Proof That If a Harmonic Function Zero on a Surface is Zero Interior to the Surface

Theorem
If

$\phi$
is a harmonic function on a volume
$V$
with surface
$S$
on which
$\phi =0$
then
$\phi =0$
on
$V$
.
Proof
By Green's First Theorem for Harmonic Functions, if
$\psi = \phi$
we have
$\int \int_V (\mathbf{\nabla \phi} ) \cdot (\mathbf{\nabla \phi} ) dV = \int \int_S \phi \frac{\partial \phi}{\partial n} dS$

By the statement in the theorem
$\phi =0$
on
$S$
.
$\int \int_V (\mathbf{\nabla \phi} ) \cdot (\mathbf{\nabla \phi} ) dV = 0$

Hence
$\phi = CONSTANT$
on
$S$
.
But
$\phi$
is continuous and twice differentiable in
$V$
, so
$\phi =0$
throughout
$V$
.