## Proof of Formula for Integral of a Gradient Over a Volume

Theorem
If
$\psi = \psi (x,y,z)$
is continuously differentiable, then
$\int \int \int_V \mathbf{\nabla} \psi \: dV = \int \int_S \psi \mathbf{n}$

Proof Let
$\mathbf{F}= \mathbf{a} \psi$
where
$\mathbf{a}$
is a constant vector. Apply the Divergence Theorem.
$\int \int \int_V \mathbf{\nabla} \cdot (\mathbf{a} \psi ) dV = \int \int_S ( \mathbf{a} \psi) \cdot \mathbf{n} dS$

We can rewrite this as
$\int \int \int_V \mathbf{a} \cdot (\mathbf{\nabla} \psi ) dV = \int \int_S ( \mathbf{a} \cdot (\psi \mathbf{n}) dS$

Since
$\mathbf{a}$
is a constant vector we can take it out as a factor, obtaining
$\mathbf{a} \cdot \int \int \int_V (\mathbf{\nabla} \psi ) dV = \mathbf{a} \cdot \int \int_S (\psi \mathbf{n}) dS$

Hence
$\int \int \int_V (\mathbf{\nabla} \psi ) dV = \int \int_S (\psi \mathbf{n}) dS$