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From the definition of divergence of a smooth vector field $F:\mathbb{R}^n\to \mathbb{R}^n$, $\mathbb{div}F:=\lim_{|V|\to 0}\frac{\int_{\partial V}F\cdot do}{|V|},$ where $\partial V$ is a piece-wise smooth $n-1$ dimensional surface, how can we directly prove $$\mathbb{div}F=\sum_{n=1}^n \frac{\partial F^i}{\partial x_i}??$$ I can verify this for special choice of $\partial V$, such as rectangles centered around a point $x$. But how can we prove this for all piece-wise smooth $\partial V$??

I know we can prove $\int_V \frac{\partial u}{\partial x_i}dV =\int_{\partial V}u v_i do$, where $v_i$ is the $i$-th component of normal unit vector on $\partial V$ and $u$ is smooth on $\overline{V}$. But that only works after we have known the formula above. Thanks.

user760
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    If you know it for “all” $V$, then you in particular know it for all rectangles, and rectangles then give you the desired formula. The difficult part is showing that from this formula for the divergence, you can actually get the full divergence theorem. – peek-a-boo Jan 07 '24 at 10:53
  • I guess the hard part is to show that you can define the divergence like that, i.e. that the limit exists and does not depend on the chosen sequence. Then you can consider cubes to obtain the formula in cartesian coordinates, spheres to obtain the formula in spherical coordinates etc. – Filippo Jan 07 '24 at 11:16
  • @peek-a-boo So we must first prove the Divergence Theorem, and then use it in the definition of divergence and take the limit and arrive at the desired formula? – user760 Jan 07 '24 at 21:22
  • See this answer of mine for more details. There I talk about the curl, but with very minor modifications you can address the divergence. The point is, if you know the limit already exists, then you can prove the given formula. The difficult part (which isn’t really the direction you’re asking about) is showing that this formula actually ensures the limit exists; for this I don’t see any other way than to prove the divergence theorem. – peek-a-boo Jan 07 '24 at 21:54
  • @peek-a-boo Thanks. But in that answer, you had an $S$ and an $M$, which are supposed to be "surfaces" in $\mathbb{R}^3$, but then this $M$ also has a boundary $\partial M$. That doesn't make much sense to me, since, from your wording, it seems $S$ is a larger surface enclosing a smaller surface $M$ which encloses the point $p$. Did I misunderstand something? – user760 Jan 08 '24 at 10:47
  • @peek-a-boo Or did you mean $S$ is a surface passing through $p$, and $M$ is a small portion of $S$ containing $p$?? – user760 Jan 08 '24 at 10:59
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    Yes $S$ is a “large surface” and $M$ is allowed to be any piece of it containing $p$ and having a boundary curve. But now it indeed seems that introducing the $S$ was superfluous and that I could have phrased things more simply (but at the same time I remember thinking carefully about this when I wrote that answer)… I’ll look more closely at it later and see if it can indeed be simplified ( also there are one or two small typos there which I just noticed…). Perhaps I did it in order to have a clean way of talking about the normal vector first? Idk I’ll think later. Anyway… it’s not wrong :) – peek-a-boo Jan 08 '24 at 11:01

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