Just to make it very clear, because these issues are a constant source of confusion for beginners in electromagnetic theory.
The most general equations are Maxwell's equations in differential form. The microscopic equations read (in a vacuum, i.e., considering all charges and currents explicitly and not making approximations to describe macroscopic electrodynamics in media)
\vec{\nabla} \times \vec{E} + \frac{1}{c} \partial_t \vec{B}=0, \quad \vec{\nabla} \cdot \vec{B}=0,<br />
\vec{\nabla} \times \vec{B} -\frac{1}{c} \partial_t \vec{E}=\frac{1}{c} \vec{j}, \quad \vec{\nabla} \cdot \vec{E}=\rho.<br />
Here \vec{E} and \vec{B} are the electric and magnetic components of the electromagnetic field, \rho the electric-charge density and \vec{j} the electric current density, and c the speed of light in a vacuum. The equations are written in Heaviside-Lorentz units. In SI units you have other constants \epsilon_0 and \mu_0 for unit conversion, which destroy the beauty of the equations. It&#039;s easy to switch from one system of units to the other. That&#039;s why I use the more natural Heaviside-Lorentz units.<br />
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The first equation ist Faraday&#039;s Law. You can use the integral theorem by Stokes to get an integral form. To that end let F be any surface with boundary curve \partial F. We assume that we have defined the surface-normal elements \mathrm{d}^2 \vec{F} and the tangent vectors on \partial F, \mathrm{d} \vec{r}, according to the right-hand rule. Then integrating Faraday&#039;s Law over the surface and using Stokes&#039;s theorem to the first term, yields<br />
\int_{\partial F} \mathrm{d} \vec{r} \cdot \vec{E}=-\frac{1}{c} \int_{F} \mathrm{d}^2 \vec{F} \cdot \partial_t \vec{B}.<br />
It is very important to write the Law in this way with the partial time derivative under the integral. One can show with some effort that you can take the time integral outside of the integral, but if the boundary is moving, there is an extra term to the naive one:<br />
int_{F} \mathrm{d}^2 \vec{F} \cdot \partial_t \vec{B} = \frac{\mathrm{d}}{\mathrm{d} t} \int_F \mathrm{d}^2 \vec{F} \cdot \vec{B} + \int_{\partial F} \mathrm{d} \vec{r} \cdot (\vec{v} \times \vec{B}),<br />
where \vec{v}=\vec{v}(t,\vec{r}) is the velocity of each point of the boundary curve of your surface. Thus, the correct Faraday Law in integral form reads<br />
\int_{\partial F} \mathrm{d} \vec{r} \cdot \left (\vec{E}+\frac{\vec{v}}{c} \times \vec{B} \right )=-\frac{1}{c} \frac{\mathrm{d}}{\mathrm{d} t} \Phi<br />
where<br />
\Phi=\int_{F} \mathrm{d}^2 \vec{F} \cdot \vec{B}<br />
is the magnetic flux through the surface. For a detailed proof, see the Wikipedia:<br />
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<a href="http://en.wikipedia.org/wiki/Faraday%27s_law_of_induction#Proof_of_Faraday.27s_law" target="_blank" class="link link--external" rel="nofollow ugc noopener">http://en.wikipedia.org/wiki/Faraday&#039;s_law_of_induction#Proof_of_Faraday.27s_law</a><br />
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The second equation is the statement that there are no free magnetic charges, i.e., any magnet appears always with a north and a south pole. If you divide the magnet in two pieces, you&#039;ll always get two magnets both with a north and a south pole but never a separate north or south pole. The integral form is easily obtained by applying Gauss&#039;s integral theorem<br />
\int_{\partial V} \mathrm{d}^2 \vec{F}=0.<br />
Here V is an arbitray volume and \partial_V it&#039;s closed boundary surface. It is very important to keep in mind that this statement holds true for closed surfaces only.<br />
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In the same way you can find integral forms of the other two inhomogeneous Maxwell equations. The easy one is Gauss&#039;s Law (the fourth equation in my ordering of the Maxwell equations). Using Gauss&#039;s Law, using the orientation of the boundary \partial V such that the surface-normal vectors all point out of the volume V under consideration:<br />
\int_{\partial V} \mathrm{d}^2 \vec{F} \cdot \vec{E}=\int_V \mathrm{d}^3 \vec{r} \rho=Q_V,<br />
where Q_V is the total charge contained in the volue V.<br />
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Finally for the Ampere-Maxwell Law you have to take into account the additional terms when taking the time derivative of the electric field outside of the integral. Then it reads<br />
\int_{\partial F} \mathrm{d} \vec{r} \cdot \vec{B} = \frac{1}{c} \frac{\mathrm{d}}{\mathrm{d} t} \int_F \mathrm{d}^2 \vec{F} \cdot \vec{E} + \frac{1}{c} \int_F \mathrm{d} \vec{F} \cdot (\vec{j}-\rho \vec{v}) + \frac{1}{c} \int_{\partial F} \mathrm{d} \vec{r} \cdot (\vec{v} \times \vec{E}).<br />
Again \vec{v}=\vec{v}(t,\vec{r}) is the velocity of each point along the boundary \partial F of the surface, and the relative orientation of the boundary and the surface-normal elements is according to the right-hand rule.
