If
\[\phi , \: \psi\]
are twice differentiable functions with domain \[V\]
bounded by a surface \[S\]
then\[\int \int \int_V (a \phi + b \psi ) dV= a \int \int_S \mathbf{\nabla} \phi \cdot \mathbf{n} dS + b \int \int_S \mathbf{\nabla} \psi \cdot \mathbf{n} dS\]
Where
\[\mathbf{n}\]
is the outward normal to \[S\]
.Proof
The Divergence Theorem states
\[\int \int_S \mathbf{F} \cdot \mathbf{n} dS = \int \int \int_V \mathbf{\nabla} \mathbf{F} dV\]
Let
\[\mathbf{F} = a \mathbf{\nabla} \phi + b \mathbf{\nabla} \psi\]
then\[\begin{equation} \begin{aligned} \int \int_S ( a \mathbf{\nabla} \phi + b \mathbf{\nabla} \psi) \cdot \mathbf{n} dS &= \int \int \int_V \mathbf{\nabla} ( a \mathbf{\nabla} \phi + b \mathbf{\nabla} \psi) dV \\ &= \int \int \int_V (a \nabla^2 \phi + b \nabla^2 \psi) dV \\ &= \int \int_S (a \mathbf{\nabla} \phi + b \mathbf{\nabla} \psi m) \cdot \mathbf{n} dS \\ &= a \int \int_S \mathbf{\nabla} \phi \cdot \mathbf{n} dS + b \int \int_S \mathbf{\nabla} \psi \cdot \mathbf{n} dS \end{aligned} \end{equation}\]