In a finite volume $V \in \mathbb{R}^3$, there is a field $\vec{J}\left(\vec{r},t\right)$. On the surface $\partial V$, $\vec{J}\left(\vec{r},t\right) = 0$.
The variable $\vec{r}$ represents position, and $t$ is, of course, time. Now say there are two time instants $t_1$ and $t_2$ such that for all $\vec{r} \in V$, $$\vec{J}\left(\vec{r},t_1\right) =\vec{J}\left(\vec{r},t_2\right)$$ Consider the following integral: $$ \int_{t_1}^{t_2} \iiint_V \iiint_V \left[ \vec{J}\left(\vec{r}_1,t\right) \cdot \frac {\partial \vec{J}\left(\vec{r}_2,t\right)} {\partial t} \right] \space dV\left(\vec{r}_1\right) \space dV\left(\vec{r}_2\right) \space dt $$ Does this integral always evaluate to zero? Any proof?
Context: comes from classical electrodynamics, $\vec{J}$ is current density.
Just so this question doesn't stay 'unanswered', here's consolidating the answer(s) given by @achillehui (all credit for the answer go to him)
First, define: $$\vec{Y}\left(t\right)= \iiint_V \vec{J}\left(\vec{r},t\right) \space dV\left(\vec{r}\right) $$ So $\vec{J}\left(\vec{r},t_1\right) = \vec{J}\left(\vec{r},t_2\right)$ for all $\vec{r}\in V$ implies $\vec{Y}\left(t_1\right) = \vec{Y}\left(t_2\right)$ also.
By Fubini's theorem, the integral in the question becomes: $$ \int_{t_1}^{t_2} \left[ \left( \iiint_V \vec{J}\left(\vec{r},t\right) \space dV\left(\vec{r}\right) \right) \cdot \left( \iiint_V \frac {\partial \vec{J}\left(\vec{r},t\right)} {\partial t} \space dV\left(\vec{r}\right) \right) \right] \space dt $$ Which becomes: $$ \int_{t_1}^{t_2} \left[ \left( \iiint_V \vec{J}\left(\vec{r},t\right) \space dV\left(\vec{r}\right) \right) \cdot \left( \frac {d} {dt} \iiint_V \vec{J}\left(\vec{r},t\right) \space dV\left(\vec{r}\right) \right) \right] \space dt $$ Which is: $$ \int_{t_1}^{t_2} \left[ \vec{Y}\left(t\right) \cdot \left( \frac {d} {dt} \vec{Y}\left(t\right) \right) \right] \space dt $$ Which is: $$ \frac{1}{2} \left[|\vec{Y}(t)|^2\right]_{t1}^{t2} $$ Which is zero because $\vec{Y}\left(t_1\right) = \vec{Y}\left(t_2\right)$