Effectivness factor for a cylinder

I would like to demonstrate the effectiveness factor of a porous cylinder, for diffusion-reaction (only radial diffusion, 1st order reaction, isothermal, diffusion coeff. is constant, steady-state), such as
$$
\eta=\frac{2 \cdot I\_1(\phi)}{\phi \cdot I\_0(\phi)} 
$$
using the info from this page [integration of bessel function](https://math.stackexchange.com/questions/17634/how-do-you-integrate-a-bessel-function-i-dont-want-to-memorize-answers-or-use).


Also, I consider $$ dV = dr \cdot r \cdot d\theta dz $$


with:


* *V* for volume
* *r* for the radius, that ranges from $0$ to *R*,
* $\theta$ for the azimuth, $0$ to $\pi$
* *z* for the length, $0$ to *h*


So I have
$$
\frac{c(Y)}{c\_{area}}=\frac{I\_0(\phi Y)}{I\_0(\phi)}
$$
with *Y* the dimensionless radius ($Y=r/l\_c, l\_c=R\_p/2$, *c* and $c\_{area}$ the concentration of the reactant in the media and at the surface of the media respectively, $\phi$ the Thiele modulus $\phi=R\_p/2\cdot\sqrt{k/D}$, and
$$
r\_p=\frac{1}{V} \iiint\_{V}{}R(c)dV
$$
$$
r\_p = \frac{1}{V}\int\_{0}^{1}-k \cdot c(r) \cdot r \cdot dr \int\_{0}^{\pi}d\theta \int\_{0}^{h}dz
$$
such as
$$
\eta=\frac{r\_p}{r\_{area}}
$$


**I would like some insight (or even clues) on how to proceed the integration of $\eta$**. Because I might be doing something wrong and I can't seem to find the issue.




By definition, the effectiveness factor is the ratio between: (1) the rate of reaction integrated along the catalytic particle, and (2) the rate of reaction integrated along the catalytic particle as if all of it is covered at the surface concentration $C\_A = C\_{As} $. So, for an isothermal operation so that the rate constant may be taken out of the integral, we have
\begin{align}
 \eta &:= \dfrac{\displaystyle\int\_V (-r\_A')(C\_A) \; dV}
 {\displaystyle\int\_V (-r\_A')(C\_{As})\; dV} 
 = \dfrac{\displaystyle\int\_0^{2\pi}\int\_0^{L}\int\_0^{R} kC\_A(r)
 \; rdrd\theta dz}
 {\displaystyle\int\_0^{2\pi}\int\_0^{L}\int\_0^{R} kC\_{As}
 \; rdrd\theta dz} \\
 &= \dfrac{2\pi Lk\displaystyle\int\_0^{R} C\_A(r) \; rdr}
 {2\pi LkC\_{As}\displaystyle\int\_0^{R} rdr}
 = \dfrac{\displaystyle\int\_0^{R} C\_A(r) \; rdr}
 {C\_{As}\dfrac{R^2}{2}}
 = \dfrac{2}{R^2}\int\_0^{R} \dfrac{C\_A(r)}{C\_{As}} \; rdr \tag{1}
\end{align}
Making the substitution $\lambda = r/R$ and $dr = Rd\lambda$, leads to the appearence of the dimensionless concentration $\psi(\lambda) = C\_A(\lambda)/C\_{As}$ ($\psi$ is for you $Y$). Thus, Eq. (1) leads to
\begin{align}
 \eta &= \dfrac{2}{R^2}\int\_0^1 \psi(\lambda) R\lambda \; d(R\lambda) \\
 \eta &= \dfrac{2R^2}{R^2}\int\_0^1 \lambda \psi(\lambda) \; d\lambda \\
 \eta &= 2\int\_0^1 \lambda \psi(\lambda) \; d\lambda \tag{2} \\
\end{align}
As you have shown there, the dimensionless concentration profile that solves the difussion reaction process of a catalytic particle in cylindrical symmetry is
$$ \psi(\lambda) = \dfrac{I\_0(\phi\lambda)}{I\_0(\phi)} \tag{3} $$
Combining Eqs. (2) and (3)
\begin{align}
 \eta &= 2\int\_0^1 \lambda \left(\dfrac{I\_0(\phi \lambda)}{I\_0(\phi)}\right) \; d\lambda \\
 \eta &= \dfrac{2}{I\_0(\phi)}\int\_0^1 \lambda I\_0(\phi \lambda)\; d\lambda \\
 \eta &= \dfrac{2}{\phi^2I\_0(\phi)}\int\_0^1 (\phi\lambda) I\_0(\phi \lambda)\; d(\phi\lambda) \tag{4} \\
\end{align}
Our mathematical friends from Stack Exchange informs us about the following relationship regarding Bessel functions, which we abuse
$$ \int\_a^b u I\_0(u) \; du = [uI\_1(u)]\vert\_a^b \tag{5} $$
And combining Eqs. (4) and (5) we have
\begin{align}
 \eta &= \dfrac{2}{\phi^2I\_0(\phi)} [\phi\lambda I\_1(\phi\lambda)]\vert\_{\lambda = 0}^{\lambda = 1} \\
 \eta &= \dfrac{2}{\phi^2I\_0(\phi)} [\phi I\_1(\phi) - 0] \\
 \eta &= \dfrac{2}{\phi^2I\_0(\phi)} \phi I\_1(\phi) \rightarrow
 \boxed{\eta = \dfrac{2I\_1(\phi)}{\phi I\_0(\phi)}} 
\end{align}


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