# Should the pressure integral of the momentum equation be in terms of surface?

I believe I found a great book on the mathematics of Computational Fluid Dynamics (CFD), OpenFOAM, etc. I just have a few questions.

The x-coordinate form of the basic momentum equation in this book is

$$\frac{\partial \rho u_{x}}{\partial t} = -\bigg(\frac{\partial [\rho u_{x}u_{x}]}{\partial x}+\frac{\partial [\rho u_{y}u_{x}]}{\partial y}+\frac{\partial [\rho u_{z}u_{x}]}{\partial z} \bigg) - \bigg(\frac{\partial \tau_{xx}}{\partial x} + \frac{\partial \tau_{yx}}{\partial y} + \frac{\partial \tau_{zx}}{\partial z} \bigg) - \frac{\partial P}{\partial x} + \rho g_{x}$$

The same equation can be written for the y and z momentum equations as well. $\rho g_{x}$ is written without a partial derivative because it was multiplied by $\Delta x \Delta y \Delta z$ and divided by the same during derivation. As such, it does not have a partial derivative.

The author then gives the vector form

$$\frac{\partial \rho \vec {U}}{\partial t} = -\nabla \bullet(\rho \vec{U} \otimes \vec {U}) - \nabla \bullet \tau - \nabla P + \rho \vec {g}$$

The integral form is then given as

$$\frac{\partial}{\partial t} \int \rho \vec {U} dV = - \oint (\rho \vec{U} \otimes \vec {U}) \cdot\vec {n} dS-\oint \tau \cdot \vec {n} dS -\int PdV + \int \rho \vec {g} dV$$

$$\frac{(P \vert_{x} - P \vert_{x + \Delta x})}{\Delta x \Delta y \Delta z} \Delta y \Delta z$$
$$\oint P \vec {n} dS$$
• You are correct: $\int_V \nabla p \,dV = \int_S p \mathbb{n} \, dS$. For a derivation see here