## Divergence Theorem for a Linear Combination of Scalar Fields

Theorem
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{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}