I'll get to the point immediately. The definition of divergence in a point (from my textbook): $$ div \bar{E} = \lim_{V \to 0} \frac{1}{V}\oint_S \bar{E}.d\bar{S}$$ (it's a surface integral)
Context (contains a figure): to show what divergence looks like in Cartesian coordinates for the vector field E, we will work with an infinitesimal volume with dimensions ∆$x$ x ∆$y$ x ∆$z$. (A cuboid or w/e) We will shrink this volume to a point ($x,y,z$). The surface integral is spread over the 6 surfaces Ⅰ,..., Ⅵ. 'Ⅰ' means the integral over the Ⅰ-side. The next part is where the problems begin.
Lets focus on sides Ⅰ and Ⅱ. (Left and right on the drawing). For a limit where ∆x, ∆y and ∆z approach 0, the integral over 'Ⅰ' approaches E$x$($x$+∆$x$,$y$,$z$)∆$y$∆$z$ and the integral over 'Ⅱ' approaches - E$x$($x$,$y$,$z$)∆$y$∆$z$.
That's what I don't understand. How do you get there? I assume it is by applying the definition. If anyone can help me understand that part, that would be awesome. The goals is that, when you apply the same logic to all sides, you get the expression: $$ div \bar{E} = \frac{\partial E_{x}}{\partial x}+\frac{\partial E_{y}}{\partial y}+\frac{\partial E_{z}}{\partial z}$$
I'm confident that once I understand that, I'll be able to figure out the rest. The 'div EⅠ+Ⅱ = ...' part is just applying the formal definition of a derivative.
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$\begingroup$A rigorous way would be to Taylor expand $E$ on each of the faces. Example for faces I and II:
$$\int_I E(x', y', z') \cdot dS' = E_x (x, y, z) \, \Delta y \, \Delta z + \partial_x E_x |_{x,y,z} \Delta x \, \Delta y \, \Delta z + \text{higher order terms}$$
and
$$\int_{II} E(x', y', z') \cdot dS' = -E_x (x, y, z) \, \Delta y \, \Delta z + \text{higher order terms}$$
When you add these contributions together, the terms proportional to $\Delta y \Delta z$ exactly cancel, the terms proportional to $\Delta x \, \Delta y \, \Delta z$ contribute terms of $O(1)$ to the limit you need to take, and everything of higher order goes to zero in the limit. (Exercise: compute some of these higher order terms to convince yourself this is the case.)
Repeat this process for the other faces, and you're done.
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