What is the current between two metal spheres held at potential difference V? We are given two metal spheres, each of radius a, immersed deep in the sea (uniform conductivity σ) and held quite far apart. The potential difference between them is $V$.
Now, I have calculated the resistance between two concentric metal spherical shells (radius a and radius b) held at potential difference $V$:

$$R = \frac{1}{4 \pi \sigma} \left(\frac{1}{a} - \frac{1}{b} \right)$$

Now if we take $b \gg a$, we get

$$ R = \frac{1}{4π\sigma a}$$

The solution manual uses this result and reads the following:

Using the previously obtained result, we note that essentially all of the resistance is in the region right around the inner sphere (radius a). Successive shells as you go out contribute less and less, because the cross-sectional area ($4 \pi r^2$) gets larger and larger. For the two submerged spheres $R = 2/4\pi \sigma a$ (one $R$ as the current leaves the first, one $R$ as it converges on the second). Therefore $I = V/R = 2\pi \sigma aV$.

I hope someone could explain this derivation to me, especially the first argument.
 A: Using Gauss's law, we can calculate the electric field between the two spheres.
$$E=\frac1{4 \pi \epsilon_0} \frac Q{r^2}$$
The current at $r=a$,
$$J=\sigma E$$
$$I = J A = \sigma E A = \sigma \frac1{4 \pi \epsilon_0} \frac Q{a^2} 4\pi a^2 = \frac{\sigma Q}{ \epsilon_0} $$
For the two concentric spheres, the potential is,
$$V= \int_a^bEdr=\int_a^b \frac1{4 \pi \epsilon_0} \frac Q{r^2} dr=\frac1{4 \pi \epsilon_0} Q (\frac 1a-\frac 1b)$$
$$R = V/I = \frac1{4 \sigma \pi  }  (\frac 1a-\frac 1b)$$

To get the current is straightforward. When the outer spheric shell's diameter is very very large. The resistance can be written,
$$R =  \frac1{4 \sigma \pi  }  (\frac 1a)$$
Knowing potential $V$, we can calculate the current,
$$I=\frac VR=V a 4 \sigma \pi $$
Here is the point, the resistance in this case is really close to the inner spheric shell. Because of the area increasing, other shells' resistance is very small.
A: If I'm understanding you correctly, your question is why the concentric-shell result (which has spherical symmetry) can be applied to the separated-sphere configuration (which doesn't).  That's a pretty cute trick.
Let's go back to the concentric spheres.  Imagine building the resistance you've calculated as a network of resistors in series, one layer at a time, where each layer has thickness $a$.  Some individual layers have resistances
\begin{align}
R_1 &= \frac1{4\pi\sigma}\left(\frac1{a} - \frac1{2a}\right) \\
R_2 &= \frac1{4\pi\sigma}\left(\frac1{2a} - \frac1{3a}\right) \\
R_3 &= \frac1{4\pi\sigma}\left(\frac1{3a} - \frac1{4a}\right) \\
\vdots \\
R_n &= \frac1{4\pi\sigma}\left(\frac1{na} - \frac1{(n+1)a}\right) 
= \frac1{4\pi\sigma a} \frac{1}{n^2+n}
\\
R_\text{symmetric} = \sum_{n=1}^{b/a} R_n &= \frac{1}{4\pi\sigma}
\left(\frac1a - \frac1b\right)
\approx \frac1{4\pi\sigma a}
\end{align}
Now suppose that the spherical symmetry is broken in the following way: let's have the first fifty of these consecutive shells have uniform conductivity, but all the spherical shells beyond be half conducting (still with conductivity $\sigma$) and half insulating.  So in the computation of these shells where the spherical symmetry is broken, the total resistance is
\begin{align}
R_\text{broken} &= 
\frac1{4\pi\sigma}\left(\frac1a-\frac1{50a}\right)
+
\frac1{4\pi\sigma}\left(\frac{2}{50a} - \frac{2}{b}\right) \\
&\approx \frac{49}{50} R_\text{symmetric} + \frac{2}{50} R_\text{symmetric}
\end{align}
This is only a 2% increase in the overall resistance, but it could be an enormous change in the conducting volume, if the space between the concentric shells is larger than the distance where we start mucking around with things.  Taking things a step further, maybe beyond $100a$ only one octant of the solid angle of the conducting shells are actually conducting.  Then the resistance is
\begin{align}
R_\text{twice broken} &= 
\frac1{4\pi\sigma}\left(\frac1a-\frac1{50a}\right)
+
\frac1{4\pi\sigma}\left(\frac{2}{50a} - \frac{2}{100a}\right) 
+
\frac1{4\pi\sigma}\left(\frac{8}{100a} - \frac{8}{b}\right) \\
&\approx \frac{49}{50} R_\text{symmetric}
+ \frac{2}{100}R_\text{symmetric}
+ \frac{8}{100}R_\text{symmetric}
\end{align}
Here's a figure showing this sphere, hemisphere, octant arrangement with $b=200a$:

What's happening is that so much of the resistance to current flow actually happens in the inner sphere that the shape of the outer layers hardly matter at all. The symmetry between the tiny inner sphere and the truncated outer sphere is quite broken, but the resistance between the two has changed by a fraction comparable to the ratio of their sizes.
In your problem with two spheres in the ocean, the actual current flow would follow the field lines of an electric dipole.  But assuming the two spheres were many radii apart, nearly all of the impedance to current flow happens near each sphere, where you can pretend it is a point charge.
