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April 18th, 2015, 08:46 AM   #1
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Volume induced by a vector field across a moving surface

We are given a vector field $\mathbf{F} \colon \mathbb{R}^3 \to \mathbb{R}^3$ and a parametric surface $\mathcal{S}$ defined by $\mathbf{r} \colon A \to \mathbb{R}^3$, where $A \subset \mathbb{R}^2$. Now the flux induced by $\mathbf{F}$ through $\mathbf{r}$ is
\begin{align}
\iint_{\mathcal{S}} \mathbf{F} \cdot \hat{\mathbf{N}} \,\mathrm{d}S &= \iint_A \Big\langle \mathbf{F}(\mathbf{r}(u, v)), \frac{\partial \mathbf{r}} {\partial u} \times \frac{\partial \mathbf{r}}{\partial v} \Big\rangle \, \mathrm{d}u \, \mathrm{d}v.
\end{align}
Now suppose that the vector field changes with time, i.e., we have $\mathbf{F} \colon \mathbb{R}^4 \to \mathbb{R}^3$, and the parametric surface $\mathcal{S}$ evolves in time as well: $\mathbf{r} \colon A \times \mathbb{R} \to \mathbb{R}^3$. If $\hat{\mathbf{N}}$ is the unit normal to surface $\mathcal{S}$, it appears to me that the area element "passes" space with rate
\begin{align}
\Big\langle \hat{\mathbf{N}}, \frac{ \partial \mathbf{r} }{ \partial t} \Big\rangle,
\end{align}
and so everything boils down to
\begin{align}
\Phi(t) &= \iint_{\mathcal{S}} \langle F, \hat{\mathbf{N}} \rangle - \Big\langle \frac{\partial \mathbf{r}}{\partial t}, \hat{\mathbf{N}} \Big\rangle \, \mathrm{d}S\\
&= \iint_{\mathcal{S}} \Big\langle \mathbf{F} - \frac{\partial \mathbf{r}}{\partial t} , \hat{\mathbf{N}} \Big\rangle \, \mathrm{d}S \\
&= \iint_A \Big\langle \mathbf{F}(\mathbf{r}(u, v, t), t) - \frac{\partial \mathbf{r}}{\partial t}, \frac{\partial \mathbf{r}}{\partial u} \times \frac{\partial \mathbf{r}}{\partial v} \Big\rangle \, \mathrm{d}u \, \mathrm{d}v.
\end{align}
Now the amount of volume flowing according to the vector field $\mathbf{F}$ through the moving surface $\mathcal{S}$ during time interval $[t_a, t_b]$ is simply
\begin{align}
V = \int_{t_a}^{t_b} \Phi(t) \, \mathrm{d}t.
\end{align}

Is this correct?

Last edited by coderodde; April 18th, 2015 at 09:38 AM.
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