This question comes from electrodynamics, but I'm stuck at the math aspect.
Consider, in any volume $V \in \mathbb{R}^3$, two fields $\rho\left(\vec{r},t\right)$ and $\vec{J}\left(\vec{r},t\right)$ related by the continuity equation $$-\frac{\partial \rho\left(\vec{r},t\right)} {\partial t} = \nabla \cdot \vec{J}\left(\vec{r},t\right)$$ The variable $\vec{r}$ represents position, and $t$ is, of course, time. Now consider two time instants $t_1$ and $t_2$ such that for all $\vec{r} \in V$, $$\rho\left(\vec{r},t_1\right) = \rho\left(\vec{r},t_2\right),\qquad \vec{J}\left(\vec{r},t_1\right) =\vec{J}\left(\vec{r},t_2\right)$$ If we define two volume integrals like this: $$ \vec{Y}\left(t\right)= \iiint_V \vec{J}\left(\vec{r},t\right) \space dV\left(\vec{r}\right) \qquad \vec{Z}\left(t\right)= \iiint_V \frac {\partial \vec{J}\left(\vec{r},t\right)} {\partial t} \space dV\left(\vec{r}\right) $$ Can the following be proved? $$ \int_{t_1}^{t_2} \vec{Y}\left(t\right) \cdot \vec{Z}\left(t\right) \space dt = 0 $$ Thanks...
As long as your volume $V$ doesn't depends on time, you can exchange the order of differentiation and integration (assuming all functions involved are regular enough). $$ \frac{d}{dt}\vec{Y}(t) = \frac{d}{dt} \iiint_{V} \vec{J}(\vec{r},t)dV(\vec{r}) = \iiint_{V} \frac{\partial \vec{J}}{\partial t}(\vec{r},t)dV(\vec{r}) = \vec{Z}(t) $$ The integral you want can now be computed as follows:
$$\int_{t1}^{t2} \vec{Y}(t)\cdot\vec{Z}(t) dt = \int_{t1}^{t2} \vec{Y}(t)\cdot\frac{d\vec{Y}}{dt}(t) dt = \frac12 \int_{t1}^{t2} \frac{d}{dt}|\vec{Y}(t)|^2 dt = \frac12 \left[|\vec{Y}(t)|^2\right]_{t1}^{t2}$$
Since you assume $\vec{J}(t_1,\vec{r}) = \vec{J}(t_2,\vec{r})$ for all $\vec{r}$, you get $\vec{Y}(t_1) = \vec{Y}(t_2)$ and hence: $$\int_{t1}^{t2} \vec{Y}(t)\cdot\vec{Z}(t) dt = \frac12 \left( |\vec{Y}(t_2)|^2 - |\vec{Y}(t_1)|^2 \right) = 0$$