The rate of change of flux from a vector field
\[\mathbf{F}\]
through a surface \[S\]
both time dependent i\[\frac{d \Phi}{dt} = \int \int_{S_t}( \frac{\partial \mathbf{F}}{\partial t} + \mathbf{v} \mathbf{\nabla} \cdot \mathbf{F}) \cdot d \mathbf{n} dS + \oint_{C_1} (\mathbf{F} \times \mathbf{v}) \cdot d \mathbf{r} \]
Proof
At any time
\[t_1\]
the vector field is \[\mathbf{F} (\mathbf{r}, t_1)\]
. Let the surface at \[t=t_1\]
be \[S_{t_1}\]
and let the boundary of \[S_{t_1}\]
be \[C_{t_1}\]
.The flux out of
\[S_{t_1}\]
is \[\Phi = \int_{S_{t_1}} \mathbf{F} (\mathbf{r}, t_1) \cdot \mathbf{n}_{t_1} dS_{t_1}\]
.\[\frac{d \Phi}{dt} = lim_{t_2 \rightarrow t_1} \frac{\Phi_{t_2} - \Phi_{t_1}}{t_2 -t_1} =lim_{t_2 \rightarrow t_1} \frac{\int \int_{S_{t_2}} \mathbf{F} (\mathbf{r}, t_2) \cdot \mathbf{n}_{t_2} dS_{t_2} - \int \int_{S_{t_1}} \mathbf{F} (\mathbf{r}, t_1) \cdot \mathbf{n}_{t_1} dS_{t_1}}{t_2 -t_1}\]
Apply the Divergence Theorem to the volume traced out by
\[S\]
between \[t_2\]
and \[t_1\]
gives\[ \begin{equation} \begin{aligned} \int \int \int_{V_{t_1}} \mathbf{\nabla} \cdot \mathbf{F}(\mathbf{r}, t_{t_1}) dV_{t_1} &= lim_{t_2 \rightarrow t_1} ( \int \int_{S_{t_2}} \mathbf{F} (\mathbf{r},t_2) \cdot \mathbf{n}_{t_2} dS_{t_2} \\ &- \int \int_{S_{t_1}} \mathbf{F} (\mathbf{r},t_1) \cdot \mathbf{n}_{t_1} dS_{t_1} + \int \int_{Sides} \mathbf{F} (\mathbf{r},t_2) \cdot \mathbf{n}_{t_2} dS_{t_2}) \end{aligned} \end{equation} \]
\[\mathbf{F}(\mathbf{r}, t_2) \simeq \mathbf{F}(\mathbf{r}, t_1) + \frac{\partial \mathbf{F}}{\partial t} dt \]
so
where
\[dt=t_2 -t_1\]
.\[\begin{equation} \begin{aligned} \int \int_{S_{t_2}} \mathbf{F} (\mathbf{r},t_2) \cdot \mathbf{n}_{t_2} dS_{t_2} - \int \int_{S_{t_1}} \mathbf{F} (\mathbf{r},t_1) \cdot \mathbf{n}_{t_1} dS_{t_1} &\simeq= \int \int_{S_{t_2}} \mathbf{F}(\mathbf{r}, t_1) dS_{t_2} \\ &+ \int \int_{S_{t_2}} \frac{\partial \mathbf{F}}{\partial t} dt dS_{t_2} - \int \int_{S_{t_1}} \mathbf{F} (\mathbf{r},t_1) \cdot \mathbf{n}_{t_1} dS_{t_1} \\ &+ \int \int \int_{V_{t_1}} \mathbf{\nabla} \cdot \mathbf{F}(\mathbf{r}, t_{t_1}) dV_{t_1} - \int \int_{Sides} \mathbf{F} (\mathbf{r},t_2) \cdot \mathbf{n}_{t_2} dS_{t_2} \end{aligned} \end{equation}\]
\[dV \simeq v dS dt\]
Hence
\[\begin{equation} \begin{aligned} \int \int_{S_{t_2}} \mathbf{F} (\mathbf{r},t_2) \cdot \mathbf{n}_{t_2} dS_{t_2} - \int \int_{S_{t_1}} \mathbf{F} (\mathbf{r},t_1) \cdot \mathbf{n}_{t_1} dS_{t_1} &\simeq= \int \int_{S_{t_2}} \mathbf{F}(\mathbf{r}, t_1) dS_{t_2} \\ &+ \int \int_{S_{t_2}} \frac{\partial \mathbf{F}}{\partial t} dt dS_{t_2} - \int \int_{S_{t_1}} \mathbf{F} (\mathbf{r},t_1) \cdot \mathbf{n}_{t_1} dS_{t_1} \\ &+ \int \int \int_{V_{t_1}} \mathbf{\nabla} \cdot \mathbf{F}(\mathbf{r}, t_{t_1}) v dS_{t_1} dt - \int \int_{Sides} \mathbf{F} (\mathbf{r},t_2) \cdot \mathbf{n}_{t_2} dS_{t_2} \end{aligned} \end{equation}\]
Now divide by
\[dt=t_2 -t_1\]
to get\[\frac{d \Phi}{dt} = \int \int_{S_t}( \frac{\partial \mathbf{F}}{\partial t} + \mathbf{v} \mathbf{\nabla} \cdot \mathbf{F}) \cdot d \mathbf{n} dS + \oint_{C_1} (\mathbf{F} \times \mathbf{v}) \cdot d \mathbf{r} \]