(adsbygoogle = window.adsbygoogle || []).push({}); The true FTC for surface integrals

Let's say that ##\vec{f}## is an exact one-form, so we have that ##\vec{f}=\vec{\nabla}f##, and ##\vec{F}## is an exact two-form, so we have that ##\vec{F}=\vec{\nabla}\times \vec{f}##.

The fundamental theorem of calculus for line integral says that: [tex]\int_{\vec{s}_0}^{\vec{s}_1}\vec{\nabla}f\cdot d\vec{s}=f(\vec{s}_1)-f(\vec{s}_0)[/tex]

Thus, by analogy, how would be the FTC for surface integral? I think that would be something more or less like:

[tex]\iint_{\vec{S}_0}^{\vec{S}_1}\vec{\nabla}\times \vec{f}\cdot d^2\vec{S}=\vec{f}(\vec{S}_1)-\vec{f}(\vec{S}_0)[/tex]

But, if my deduction is correct, so what is ##\vec{S}##? Well, ##\vec{S}## can't be a vector because a vector have two or three scalar components and your variation, ##\vec{s}_1 - \vec{s}_0## describe a unidimensional manifold (a curve). As we want that the variation of ##\vec{S}## represents a bidimensional manifold (a surface), so is necessary that your components be 2 vectors because two points (two vectors) varying simultaneously describe a surface. This ideia is consistent with the d² of double integral above, because ##\frac{df(x)}{dx} = \frac{f(x_1) - f(x_0)}{dx}##, ##\frac{df(x,y)}{dx} = \frac{f(x_1,y) - f(x_0,y)}{dx}##, ##\frac{d^2f(x,y)}{dxdy} = \frac{f(x_1, y_1) - f(x_0, y_0)}{dxdy}##, so ##d^2 f(x,y) = f(x_1, y_1)-f(x_0, y_0)##. If ##\vec{S}## represents a set of vectors, so ##\vec{S} = (\vec{s}, \vec{t})##. The integral of surface becomes: [tex]\iint \limits_{\vec{s}_0\vec{t}_0}^{\vec{s}_1\vec{t}_1}\vec{\nabla}\times \vec{F}\cdot d^2\vec{S}=\vec{F}(\vec{s}_1,\vec{t}_1)-\vec{F}(\vec{s}_0,\vec{t}_0)[/tex] If all this ideia is correct, thus we have a lot of questions! 1st, which is the geometric interpretation for this integral? You cant give the same geometric interpretation as in: [tex]\oint\!\! \oint_{S} \vec{F}\cdot d^2\vec{S} = \iiint_{V} \vec{\nabla}\times \vec{F}d^3V[/tex] because this interpretation is the analogous to [tex]\oint_{s} \vec{f}\cdot d\vec{s} = \iint_{S} \vec{\nabla}\cdot \vec{f}\cdot d^2\vec{S}[/tex] and not represents the FTC.

2nd place, what hell is actually ##\vec{S}## ? This isn't a vector! Should be a bivector or a tensor.

I think that all is quite something. What you think about?

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# The true TFC for surface integrals

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