[math]\int \int \int_V \nabla \times F \cdot dV = \int \int_S F \cdot dS[/math]
Stokes theorem states
[math]\int \int_{S} \nabla \times F \cdot dS = \oint_{\partial S} F \cdot dr[/math]
Greens theorem is a special case of Stokes theorem.
force is
[math]F = \frac{\partial U(r)}{\partial r}[/math]
substitution gives
[math]\int \int_{S} \nabla \times F \cdot dS = \oint_{\partial S} \frac{\partial U(r)}{\partial r} \cdot dr[/math]
As you can see it has dimensions of energy.
Consider the magnetic field now
[math]B = \frac{\hbar}{emc^2} \frac{1}{r}\frac{\partial U(r)}{\partial r}[/math]
Plugging in only part of the magnetic field expression [math]\frac{\hbar}{emc^2} \frac{1}{r}[/math]
[math]\int \int_{S} \frac{\hbar}{emc^2} \frac{1}{r} \cdot [\nabla \times F \cdot dS] = \oint_{\partial S} \frac{\hbar}{emc^2} \frac{1}{r} \frac{\partial U(r)}{\partial r} \cdot dr[/math]
working from the last equation, performing the integral over the volume this time will yield
[math]\int \int \int_{V} \frac{\hbar}{emc^2} \frac{1}{r} \cdot [\nabla \times F \cdot dV] = \oint_{\partial S} \frac{\hbar}{emc^2} \frac{1}{r} \frac{\partial U(r)}{\partial r} \cdot dS[/math]
Since this is just essentially a magnetic field times a surface, we have a magnetic flux
[math]\Phi = \int \int_S B \cdot dS = \int \int \int_{V} \frac{\hbar}{emc^2} \frac{1}{r} \cdot [\nabla \times F \cdot dV] = \oint_{\partial S} \frac{\hbar}{emc^2} \frac{1}{r} \frac{\partial U(r)}{\partial r} \cdot dS[/math]
we should obtain a quantization
[math]\hbar = e\Phi = \frac{e}{2\pi} \int \int_S B \cdot dS = \frac{1}{2\pi} \int \int \int_{V} \frac{\hbar}{mc^2} \frac{1}{r} \cdot [\nabla \times F \cdot dV] = \frac{1}{2\pi}\oint_{\partial S} \frac{\hbar}{mc^2} \frac{1}{r} \frac{\partial U(r)}{\partial r} \cdot dS[/math]
