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.