explaining the rate of change of volume with divergence Now I have to solve this problem below:

Show that for every $t$,
  $$\dfrac{d}{dt}\text{Vol}(D_t)=\iiint_{D_t}\nabla\cdot\mathbf FdV$$

Here, for a region $D\subset\mathbb R^3$ and a vector field $\mathbf F$, $D_t$ is the image of a mapping $\Phi_t:D\rightarrow D_t\subset\mathbb R^3$, while $\Phi_t$ satisfies the condition:
$$\dfrac{d}{dt}\Phi_t(X)=\mathbf F(\Phi_t(X)),\;\Phi_0(X)=X$$
$\text{Vol}(D_t)$ means the volume of $D_t$.
I think this should be related to the divergence theorem $\displaystyle\iint_{\partial D_t}\mathbf F\cdot\mathbf ndS=\iiint_{D_t}\nabla\cdot\mathbf FdV$, but the thing is that I did not learn the divergence theorem in 3 dimensions since I didn't learn about surface integrals. I also know for the fact that if a square matrix $A(t)$ is the identity matrix when $t=0$, then $\left.\dfrac{d}{dt}\right|_{t=0}\det A(t)=\text{trace}\left.\dfrac{d}{dt}\right|_{t=0}A(t)$. Also the definition of divergence I learned is if $\mathbf F=(f_1,f_2,f_3),\nabla\cdot\mathbf F=\dfrac{\partial f_1}{\partial x}+\dfrac{\partial f_2}{\partial y}+\dfrac{\partial f_3}{\partial z}$.
Any help would be appreciated.
 A: By Reynolds transport theorem you have that 
$\frac{d}{dt}\left( \int_{\Omega(t)} \mathbf{f}\,dV\right)
= \int_{\Omega(t)} \left( \frac{\partial \mathbf{f}}{\partial t} + \boldsymbol{\nabla} \mathbf{f}\cdot\mathbf{v} + \mathbf{f}\,\boldsymbol{\nabla} \cdot \mathbf{v}\right)\,dV$, so using $f=1, v=\dfrac{d}{dt}\Phi_t(X)=\mathbf F(\Phi_t(X)),\Omega(t)=D_t$ you get what you want.
$$\dfrac{d}{dt}\text{Vol}(D_t)=\frac{d}{dt}\left( \iiint_{D_T}dV\right)
= \iiint_{D_T} \left( 0 + 0 +\boldsymbol{\nabla} \cdot \mathbf{\dfrac{d}{dt}\Phi_t(X)}\right)\,dV=\iiint_{D_t}\nabla\cdot\mathbf FdV$$

By Jacobi's formula we have  $\frac{d}{dt} \det A =\mathrm{tr} ( \mathrm{adj}\ A \; \frac{dA}{dt})$ or in  the equivalent form $\frac{d}{dt} \det A = \det A \; \mathrm{tr} \left(A^{-1} \frac{dA}{dt}\right)$ and for $\mathbf F$ we have that + $\frac{\mathbf dF}{dt}=F\nabla\varphi$ so putting everything together
$$\frac{d}{dt} \det \mathbf F =\mathrm{tr}( \mathrm{adj}\ \mathbf F \; \frac{d\mathbf F}{dt})=\det \mathbf F \; \mathrm{tr}(\mathbf F^{-1} \frac{d\mathbf F}{dt})=\det \mathbf F \; \mathrm{tr}(\nabla\varphi)=\det \mathbf F \; \nabla\varphi$$
and we recover $\frac{d}{dt} \det \mathbf F =\det \mathbf F \; \nabla\varphi$ that is how it is used in the first link.
Hope it helps
