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

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

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
\[\begin{equation} \begin{aligned} \int \int_S \phi \mathbf{F} \cdot \mathbf{n} dS &= \int \int \int_V \phi \mathbf{\nabla} \cdot \mathbf{F} + \mathbf{F} \cdot (\mathbf{\nabla} \phi) dV \\ &= \int \int \int_V \phi \mathbf{\nabla} \cdot \mathbf{F} dV+ \int \int \int_V \mathbf{F} \cdot (\mathbf{\nabla} \phi) dV \end{aligned} \end{equation}\]