I am solving a reaction-diffusion problem in one dimension for a catalyst particle to get the internal effectiveness factor ($\eta$),as given below:
$$ \eta = \frac{\int_0^{V_p}{R_i\ dV}}{R_i^{surf}\cdot V_p}$$
Where the reaction rate of the limitting reactan is $R_i$, $V_p$ is the particle volume, and $V$ is the volume array, discretized along the same spatial points as the dimension of solution.
For a spherical particle $V$ is quite obvious, as using a system of spherical coordinates, one assumes angular symmetry, and solves along the radial domain ($r$): $V=\frac{4}{3}\pi r^3$.
But moving to slightly more complex shapes, lets say a cuboid, solved in cartesian coordinates, where all three edges have distinct lengths. The solution is computed along the dimension of shortest length, that is because mass flux across a plane perpendicular to this dimension is far greater than at its edges, thus the species concentration will be more uniform along this plane, and so edge effects apply. Where the last statement is more aplicable at $L_x<\!<L_y,\ L_z $, where $L$ is the corresponding edge length to each cartesian dimension.
Going back to the volume discretization, the volume of a cuboid will be given by: $V_p=L_x \cdot L_y \cdot L_z$. If the solution is computed along the domain of $x$, where the effective solution domain is discretized from the centre of the particle, (assuming central symmetry where the boundary condition of the reaction diffusion balance: $\frac{\partial c_i(x=0)}{\partial x}=0$). Then when integrating the solved reaction rate ($R_i$) at the discrete points of $x$ to estimate $\eta$, how does one compute the volume array ($V$) with discretized volume elements along $x$? Intuitively, $ V=2\cdot x\cdot L_y \cdot L_z$, where $x$ is a discrete array ranging from 0 to $L_x/2$, and $L_y$ and $L_z$ are both scalars. But should $x$ be multiplied by $L_y$ & $L_z$ directly? Or a discretized version of those, with an equal number of elements as $x$? Because the solution is vastly different.
If I had to guess it would be that the volume array is in fact $ V=2\cdot x\cdot L_y \cdot L_z$, because the solution is constant along $y$ and $z$, but I cannot logically rationalize it.
How would this hold up in the case of a shape where the edges are not so clearly defined, such as in an ellipsoid? Taking the volume of an ellipsoid: $V_p=\frac{4}{3} \pi\cdot a \cdot b \cdot c$; would the same approach apply?
