If E(t,x,y,z) and B(t,x,y,z)represent the electric and magnetic fields at point (x,y,z) at time t, a basic principle of electromagnetic theory says th

Given information:
The functions E(t,x,y,z) nad B(t,x,y,z) represent the electric and magnetic fields at point (x,y,z) at time t.
The Faraday's law is ${\displaystyle {\oint }_{C}E\cdot dr=-\frac{\partial }{\partial t}\int {\int }_{S}B\cdot nd\sigma }$.
Here, C represents a wire loop through which current flows counter clockwise with respect to the surface's unit normal n, giving rise to the voltage ${\displaystyle {\oint }_{C}E\cdot dr}$
The basic principle of the electromagnetic theory is ${\displaystyle \mathrm{\nabla }\times E=-\frac{\partial B}{\partial t}}$.
Stokes' Theorem:
Let S be a piecewise cmooth oriented surface having a piece smooth boundary curve C. Let $F=Mi+Nj+Pk$ be a vector field whose components have continuous first partial derivatives on an open region containing S. Then the circulation of F around C in the direction counter clockwise with respect to the surface's unit normal vector n equals the integral of the curl vector field ${\displaystyle \mathrm{\nabla }\times F}$ over S:
${\displaystyle {\oint }_{C}F\cdot dr=\int {\int }_S(\mathrm{\nabla }\times F)\cdot nd\sigma }$
Calculation:
Assume that S is a surface in which the Stoke's theorem holds. Then ${\displaystyle {\oint }_{C}E\cdot dr=\int {\int }_S(\mathrm{\nabla }\times E)\cdot nd\sigma }$ ...(1)
By the basic principle of the electromagnetic theory, ${\displaystyle \mathrm{\nabla }\times E=-\frac{\partial B}{\partial t}}$.
Then equation(1) becomes,
${\displaystyle {\oint }_{C}E\cdot dr=\int {\int }_S(\mathrm{\nabla }\times E)\cdot nd\sigma }$
${\displaystyle =\int {\int }_S-\frac{\partial B}{\partial t}\cdot nd\sigma }$
${\displaystyle =-\frac{\partial }{\partial t}\int {\int }_{S}B\cdot nd\sigma }$
Then the voltage around a loop equals the negative of the rate of change of magnetic flux through the loop.
Therefore, the Faraday's law ${\displaystyle {\oint }_{C}E\cdot dr=-\frac{\partial }{\partial t}\int {\int }_{S}B\cdot nd\sigma }$ is derived by using Stoke's theorem.