How is obtained the formula $\int_\Omega \Psi \text{div} F dx =\int_{\partial \Omega }\Psi F \cdot \nu dS - \int_\Omega \nabla \Psi \cdot Fdx $

Question

How to show that this inequality holds?

$\int_\Omega \Psi \text{div} F dx =\int_{\partial \Omega }\Psi F \cdot \nu dS - \int_\Omega \nabla \Psi \cdot Fdx $

Where $\psi$ is a scalar function and $F$ is a vectorial function

Answer 1

Apply the definition of divergence of the product:

$\text{div}(\Psi \mathbf{F})= \Psi\text{div}\mathbf{F}+\nabla\Psi\cdot \mathbf{F}$

$\int_\Omega \text{div}(\Psi \mathbf{F})= \int_\Omega\Psi\text{div}\mathbf{F}+\int_\Omega \nabla\Psi\cdot \mathbf{F}$

$$\int_\Omega \text{div}(\Psi \mathbf{F})= \int_{\partial \Omega}\Psi \mathbf{F}\cdot v $$

Replacing: We obtain the result.

$\int_{\partial \Omega}\Psi \mathbf{F}\cdot v = \int_\Omega\Psi\text{div}\mathbf{F}+\int_\Omega \nabla\Psi\cdot \mathbf{F}$