# The true TFC for surface integrals

1. Mar 29, 2014

### 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 cant 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?

Last edited: Mar 29, 2014
2. Mar 29, 2014

### 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!

3. Mar 30, 2014
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