I am engineer and not so deep in mathematics and i need your help please.
Let $\textbf{x}=[x,y,z]^T$ denotes a position vector of a point in a domain $\Omega_t$ say for example $\Omega_t = $ a$~$deforming sponge. Since the sponge is deforming, the domain shape will shange with time and will be a function of $\textbf{x}(t)$
Let $f = f(\textbf{x}, \ d\textbf{x}/dt, \ t)\in R$ be continuous function (which could be the mass or energy) defined over the domain $\Omega_t$ .
Engineers always say in their books that it is well known that if for arbitrary $\Omega_t$
$\int_{\Omega_t} f(\textbf{x}, \ d\textbf{x}/dt, \ t) \ d v= 0$
is true. then
$f(\textbf{x}, \ d\textbf{x}/dt, \ t) = 0$
how to prove this statement?
Lets make life a little bit easier and suppress all the non-important arguments. Then, I think your question is:
Let $g : \Omega \to \mathbb{R}$ be a continuous function, such that for all $U \subset \Omega$ we have $\int_U g(v) \, \mathrm{d} v = 0$. Then, $g(v) = 0$ for all $v \in \Omega$.
Let me sketch a proof for this statement (actually, I will supress some technicalities).
Assume, that $g(v) = 0$ for all $v \in \Omega$. Then, there must be a point $v$ with $g(v) > 0$ (in case $g(v) < 0$, you can argue similarly).
Since $g$ is continuous, there is a small ball $U$ around $v$, such that $g(u) > 0$ for all $u \in U$. But then, $\int_U g(u) \, \mathrm{d}u > 0$. This is a contradiction to your assumptions.