm{d}v = 0. $$
Therefore $$\iint_{S+S_{aux}}\nabla\times\vec F \mathrm{d}s = 0, $$ and since $$\iint_{S+S_{aux}}\nabla\times\vec F \mathrm{d}s = \iint_S\nabla\times\vec F \mathrm{d}s + \iint_{S_{aux}}\nabla\times\vec F \mathrm{d}s = 0$$ we have $$\iint_S\nabla\times\vec F \mathrm{d}s = -\iint_{S_{aux}}\nabla\times\vec F \mathrm{d}s. $$

Could you please tell me if my thinking is correct? Thank you very much!

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  • $\begingroup$ $S=\{(x,y,z)|x^2+2y^2+3z^2=10,y≥10\}?$ If $y\ge10,$ then $x^2+3z^2\le -190$ Wait. Reading further, maybe you meant $y\ge0?$ $\endgroup$
    – saulspatz
    Dec 2, 2018 at 22:11
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    $\begingroup$ Couldn't you just use Stokes' theorem? $\endgroup$
    – saulspatz
    Dec 2, 2018 at 22:16
  • $\begingroup$ Yes, I meant y≥0. Thanks. Let me change that now $\endgroup$ Dec 2, 2018 at 22:21

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I'm certain that the point of the problem is to use Stokes' theorem. $\partial S$ is the intersection of the plane $y=0$ with $S$ so it's the ellipse $x^2+3z^2=10, y=0.$ On this curve, $F$ simplifies to $(x^2, e^x, 0)$ When you compute the circulation of $F$ over $\partial S$ there will only be a contibution from the first coordinate.

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  • $\begingroup$ I have just tried that by parametrizing the ellipse as $\alpha(\theta):(\sqrt{10}cos{\theta},0,\sqrt{\frac {10}3}sin{\theta})$, then calculated the line integral which ended up giving me zero as well. Does that mean that what I did was valid? $\endgroup$ Dec 2, 2018 at 23:16
  • $\begingroup$ Yes, I think what you did was correct, but it's been a long time since I worked with this stuff, so I'm not certain. $\endgroup$
    – saulspatz
    Dec 3, 2018 at 0:01

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