If q is a smooth scalar field on R \times D, then:
\frac{d}{dt} \int_{V_t} q(t,{\bf x}) \ dVol({\bf x}) = \int_{V_t} \left( \frac{\partial {\bf q}}{\partial t} + {\bf \nabla} \cdot (q{\bf u}) \right)_{(t,{\bf x})} dVol({\bf x})
Proof
Let f:D \rightarrow R be a smooth function which vanishes outside V and on \partial V. Define \chi on R \times D by \chi (t,{\bf R}_t({\bf y})) = f({\bf y})
By definition \frac{D\chi}{Dt} = 0 \ \ \ \ {\text so} \ \ \ \ \frac{\partial \chi}{\partial t} + {\bf u}\cdot {\bf \nabla}\chi = 0
moreover, \chi (t, {\bf y}) = 0 for {\bf y} \ \epsilon \ V_t or {\bf y} \ \epsilon \ \partial V_t
\frac{d}{dt} \int_{V_t} q\chi \ dVol = \frac{d}{dt} \int_D q\chi \ dVol \ \ \ \ \ \ \left[ \ as \ \chi (t, {\bf y}) = 0 \ for \ {\bf y} \ \epsilon D \ \backslash V_t \right]
= \int_D \frac{\partial}{\partial t}(q\chi ) \ dVol
= \int_D \left( \frac{\partial q}{\partial t} \chi + q \frac{\partial \chi}{\partial t} \right) \ dVol
= \int_D \left( \frac{\partial q}{\partial t} \chi - q{\bf u} \cdot {\bf \nabla} \chi \right) \ dVol
Now note that q{\bf u}\cdot {\bf \nabla} \chi = {\bf \nabla} \cdot (q {\bf u} \chi) - \chi {\bf \nabla} \cdot (q{\bf u}) so:
\int_D q{\bf u}\cdot {\bf \nabla} \chi \ dVol = - \int_D \chi {\bf \nabla} \cdot (q {\bf u}) \ dVol + \int_{\partial D} \chi q{\bf u}\cdot {\bf n} \ dS
Since \chi = 0 \ on \ \partial D, \ \ \ \ \ \int_{\partial D} \chi q {\bf u}\cdot {\bf n} \ dS = 0 \ \ \ , thus.
\frac{d}{dt} \int_{V_t} q\chi \ dVol = \int_D \chi \frac{\partial q}{\partial t} + {\bf \nabla}\cdot (q{\bf u}) \ dVol = \int_{V_t} \chi \frac{\partial q}{\partial t} + {\bf \nabla}\cdot (q{\bf u}) \ dVol
Now let f tend to the characteristic function of V in the limit.
\frac{d}{dt} \int_{V_t} q \ dVol = \int_{V_t} \left( \frac{\partial q}{\partial t} + {\bf \nabla} \cdot (q{\bf u}) \right) \ dVol
As required.