Explain the Stokes -theorem from differential from into Integral form

Question

I want to understand the Stokes -theorems deeper. I am trying to understand the operation from

$$\nabla \times \mathbf{E} = -\frac{\partial \mathbf{B}} {\partial t}$$

to

$$\oint_{\partial \Sigma} \mathbf{E} \cdot d\boldsymbol{\ell} = - \int_{\Sigma} \frac{\partial \mathbf{B}}{\partial t} \cdot d\mathbf{A}$$

relating to Maxwel-Faraday's law here.

You have there surface-integral, closed line-integral, curl and the rate of change. Explain the transition from the differential from into the integral form.

Bonus points and Puzzles

1. Notation? I am unsure whether $\int$ is just physical convention, instead of $\int\int$ for the surface integral like the Kelvin-Stokes -theorem $\oint_{\Gamma} \mathbf{F}\, d\Gamma = \iint_{\mathbb{S}} \nabla\times\mathbf{F}\, d\mathbb{S}$, the two forms of notation confuses me greatly -- mathematical notation and physical notation?

Answer 1

If I undertstand your query correctly, all you really need to complete the thought are Stoke's and Gauss' Theorems: $$ \iint_{\partial E} \vec{F} \cdot d\vec{S} = \iiint_E \nabla \cdot \vec{F} dV $$ $$ \int_{\partial M} \vec{F} \cdot d\vec{r} = \iint_{M} \nabla \times \vec{F} \cdot d\vec{S}$$ Here we suppose that $E$ is a simple solid region with boundary $\partial E$ and $M$ is a simply connected surface with boundary $\partial M$. The boundaries must be consistent with the interiors of the integration regions. This means $\partial E$ has outward pointing normal whereas a trip around $\partial M$ finds the interior of $M$ always on the left of your journey.

Begin with $\nabla \times \vec{E} = -\frac{\partial \vec{B}}{\partial t}$ and integrate over some surface $M$ at time $t$. Apply Stokes' theorem to convert the flux integral of the curl to the line integral of the boundary: $$ \iint_{M} -\frac{\partial \vec{B}}{\partial t} \cdot d\vec{S}=\iint_{M} \nabla \times \vec{E} \cdot d\vec{S}=\int_{\partial M} \vec{E} \cdot d\vec{r} = \mathcal{E}_{induced}$$ Moreover, the term on the l.h.s. of the above can be expressed as $\frac{\partial}{\partial t} \iint_{M} \vec{B} \cdot d\vec{S} = \partial_t \Phi_B$. We thus derive the integral form of Faraday's Law; the voltage induced around a closed loop is proportional to the change in the magnetic flux through the loop.

Answer 2

You need to use the theories below. The notation with integrals between the two cases in KS -threorem look different: $\int\int_S$ and $\int_\Sigma$. I think they are still meaning the same thing. The $\partial\Sigma$ means the boundary of the manifold $\Sigma$.

I am unable to get inside this theorem so I am unable to explain how to get from the differential from into the integral form, anyway below some working, perhaps someone could continue here.

Trials

I. Suppose I integrate $\nabla \times \mathbf{E} = -\frac{\partial \mathbf{B}} {\partial t}$ with respect to the area so

$$\int_\Sigma\left(\nabla\times \bf E \right)\cdot d\bf A=-\int_\Sigma \frac{\partial \mathbf{B}} {\partial t}\cdot d\bf{A}:=RHS.$$

II. Suppose I integrate it with respect to the line so

$$\oint_{\partial \Sigma} \left(\nabla\times \mathbf{E}\right) \cdot d\boldsymbol{\ell} = \oint_{\partial \Sigma} \left(-\frac{\partial \mathbf{B}} {\partial t}\right) \cdot d\boldsymbol{\ell}$$

err no LHS emerging.

III. Some other way? I am clearly not getting sides equal like that, at least easily.

Theories

1. Stokes theorem $\int_{\partial \Omega} w =\int_\Omega dw$ (sthing about boundaries and manifolds here).

2. Kelvin-Stokes -theorem $$\oint_{\Gamma} \mathbf{F}\, d\Gamma = \iint_{\mathbb{S}} \nabla\times\mathbf{F}\, d\mathbb{S}$$ (source) or $$\int_{\Sigma} \nabla \times \mathbf{F} \cdot d\mathbf{\Sigma} = \oint_{\partial\Sigma} \mathbf{F} \cdot d \mathbf{r}$$ (source).

Perhaps related here.