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I'm attempting to prove that for a closed, path-connected, volume $V$ of $\mathbb{R}^3$,

$$\iiint_{V} (\operatorname{div} F) \, \text{d}V = \iint_{\partial V} F \cdot \mathbf{\hat n} \, \text{d}S$$

Where $F : \mathbb{R}^3 \to \mathbb{R}^3$ is any differentiable function.

Is there anything wrong with my proof? Can I make it simpler?


Consider a box $B$ containing $V$. Partition $B$ into nonoverlapping boxes $B_1, B_2, \ldots B_m$, such that diameter of the partition is $< 1/n$. Now consider "chunks" $C_k = B_k \cap V$. Clearly,

$$\sum_{k} \left (\iint_{\partial C_k} F \cdot \mathbf{\hat n} \, \text{d}S \right) = \iint_{\partial V} F \cdot \mathbf{\hat n} \, \text{d}S$$ Moreover, by my generalized mean value theorem, there exists some $p_k \in C_k$ such that

$$\iint_{\partial C_k} F \cdot \mathbf{\hat n} \, \text{d}S = \operatorname{div} F|_{p_k}\cdot {\rm m}(C_k)\ .$$

Pick all of these $p_k$ and use them to form a Riemman sum

$$S_n = \sum_k \operatorname{div} (F|_{p_k}) \cdot \operatorname{m}(C_k) = \sum_{k} \left (\iint_{\partial C_k} F \cdot \mathbf{\hat n} \, \text{d}S \right) = \iint_{\partial V} F \cdot \mathbf{\hat n} \, \text{d}S$$

As we assumed $\operatorname{div} F$ was integrable, $$\lim_{n \to \infty} S_n = \iiint_{V} (\operatorname{div} F) \, \text{d}V$$

But $S_n$ is a constant sequence! Namely,

$$S_n = \iint_{\partial V} F \cdot \mathbf{\hat n} \, \text{d}S$$

Thus, it is the only possibility that

$$\iiint_{V} (\operatorname{div} F) \, \text{d}V = \iint_{\partial V} F \cdot \mathbf{\hat n} \, \text{d}S$$

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  • $\begingroup$ I'm assuming that the $m(C_k)$ are the volumes of the $C_k$'s? $\endgroup$ – D.B. May 19 '18 at 15:59
  • $\begingroup$ Yes, it is the measure. $\endgroup$ – Jeffery Opoku-Mensah May 19 '18 at 16:02
  • $\begingroup$ @ChristianBlatter You literally just changed \text to \rm. Why? $\endgroup$ – Jeffery Opoku-Mensah May 19 '18 at 18:12
  • $\begingroup$ You had forgotten the factor ${\rm m}(C_k)$. I have inserted it. $\endgroup$ – Christian Blatter May 19 '18 at 18:16
  • $\begingroup$ Oh, ok. Thanks! $\endgroup$ – Jeffery Opoku-Mensah May 19 '18 at 18:19

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