Euler equation of fluid dynamics

Question

I'm trying to obtain Euler equation for a perfect fluid in laminar or stationary flow. A particle fluid is submitted at volume forces and surface force. The fist, in my case, is giving only by gravity and the second by pressure. By Newton's second law I obtain:

$$\vec{F}_V + \vec{F}_s = m\frac{d\vec{v}}{dt}.$$

An element of volume force is given by $$d\vec{F}_V = dm\vec{g}=\rho d\omega\vec{g}$$ and an element of surface force is given by $$d\vec{F}_S = -pd\vec{S}.$$

Integrating I obtain

$$ \int_V \rho \,d\omega\vec{g} - \int_S p\,d\vec{S} = \frac{d\vec{v}}{dt}\int_V \rho\, d\omega$$.

Now Euler equation is written in local form as $$\rho\vec{g} - \nabla p = \rho \frac{d\vec{v}}{dt}.$$

My question is this: where the gradient of $p$ comes from? I must have the following identity $$-\int_S pd\vec{S} = -\int_V \nabla p\,d\omega.$$

Why the transformation from a surface integral to a volume integral is given by the gradient and not by the divergence? I'm doing something wrong in the previous calculations?

Answer 1

This confusion is caused by vector calculus. You should treat each component separately, and then it is obvious. For example, for the x component:

$$ \int_V \partial_x p dx dy dz = \int_{\partial V} p(x) dy dz = \int_{\partial V} p dS_x $$

by the fundamental theorem of calculus (do the x integral first). Likewise for the other components. You can make up a proof for this from the divergence theorem by introducing the fictitious vector field

$$ Q = (p,0,0) $$

And then the divergence of Q is the left hand side, while the right hand side is $Q\cdot dS$. But it's really just the fundamental theorem of calculus.

Answer 2

Because the Gauss-Ostrogradski theorem says that $$\iiint_{V}\nabla\cdot\mathbf{C}dv=\iint_{\partial V}\mathbf{C}\cdot\mathbf{n}da$$ Where $\mathbf{C}$ is a vector field. Here you don't have a vector field inside the integral. So, why do you expect that the G-O theorem is applied in this case?? By the way, the last equality that you wrote is correct. I don't know how to prove it, but I'm pretty sure that I saw that kind of theorem in Jackson's book of Electrodynamics.