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