Relating the Minimizing Integral to the Capacity of a Cube

[Mathmos6]

Homework Statement

The capacity C of an object is the integral over its surface
$-\int_S \frac{\partial \phi}{\partial n} dA$,
where the potential φ(x) satisfies Laplace’s equation in the volume outside the object, $\phi = 1$ on S and $\phi \to 0$ at $\infty$. Show that the capacity of a sphere of radius R is $4\pi R$. (I've done that bit).

Now I need to show that the capacitance of a cube is s.t. $2 \pi a < C < 2\sqrt{3} \pi a$. The hint says I need to "relate the minimizing integral (below) to the capacity. Then for the lower bound, use the volume outside the inscribing sphere and take w equal to the solution to Laplace’s equation outside the cube which is extended by w=1 in the gap between the sphere and the cube.".

Homework Equations

The 'minimising integral' is (I've proven)

$\int_V |\nabla w|^2 dV \geq \int_V |\nabla u|^2 dV$ where u and w are both equal to f on 'S' enclosing 'V', w has continuous first partial deriv.s and u is a solution to Laplace's equation.

The Attempt at a Solution

We know $\phi$ is going to be a function of (r) by symmetry, but I can't really even see how to begin the second part - relating the minimising integral to the capacity. I've played around with a number of identities to try and make the surface integral look like the volume one, but to no avail... help!

 

[Mathmos6]

Right, i think I've got a little further:

So you can consider $\nabla \cdot (\phi \nabla \phi) = (\nabla \phi)^2 + \phi \nabla ^2 \phi$, so since $\phi = 1$ on the relevant surfaces, $\int_S \nabla \phi \cdot n dA = \int_S \phi \nabla \phi \cdot n dA$? In which case by divergence theorem capacity = $-\int_V \nabla \cdot (\phi \nabla \phi) dV = -(\int_V (\nabla \phi)^2 + \phi \nabla ^2 \phi dV)$? At which point you'd want the integral for the volume outside the insphere = integral of (volume between insphere & cube + volume outside cube)?