For a calm fluid of uniform density $\rho_0$, that occupies the space $W \subset \mathbb{R}^3$, and is subject to massive forces (per unit of mass) $\overrightarrow{b}(\overrightarrow{x})$, write the integral form of the balance equation and derive from that the corresponding differential balance equation. From the last equation, for a calm fluid that is subject only to the forces of gravity, derive the law of hydrostatic pressure.
I have done the following:
The total action-at-a-distance force at the volume $\delta V$ of the element of the fluid is $$\int_{\delta V} \rho (\overrightarrow{x}, t) \overrightarrow{b}(\overrightarrow{x}, t)dV$$ and the total contact force is $$-\int_{\delta A}p(\overrightarrow{x}, t)\overrightarrow{n}(\overrightarrow{x})dA$$ where $\delta A$ is the boundary of the volume $\delta V$.
The equation of the balance of forces is $$\int_{\delta V} \rho \overrightarrow{b}dV=\int_{\delta A}p\overrightarrow{n}dA=\int_{\delta V}\nabla pdV \Rightarrow \int_{\delta V}(\rho \overrightarrow{b}-\nabla p)dV=0$$
So, this is the integral form of the balance equation, right??
The differential form of the alance equation is $$\rho \overrightarrow{b}=\nabla p$$ right??
Could you give me some hints how we can derive from this equation the law of hydrostatic pressure??