Resistance of hollow metal sphere 
A hollow metallic sphere has inner and outer radii $a$ and $b$ respectively. How to calculate its resistance between two a points $A$ (on the inner surface) and a point $B$ (on the outer surface)? The resistivity of the metal is $r$.

What is the simplest approach to this problem?Is it possible to calculate without using integration?How to apply integration to this problem? Please help!!
 A: Since Michael has already pointed out that the problem as stated has no answer, I will answer a different question instead: if we have a resistive spherical shell with inner radius $a$, outer radius $b$, and bulk resistivity $\rho$, and the surfaces of this shell are coated with a conductive layer, what is the resistance between the inner and outer surface?
Now we can break the problem into a simple integration. Imagine the shell to be made up of infinitesimally thick shells at radius $r$ and with thickness $\delta r$. The same current $I$ has to flow through each consecutive shell (conservation of charge) so we should be able to compute the resistance of this shell. As all the shells are in series, we can integrate the expression to give the total resistance. Here we go:
Area of shell:
$$A = 4\pi r^2$$
Resistance
$$\delta R = \frac{\rho \delta r}{A} = \frac{\rho \delta r}{4\pi r^2}$$
Total resistance:
$$R = \int_a^b \frac{\rho dr}{4\pi r^2} = \frac{\rho}{4\pi}\left(\frac{1}{a}-\frac{1}{b}\right)$$
Incidentally you can see that if $a$ goes to zero, this resistance becomes infinite - just as it does when the contact surface of the point is truly a "point" as it appears to be in your question.
A: If you really mean "points", see the answer to this question.  Basically, the logic is as follows:

*

*If you try to inject a finite current at a "point" in a bulk, it will necessarily lead to a divergent current density $\vec{J}$ in a neighborhood of that point, proportional to $r^{-2}$ (where $r$ is the distance from the injection point.)

*A divergent current density $\vec{J}$ implies a divergent electric field $\vec{E}$, via the microscopic Ohm's Law $\vec{J} = \sigma \vec{E}$.

*$\vec{E} = - \nabla V$.  Since the electric field is proportional to $r^{-2}$, the potential will diverge as we go to the injection point (proportionally to $r^{-1}$).

*Thus, the potential difference $V$ between the points is infinite for a finite current $I$.

*Thus, $R = V/I$ is infinite.

To avoid this, you need to specify the size of the electrodes, and your answer will depend on the exact geometry.  The answer is pretty easy when the "electrodes" are the entire inner and outer surfaces (as @Floris has shown);  but any other configuration is much harder to address.
Note that the above argument implies that for sufficiently small electrodes, the resistance will be in general be inversely proportional to the electrode size.  (This is because the potential difference will scale inversely proportionally to the electrode size.)  The meaning of "sufficiently small" here will, of course, depend quite a lot on the precise geometry in question.
The above argument can also be extended to other dimensionalities.  For example, the same chain of arguments implies that the resistance between "sufficiently small" electrodes in 2-D will also be divergent.  The only difference is that the divergence will be logarithmic in this case, rather than inversely proportional to the electrode size.
