I am trying to derive the error for approximatively computing the divergence of a vector field $\mathbf{a}$. The Gauss divergence theorem states
$\int_V \nabla \cdot \mathbf{a} dx = \oint_{\partial V} \mathbf{a} \cdot \mathbf{n} ds$
For a cube of volume $V = dx_1 dx_2 dx_3$ with a center $\mathbf{x} = (x_1, x_2, x_3)$ in a Cartesian coordinate system we can write
(1) $\lim_{V\rightarrow 0}\oint_{\partial V} \mathbf{a} \cdot \mathbf{n} ds = \sum_{i = 1, i\ne j\ne k}^3 \left[ \mathbf{a}(x_i + \dfrac{dx_i}{2}) - \mathbf{a}(x_i - \dfrac{dx_i}{2}) \right]\mathbf{e}_i dx_j dx_k$
$= \sum_{i = 1, i\ne j\ne k}^3 \left[ a_i(x_i + \dfrac{dx_i}{2}) - a_i(x_i - \dfrac{dx_i}{2}) \right]dx_j dx_k$
If $\mathbf{a}$ is an analytical function we can develop Taylor series around $\mathbf{x}$:
(2) $a_i (x_i + \dfrac{dx_i}{2}) = a_i(x_i) + \dfrac{da_i}{dx_i}\dfrac{dx_i}{dx_i 2} + \dfrac{d^2a_i}{dx_i^2}\dfrac{dx_i^2}{ 4} + \dfrac{d^3a_i}{dx_i}\dfrac{dx_i^3}{ 12} + O(\dfrac{dx_i^4}{2\cdot4!})$
(3) $a_i (x_i - \dfrac{dx_i}{2}) = a_i(x_i) - \dfrac{da_i}{dx_i}\dfrac{dx_i}{2} + \dfrac{d^2a_i}{dx_i^2}\dfrac{dx_i^2}{4} -\dfrac{d^3a_i}{dx_i}\dfrac{dx_i^3}{ 12} + O(\dfrac{dx_i^4}{2\cdot4!})$
With $\mathbf{a}\cdot \mathbf{e}_i = a_i$, we get
(4) $\sum_{i = 1, i\ne j\ne k}^3 \left[ a_i(x_i + \dfrac{dx_i}{2}) - a_i(x_i + \dfrac{dx_i}{2}) \right]dx_j dx_k$
Using (2) - (3) in (4), (1) will give
$\lim_{V\rightarrow 0}\oint_{\partial V} \mathbf{a} \cdot \mathbf{n} ds = \sum_{i = 1, i\ne j\ne k}^3 \dfrac{da_i}{dx_i} dx_i dx_j dx_k + \sum_{i = 1, i\ne j\ne k}^3 \dfrac{d^3a_i}{dx_i^3} \dfrac{dx_i^2}{6} dx_i dx_j dx_k + \sum_{i = 1, i\ne j\ne k}^3 O(dx_i^5) dx_j dx_k$
and since $dx_i dx_j dx_k = V$
(5) $\lim_{V\rightarrow 0} \dfrac{1}{V} \oint_{\partial V} \mathbf{a}\cdot\mathbf{n} ds = \nabla \cdot \mathbf{a} + \sum_{i = 1}^3 \dfrac{d^3a_i}{dx_i^3} \dfrac{dx_i^2}{6} + \sum_{i = 1}^3 O(\dfrac{dx_i^5}{5!}) $
With the limit applied, the second and third terms on the r.h.s. tend faster to zero than the first one, and the definition of the divergence is obtained. However, if $V$ doesn't tend to zero, but to a finite small number, can I use the (4) to claim that the approximation is second order accurate because of the first term in the truncation error:
$\sum_{i = 1}^3 \dfrac{d^3a_i}{dx_i^3} \dfrac{dx_i^2}{6}$
My quetion is: is this the correct way to derive the error? Is there anything missing / wrong in the derivation?