Circuit Diagrams Ignore The Orientation Of The Resistor. Is This An Oversight?

We know that: $R=\dfrac{\rho l}{A}$ where $R=\text{Resistance}, \hspace{.25cm} \rho = \text{Specific Resistance}, \hspace{.25cm} l = \text{length}, \hspace{.25cm} A = \text{Area}$ .
We can rewrite the equation by multiplying $I$ so that both sides become: $IR=\dfrac{\rho lI}{A}$ From Ohm's law $V=IR$ therefore: $V=\dfrac{\rho lI}{A} \longrightarrow I=\dfrac{VA}{\rho l}$ For minimum current $A$ is minimum and length is maximum: Hence $A=2\times1$ and $l=3$ $1=\dfrac{2V}{3\rho}$ $\dfrac{V}{\rho}=\dfrac{3}{2}$ For maximum current $A$ is maximum and $l$ is minumum: Hence $A=2\times3$ and $l=1$ : $I=\dfrac{6V}{\rho} = 6\times\dfrac{3}{2} = 9\text{mA}$

Considering that R=ρL/A, The three possible resistances in the solid are 2ρ/3 , 3ρ/2 and ρ/6. The minimum current flows through the maximal resistance, and V=RI. So: V=3p/2. We can rewrite this as V/ρ = 3/2. Now we use this result at the same equation but with the minimal resistance: I=V/(ρ/6) and we will end up with I=0.009 but we need to remember that this value is in miliamperes and multiply by 10⁻³.