I have the next problem:
If $X:R^n\rightarrow R^n$ is a $C^2$-vector field with associate flux $\phi:D\subset R\times R^n\rightarrow R^n$ and $B\subset \pi_2(D)$ a bounded open subset, defining $\phi_t(x)=\phi(t,x)$ and $v(t)=vol[\phi_t(B)]$ prove that $v'(t)=\int_{\phi_t(B)}divX$.
I tried use the Gauss divergence theorem but I can't find relation with $vol[\phi_t(B)]$.
One approach I can think of is to make use of the general formula given in this question (the proof is also given in my answer there).
Now, \begin{align} v(t) &:= \text{vol}[\phi_t(B)] \\ &= \int_{\phi_t(B)}1 \, d^nx \\ &= \int_B |J(t,x)| \, d^nx \end{align} where $J(t,x) = \det\bigg( D(\phi_t)_x\bigg)$. So far, this has been nothing but the definition of volume, and in the final equality, the change of variables formula for multivariable integrals in $\Bbb{R}^n$. Now, note that $J(t,x)$ is a smooth, real valued function of $(t,x)$, which is nowhere vanishing (because we're taking the determinant of an invertible linear transformation $D(\phi_t)_x$). If we assume that the domain of the flow, is connected, then by the intermediate value theorem (i.e image of connected sets is connected), it follows that $J(t,x)$ maintains a constant sign everywhere.
So, we can write $|J(t,x)| = \sigma \cdot J(t,x)$, where $\sigma$ is the sign of $J(t,x)$ (positive or negative), and this is indeed a constant which doesn't depend on $(t,x)$, by my argument above. So, \begin{align} v(t) &= \int_B \sigma \cdot J(t,x) \, d^nx \end{align} Now, given sufficient regularity, when differentiating with respect to $t$, we may apply the Leibniz integral rule to interchange the derivative and integral: \begin{align} v'(t) &= \int_B \sigma \cdot\dfrac{\partial J}{\partial t}\bigg|_{(t,x)} \, d^nx \\ &= \int_B \sigma \cdot (\text{div }X)(\phi_t(x)) \cdot J(t,x) \, d^nx \tag{see the link}\\ &= \int_B (\text{div }X)(\phi_t(x)) \cdot |J(t,x)| \, d^nx \\ &= \int_{\phi_t(B)} (\text{div} X) \end{align} where in the last line, I used the change of variables formula in reverse.
