Proof of Formula for Integral of a Gradient Over a Volume
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
\[ \int \int \int_V (\mathbf{\nabla} \psi ) dV = \int \int_S (\psi \mathbf{n}) dS \]
