# Confusion with representing the negative rate of change of magnetic flux in Faraday's law

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?

• $t\neq t(s)$, so you are good to go. – Aaron Stevens Jul 25 '19 at 11:36
• 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. – Puk Jul 25 '19 at 18:29

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?