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I want to prove the following two equalities: $$ \oint_{\sigma}(f\nabla g)\cdot T\ dS=\iint_{\Sigma}(\nabla f\times \nabla g)\cdot N\ dA \\ \oint_{\sigma}(f\nabla g+g\nabla f)\cdot T\ dS=0$$

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For the first one:

From the Stokes theorem we have the following: $$\oint_{\sigma}(f\nabla g)\cdot T\ dS=\iint_{\Sigma}(\nabla \times (f\nabla g))\cdot N\ dA$$ So, we have to show that $$\nabla \times (f\nabla g)=\nabla f\times \nabla g$$

We have that $$\nabla \times (f\nabla g)=\text{curl} (f\nabla g)=\text{curl} \left (f\left (g_x, g_y, g_z\right )\right )$$ How can we continue?

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    $\begingroup$ hint: product rule for the second equality $\endgroup$
    – Secret
    Dec 6, 2017 at 11:16
  • $\begingroup$ You mean $\nabla (fg)=f(\nabla g)+g(\nabla f)$, right? So we get $\oint_{\sigma}(f\nabla g+g\nabla f)\cdot T\ dS=\oint_{\sigma}(\nabla (fg))\cdot T\ dS=\iint_{\Sigma}(\nabla \times (\nabla (fg))\cdot N\ dA$, right? @Secret $\endgroup$
    – Mary Star
    Dec 6, 2017 at 11:35
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    $\begingroup$ right, now what is curl of grad (for anything that is continuous in first derivatives) ? $\endgroup$
    – Secret
    Dec 6, 2017 at 12:37
  • $\begingroup$ It is equal to 0, right? @Secret $\endgroup$
    – Mary Star
    Dec 6, 2017 at 13:37
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    $\begingroup$ you answered your own question $\endgroup$
    – Secret
    Dec 6, 2017 at 15:05

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