Resistance of a conductor shaped like a ball As we know, there is an equation for the resistance of conductors given their dimensions, which is:   where $l$ is the length of the conductor, measured in metres [$m$], $A$ is the cross-section area of the conductor measured in square metres [$m^2$], and $\rho$ is the electrical resistivity of the material, measured in ohm-metres ($\Omega m$).  And that formula requires that the cross-sectional area $A$ be constant, by wich I mean that the cross-sectional are of any part of the conductor should be the same.  Now, I am asking how can we deal with conductors shaped like a ball or a pyramid, which have different cross-sectional areas from one part of them and other ones.  Can anyone help me with that problem, please?
General integral to find resistance My question is: is there a simple and truly general equation for the resistance between two electrical equipotential surfaces?.  Obviously, if so, what is it, and if not, why?  It would be very difficult to solve, granted, but I just want to see a calculus equation that is fully descriptive.  I have two frameworks under which this could be entertained, I'll write those out and then explain the motivation.  To start with, we need propose that the volume separating the two surfaces has a volumetric resistivity, $\rho$ in units of $(\Omega m)$.  Single Volume Framework  We can limit the discussion to a defined volume, then the surfaces reside in that volume or on the surface of it.  This volume may have a constant resistivity $\rho$ while everywhere outside the volume is completely electrically insulating.  Infinite Volume Framework  An alternative to the above approach that might make the task more or less difficult would be to replace a constant resistivity with a spatial dependence $\rho(\vec{r})$ and no longer require a boundary condition.  In that case we only have 3 mathematical inputs to the problem, which is the resistivity defined for all $\vec{r}$ and a definition of the two surfaces, $S_1$ and $S_2$.  Known Algebraic Analogs  The  that I find insufficient is:  $$R = \rho \frac{\ell}{A}$$  Where $l$ is the length of the restive material that is any shape which has translation symmetry over that length, and $A$ is the cross-sectional area.  Obviously, this is a rather simple equation that won't apply to more complicated geometry.  Even more  seem to give equations that fall short of what I'm asking.  For example:  $$R = \rho \int_0^l \frac{1}{A(x)} dx$$  I think it's obvious that an equation such as this is built upon a myriad of assumptions.  For a thought experiment, imagine that the area starts out as very small and then pans out to very large quickly.  Well, accounting for the larger area in the above sense underestimates the resistance, because the charge has to diffuse out perpendicular to the average direction of flow as well as parallel to it.  I have some reasons to suspect this might actually be rather difficult.  A big reason is that all the approaches I'm familiar with require the flow paths to be established beforehand, which can't be done for what I'm asking.  So maybe this will result in two interconnected calculus equations.  Motivation  I had an interest in , and it occurred to me that I can't quickly and simply write down the equation for resistance between two points.  The unique thing about Squishy Circuits is that it calls for two types of dough, one that conducts and one that is mostly insulating.  However, the recipes aren't perfect and because of that, the young children who play with these circuits regularly encounter the limits of conductor and insulator definitions.  If you make your conductor dough too long and/or too thin, you will encounter dimming of the light you connect with it.  Similarly, a thin insulator layer will lead to a lot of leakage current which also dims the light.
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