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So Faraday's law states in differential form that $$ \nabla \times \vec{E} = -\frac{\delta H}{\delta t} $$ Using Stoke's theorem, the right hand side (the magnetic flux rate of change) is expressed as $$ - \frac{\delta}{\delta t}\iint_S H \boldsymbol{\cdot} ds = \iint_S -\frac{\delta H}{\delta t} \boldsymbol{\cdot} ds $$ For the left side, many textbooks say to assume that we can move $\frac{\delta}{\delta t}$ under the integral, but why is that? Is the time a constant in this case? And what's the point for representing it like that? Is there a conceptual reason for this?

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    $\begingroup$ $t\neq t(s)$, so you are good to go. $\endgroup$ – Aaron Stevens Jul 25 '19 at 11:36
  • $\begingroup$ Your second equation only holds in general when $S$ is not time varying. If you have a surface $S$ moving/deforming in a static field $\vec{H}$, the left hand side is not zero in general while the right hand side is clearly zero. $\endgroup$ – Puk Jul 25 '19 at 18:29
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Think of the circumstances where you cannot move the time derivative under the integral sign. For example consider Liebnitz rule: $$ \frac d{dt} \int_{a(t)}^{b(t)} f(t,x)dx= \int_{a(t)}^{b(t)} \frac{\partial}{\partial t} f(t,x)dx + \frac {da(t)}{dt}f(t,a) - \frac {db(t)}{dt}f(t,b) $$ Is there anything like the this going on in your example? For example how would your second equation change the surface $S$ was moving or changing shape?

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