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As I understand it, when doing a surface integral we have,

$$\iint_S F\cdot ndS=\iint_D r~\frac{r_a \times r_b}{|r_a \times r_b|}|r_a \times r_b|dA$$ and this is true because $$ndS=\frac{r_a \times r_b}{|r_a \times r_b|}|r_a \times r_b|dA$$ (A unit normal vector multiplied by the differential of the surface).

But looking online I see $$\iint_S F\cdot ndS=\iint_Dr~\frac{\nabla f}{|\nabla f|}|\nabla f|dA$$ why is this true though? I don't see how we switch from a surface to a double integral in the second case.

I suppose I should link the problem where I saw the second method. It is the last example on this page: http://tutorial.math.lamar.edu/Classes/CalcIII/StokesTheorem.aspx

Josh I
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  • What is the relationship between $S$ and $f$ here? – Michael Albanese Jul 26 '15 at 21:16
  • I believe that $f$ is a parametrisation of $S$ but I'm not entirely sure. I suppose I should link the problem where I saw the second method. It is the last example on this page: http://tutorial.math.lamar.edu/Classes/CalcIII/StokesTheorem.aspx – Josh I Jul 26 '15 at 21:17
  • @JoshI The link does not display well on my screen and I cannot read it well enough. Suffice to say that the gradient of a function $f$ points in the direction normal to a level surface. Alternatively, if the surface is parameterized as $\vec r=\vec r(u,v)$, then a surface normal is given by $\left|\frac{\partial \vec r}{\partial u}\times\frac{\partial \vec r}{\partial v}\right|$. – Mark Viola Jul 26 '15 at 22:46

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