I was watching these videos from MIT's series: Green, Stokes.
And I didn't understand the justification: their "extended version" of the theorems.
I looked up on google and couldn't find many references (or they explained things with differential forms, which haven't yet studied).
Does anyone know some references about these versions? I'm not looking for very rigorous/proofy style notes, just some explanation and examples.
On
I'm going to share my approach to thinking intuitively about it. Correct me if you spot anything wrong.
You can think of $C_{1}$ as equal to the double integral of $curl$ along some other surface which has $C_{1}$ as its boundary and is "glued" to $S$ from below. In other words, instead of closing $S$ with a curve (and calculating line integral along it), you close it with another surface, thus creating one large closed surface.
So, let's have the following setup (similar to the video, but I've adapted it slightly to make the explanation more clear):
We want to find the work done along $C$, but it is wavy and curly and in general we would like to find easier way to handle it. We can use the extended Stokes' theorem and instead calculate
$$
\iint_S (\nabla \times \vec{F}) \cdot \hat n \,dS + \oint_{C_{1}} \vec{F} \cdot d \vec{r} = \oint_{C} \vec{F} \cdot d \vec{r}
$$
where $C_{1}$ is the circle on the $xy$ plane.
Why this works? Let's have another region, $S_{1}$, which is bounded above by the $xy$ plane. I've drawn it as a paraboloid, but it can be anything. The important thing is that it (we assume) and $S$ make a continuous surface along their boundary (where $C_{1}$ is). In other words, we "extend" $S$ with $S_{1}$. If this is so, then we have: $$ \iint_S (\nabla \times \vec{F}) \cdot \hat n \,dS + \iint_{S_{1}} (\nabla \times \vec{F}) \cdot \hat n \,dS = \oint_{C} \vec{F} \cdot d \vec{r} $$ This is the standard Stokes' theorem, not the extended one (by assumption on the relation between $S$ and $S_{1}$).
But... from the standard Stokes' theorem we know that work done along the circle on the $xy$ plane is the same as the total curl on the paraboloid below it. Or, which is the same: $$ \iint_{S_{1}} (\nabla \times \vec{F}) \cdot \hat n \,dS = \oint_{C_{1}} \vec{F} \cdot d \vec{r} $$
Substituting the third equality into the second, we get the first.
On
I really like cyau's explanation (it helped me wrap my head around the theorem), though I believe the correct equality should be: $$\iint_S (\nabla \times \vec{F}) \cdot \hat n \,dS + \iint_{S_{1}} (\nabla \times \vec{F}) \cdot \hat n \,dS = - \oint_{C} \vec{F} \cdot d \vec{r}$$ This can be found by using various methods Prof. Aurox explained in lecture, one of which being pointing the thumb along $\mathit{C}$ and then the index towards the surface (which would be "down" in this case). The middle finger would then point inward the surface, so a negative is needed to match the positive outward orientation. With this correction, the equality derived in the video is matched with this approach: $$\iint_S (\nabla \times \vec{F}) \cdot \hat n \,dS = \oint_{C_1} \vec{F} \cdot d \vec{r} - \oint_{C} \vec{F} \cdot d \vec{r}$$
I would've commented on cyau's post, but I unfortunately do not have enough reputation to do so, and I felt this was an important correction to make.
For Green's theorem, this page has a good explanation of the technique and a good way to think about the multiple boundaries. And this page goes into more detail about why the technique works. The orientation of the curves is positive if the region is always to the left of the curve in the direction of travel, and you sum the positive line integrals (or negate the terms with negative orientation) to get the integral over the region.
Something similar occurs for Stokes' theorem; I can't find any good references apart from some course notes I don't have permission to reproduce. In general, the positive orientation of the curve is the one where you traverse the curve with your head pointing in the direction of the surface normal and the surface is on your left, and again you sum the positive orientation line integrals (or negate the negative orientation line integrals) to get the integral over the surface.