The true TFC for surface integrals

[Jhenrique]

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: $$\int_{\vec{s}_0}^{\vec{s}_1}\vec{\nabla}f\cdot d\vec{s}=f(\vec{s}_1)-f(\vec{s}_0)$$
Thus, by analogy, how would be the FTC for surface integral? I think that would be something more or less like:
$$\iint_{\vec{S}_0}^{\vec{S}_1}\vec{\nabla}\times \vec{f}\cdot d^2\vec{S}=\vec{f}(\vec{S}_1)-\vec{f}(\vec{S}_0)$$
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: $$\iint \limits_{\vec{s}_0\vec{t}_0}^{\vec{s}_1\vec{t}_1}\vec{\nabla}\times \vec{F}\cdot d^2\vec{S}=\vec{F}(\vec{s}_1,\vec{t}_1)-\vec{F}(\vec{s}_0,\vec{t}_0)$$ If all this ideia is correct, thus we have a lot of questions! 1st, which is the geometric interpretation for this integral? You can't give the same geometric interpretation as in: $$\oint\!\! \oint_{S} \vec{F}\cdot d^2\vec{S} = \iiint_{V} \vec{\nabla}\times \vec{F}d^3V$$ because this interpretation is the analogous to $$\oint_{s} \vec{f}\cdot d\vec{s} = \iint_{S} \vec{\nabla}\cdot \vec{f}\cdot d^2\vec{S}$$ 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?

 

[chogg]

You seem very interested in vector calculus from a Geometric Algebra perspective.

So am I! Fortunately, a helpful book was just released on the subject. I'm reading through it now.
http://faculty.luther.edu/~macdonal/vagc/

This book promises to show you what the fundamental theorem of calculus looks like in geometric algebra, though I'm not that far in it. So far it's very interesting though!

 

[Jhenrique]

Happens that this book don't talk about the fundamental theorem of calculus for surface integral, btw, no one coment about this!

 

[chogg]

Are you sure?

Chapter 10 is called The Fundamental Theorem of Calculus. Looks to me like it discusses surface integrals rather extensively.

 

[blue_raver22]

I think your deduction and analysis are on the right track. The fundamental theorem of calculus for surface integrals is a bit more complicated than the one for line integrals, as it involves a two-dimensional surface instead of a one-dimensional curve. The notation and interpretation can be a bit confusing, but your understanding of the components of $\vec{S}$ and the need for it to represent a bidimensional manifold is correct.

As for the geometric interpretation, it is not as straightforward as the one for line integrals, as you have mentioned. It involves the flux of the two-form $\vec{F}$ through the surface $\vec{S}$. The specific interpretation would depend on the physical situation and the specific form of $\vec{F}$.

In terms of what $\vec{S}$ actually is, it can be thought of as a surface element or a two-dimensional infinitesimal area. It can also be represented as a bivector or a tensor, as you have mentioned.

Overall, your understanding and analysis of the true FTC for surface integrals is correct. However, this is still an area of active research and there may be further developments and refinements in the future. Keep up the good work in exploring and understanding these concepts!