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Consider a vector field $\mathbf B:\mathbb R\times \mathbb R^3\to\mathbb R^3$. Let there exists some $\mathbf A:\mathbb R\times \mathbb R^3\to\mathbb R^3$ for which $\mathbf B = \nabla \times\mathbf A$ where the curl is taken over the "$\mathbb R^3$" argument.

Now for each $t\in\mathbb R$, let a surface element $\Sigma_t$ with the topology of a disk be given. Let $\gamma:\mathbb R\times[0,1]\to\mathbb R^3$ parameterize $\partial \Sigma_t$. In particular, for each $t\in\mathbb R$, let $\gamma(t, \cdot)$ traverse $\partial\Sigma_t$ exactly once. Then, using the convention that repeated indices are summed from $1$ to $3$ we have \begin{align} \frac{d}{dt}\int_{\Sigma_t}\mathbf B\cdot d\mathbf a &= \frac{d}{dt}\int_{\partial\Sigma_t} \mathbf A\cdot d\boldsymbol \ell \\ &= \frac{d}{dt}\int_0^1 A_i \frac{\partial\gamma^i}{\partial \lambda}d\lambda \\ &= \int_0^1 \frac{\partial A_i}{\partial t}\frac{\partial\gamma^i}{\partial \lambda} d\lambda +\int_0^1\left(\frac{\partial\gamma^j}{\partial t}\frac{\partial A_i}{\partial x^j}\frac{\partial\gamma^i}{\partial\lambda} +A_i\frac{\partial^2\gamma^i}{\partial t\partial\lambda}\right)d\lambda\\ &= \int_{\Sigma_t} \frac{\partial\mathbf B}{\partial t}\cdot d\mathbf a+\int_0^1\left(\frac{\partial\gamma^j}{\partial t}\frac{\partial A_i}{\partial x^j}\frac{\partial\gamma^i}{\partial\lambda} +A_i\frac{\partial^2\gamma^i}{\partial t\partial\lambda}\right)d\lambda\\ \end{align} On the other hand, for a divergence-free vector field $\mathbf B$ (which we have here since $\mathbf B$ is the curl of $\mathbf A$), the Leibniz integral rule gives \begin{align} \frac{d}{dt}\int_{\Sigma_t}\mathbf B\cdot d\mathbf a &= \int_{\Sigma_t} \frac{\partial\mathbf B}{\partial t}\cdot d\mathbf a - \int_{\partial\Sigma_t} \mathbf v\times \mathbf B\cdot d\boldsymbol \ell \end{align} Question. Presumably, the formula I derived above should agree with this expression, but I can't see how it does.

As far as I can tell, the second term on the right of the Leibniz integral formula can be written as follows. Firstly, as far as I can tell the notation can be translated as $$ \mathbf v = \frac{\partial\gamma}{\partial t}, \qquad d\boldsymbol \ell = \frac{\partial\gamma}{\partial\lambda}d\lambda $$ Secondly, we note the following identity $$ \mathbf v\times\mathbf B = (v^j\partial_i A_j - v^j\partial_jA_i)\mathbf e^i $$ Putting these facts together gives $$ -\int_{\partial\Sigma_t} \mathbf v\times\mathbf B\cdot d\boldsymbol\ell = \int_0^1\left(\frac{\partial\gamma^j}{\partial t}\frac{\partial A_i}{\partial x^j}\frac{\partial\gamma^i}{\partial\lambda} -\frac{\partial\gamma^j}{\partial t}\frac{\partial A_j}{\partial x^i}\frac{\partial\gamma^i}{\partial\lambda}\right)d\lambda $$ The second term in the integrand does not seem to match the second term in the integrand at the end of the first computation above.

Have I interpreted the integral formula incorrectly? Perhaps my first computation has an error in it? Thanks for the help.

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As your curve is closed, integration by parts gives $$ \int\limits_{0}^{1} A_{i} \frac{\partial^{2}\gamma^{i}}{\partial t \partial \lambda} d\lambda = \underbrace{A_{i} \frac{\partial\gamma^{i}}{\partial t} \Big\vert_{0}^{1}}_{=0}-\int\limits_{0}^{1} \frac{A_{i}}{\partial x^{j}}\frac{\partial\gamma^{j}}{\partial \lambda } \frac{\partial\gamma^{i}}{\partial t }d\lambda.$$ Combining this with the stuff you already wrote down and flipping the indices $i$ and $j$ you are fine.

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    $\begingroup$ Ahhhhhhhhhhhhhhhhh. How did I not see this!? Leave it to integration by parts to ruin your night. Thank you very very much Dominik. $\endgroup$ – joshphysics Jul 27 '13 at 8:32

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