HINT:
You can write,
$$I_1 = I_2 + I_3$$
in terms of
$$\frac{\varepsilon}{R_1}-\frac{V_c}{R_1}=\frac{V_c}{R_2}+C\frac{dV_c}{dt}$$
This is a first order linear diff. equation:
$$\frac{dV_c}{dt}+\frac 1 C (\frac{R_1+R_2}{R_1R_2})V_c=\frac{\varepsilon}{R_1C}$$
And the charge is:
$$q=CV_c$$
The required region is constituted of $3$ parts.
My notation: $\theta$ = Polar angle, where angle is taken with respect to the vector joining the two spheres, pointing two origin, $\phi$ = Azimuthal angle.
Parametrization of upper spherical cap: $$ \begin{pmatrix} &2 \sin(\theta) \cos(\phi) \\ & 2 \sin(\theta) \sin(\phi)\\ &2 \cos(\theta) \end{pmatrix}$$
with limits $ \theta \in \bigg ( \dfrac{\pi}{2} ,\dfrac{3 \pi}{2} \bigg) $,$ \phi \in (0, 2 \pi )$
Parametrization of lower spherical cap: $$ \begin{pmatrix} &G \sin(\theta) \cos(\phi) \\ & G \sin(\theta) \sin(\phi)\\ &G \cos(\theta) \end{pmatrix}$$
Where $$G = 2 \cos \theta - \sqrt{4 \cos^2 \theta -3}$$
with limits: $ \theta \in \bigg ( 0 , \arccos \bigg( \cfrac{ 2 - 2^{-1/2}}{\sqrt{5 - 2 \sqrt{2} } } \bigg) $,$ \phi \in (0, 2 \pi )$
Parametrization of conical section: $$ \begin{pmatrix} &H \sin(\theta) \cos(\phi) \\ & H \sin(\theta) \sin(\phi)\\ &H \cos(\theta) \end{pmatrix}$$
Where $$H = \cfrac{2}{\cos \theta + \sin \theta} $$
with limits: $ \theta \in \bigg ( \arccos \bigg( \cfrac{ 2 - 2^{-1/2}}{\sqrt{5 - 2 \sqrt{2} } }\bigg) , \dfrac{ \pi}{2} \bigg) $,$ \phi \in (0, 2 \pi )$
Best Answer
The $r(x)$ given in your question seems like a square of a linear term in $x$
So, for the sake of simplicity, say $r(x)=(ax+b)^2$
Now the integral calculating the resistance becomes:
$$R=\rho \int_0 ^L \frac{1}{\pi (ax+b)^2} \,dx$$
$\frac{1}{linear^n}$ has an easy indefinite integral
$$R=\frac{\rho}{\pi a} . \left[ \frac{-1}{3(ax+b)^3} \right]_0^L $$
Just plug in the limits to get the answer.