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This comes up in a physics note: ($S$ is a surface, $\vec{S}$ is a normal vector, $\nabla V$ is a gradient of $V$, and $\frac{\partial}{\partial n}$ is normal derivative) $$\nabla V \cdot {\rm d}\vec{S}=\frac{\partial V}{\partial n}{\rm d}S$$ It's probably super obvious, but I have very shaky background in differentials and mostly just "shut up and calculate." I don't even know what the equation is supposed to mean. What exactly is $dS$ here (have done surface integral before but never ask this)? In case this is somehow ill-defined, I just want to know why is $$\int_{S}\nabla V \cdot {\rm d}\vec{S}=\int_{S}\frac{\partial V}{\partial n}{\rm d}S$$ super obvious?

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I think the notation $\vec{S}$ by itself is awkward, but... here ${\rm d}S$ is the area form in the surface $S$, so that $$\int_S{\rm d}S = {\rm area}(S),$$and ${\rm d}\vec{S} = \vec{n}\,{\rm d}S$, where $\vec{n}$ is a unit normal field along $S$. For example, the flux of a vector field $\vec{F}$ tangent to $\Bbb R^3$ along $S$ is $$\int_S \vec{F} \cdot {\rm d}\vec{S} = \int_S \vec{F}\cdot \vec{n}\,{\rm d}S.$$If $V$ is a scalar field on $\Bbb R^3$, $\partial V/\partial \vec{n}$ is the directional derivative of $V$ in the direction of $\vec{n}$, and it equals $\nabla V \cdot \vec{n}$. It is really just a matter of understanding the notation. We have $$\int_S \nabla V\cdot {\rm d}\vec{S} = \int_S \nabla V\cdot \vec{n}\,{\rm d}S = \int_S \frac{\partial V}{\partial\vec{n}}\,{\rm d}S.$$

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  • $\begingroup$ Thank you so much! I can’t believe I don’t understand this earlier lol. I hate physics convention... $\endgroup$ – Ahmbak Jun 8 at 17:44
  • $\begingroup$ Don't we all? :P $\endgroup$ – Ivo Terek Jun 8 at 17:49

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