Faraday's Law of Electromagnetic Induction states that
\[\oint_C \mathbf{E} d \mathbf{r} =- \frac{1}{c} \frac{\partial}{\partial t} \int \int \mathbf{H} \cdot \mathbf{n} dS\]
From Stoke's Theorem
\[\oint_C \mathbf{F} \cdot d \mathbf{r} = \int \int_S (\mathbf{\nabla} \times \mathbf{F}) \cdot \mathbf{n} dS\]
for a vector field \[\mathbf{F}\]
with continuous derivatives.If
\[\mathbf{H}\]
varies smoothly with time then we can write \[\frac{1}{c} \frac{\partial}{\partial t} \int \int \mathbf{H} \cdot \mathbf{n} dS = \frac{1}{c} \int \int \frac{\partial \mathbf{H}}{\partial t} \cdot \mathbf{n} dS\]
Hence
\[ \int \int_S (\mathbf{\nabla} \times \mathbf{E}) \cdot \mathbf{n} dS= \int \int_S - \frac{1}{c} \frac{\partial \mathbf{H}}{\partial t} \cdot \mathbf{n} dS \]
Equating the integrands gives
\[\mathbf{\nabla} \times \mathbf{E} \cdot \mathbf{n}= - \frac{1}{c} \frac{\partial \mathbf{H}}{\partial t} \cdot \mathbf{n} \]