Divergence Theorem Proof

Consider a region $A\subset\mathbb{R}^2$ whose boundary $\partial A$ is a rectifiable simple closed curve. Partition $A$ into a set of sufficiently small rectangles $\delta A_i$ (near the boundary, the partition elements are not necessarily rectangular). Let $\vert\delta A_i\vert$ denote the area of $\delta A_i$. For each of the $\delta A_i$, let $(x_i, y_i)$ be a point in the interior of $\delta A_i$.

Figure: Region $A$, its boundary $\partial A$, and the partition.

From the definition of divergence, for the point $(x_i, y_i)$ in the partition element $\delta A_i$,

   div$$(\boldsymbol u)(x_i,y_i) = \lim_{\vert\delta A_i\vert\rightarrow ... ...delta A_i}\boldsymbol u\cdot\boldsymbol n\ ds + \mbox{o}(\vert\delta A_i\vert)$$

where the error term o$(\vert\delta A_i\vert)$ goes to zero sufficiently fast as $\vert\delta A_i\vert\rightarrow 0$.

Now, summing over all of the partition elements,

$$\sum_i$$div$$(\boldsymbol u)(x_i,y_i)\vert\delta A_i\vert = \sum_i\int_{\partial\delta A_i}\boldsymbol u\cdot\boldsymbol n\ ds +$$   O$$(\vert\delta A_i\vert) = \int_{\partial A}\boldsymbol u\cdot\boldsymbol n\ ds +$$   O$$(\vert\delta A_i\vert)$$

where the second equality is due to the cancellation of line integrals on the boundaries of adjacent partition elements (cf. Figure (b)).

Finally, taking the limit as $\vert\delta A_i\vert\rightarrow 0$, one finds

$$\int\hspace{-0.33cm}\int_{\hspace{-0.76cm}\ _{\ _{\ _A}}}$$   div$$(\boldsymbol u) dA = \lim_{\vert\delta A_i\vert\rightarrow 0}\sum_i$$div$$(\boldsymbol u)(x_i,y_i)\vert\delta A_i\vert = \int_{\partial A}\boldsymbol u\cdot\boldsymbol n\ ds$$

Reference: P.C. Mathews, Vector Calculus, Springer-Verlag, London, 1998.

Copyright, 2001, Joseph D. Fehribach