## Proof of Integral Identity for Dot Product of Normal With Product of a Scalar With a Vector Over a Surface

Theorem
If
$\phi$
is a differentiable scalar function and
$\mathbf{f}$
is a differentiable vector field then
$\int \int_S \phi \mathbf{F} \cdot \mathbf{n} dS = \int \int \int_V \mathbf{F} \cdot (\mathbf{\nabla} \phi) dV +\int \int \int_V \phi \mathbf{\nabla} \cdot \mathbf{F} dV$

Proof
Apply the Divergence Theorem to the vector field
$\phi \mathbf{F}$
to obtain
$\int \int_S \phi \mathbf{F} \cdot \mathbf{n} dS = \int \int \int_V \mathbf{\nabla} \cdot (\phi \mathbf{F}) dV$

Now use the identity
$\mathbf{\nabla} \cdot (\phi \mathbf{F}) = \phi \mathbf{\nabla} \cdot \mathbf{F} + \mathbf{F} \cdot (\mathbf{\nabla} \phi)$
to obtain
\begin{aligned} \int \int_S \phi \mathbf{F} \cdot \mathbf{n} dS &= \int \int \int_V \phi \mathbf{\nabla} \cdot \mathbf{F} + \mathbf{F} \cdot (\mathbf{\nabla} \phi) dV \\ &= \int \int \int_V \phi \mathbf{\nabla} \cdot \mathbf{F} dV+ \int \int \int_V \mathbf{F} \cdot (\mathbf{\nabla} \phi) dV \end{aligned}

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