Talk:Kelvin-Stokes Theorem
Filling in the details
I appreciate what is being done here, but I wonder whether it would be better to extract the complexity out into another page (where we may be able to invoke some already-documented vector-calculus identities) and hence structure it in a more easily-digested form (e.g. extracting the sub-calculations as building blocks). At the moment there is just too much going on in any given line, and as a result, not only does it not fit on a page, but also it's difficult to comprehend what is actually being done. --prime mover (talk) 10:51, 4 July 2019 (EDT)
- I am of the same mind generally, though I am am having difficultly thinking about/finding relevant identities themselves, other than Derivative of Dot Product of Vector-Valued Functions, and naming them if they do not yet have pages. In the mean time, I could try re-adding some comments I removed (but as new lines within the final equation) and then remove the work in progress tag so that someone with more experience breaking articles into new pages could work on this. The issue I encountered is that by keeping it structured as one equation, all the comments I had added were pushed too the far right and were basically unreadable. Formatting the final equation without using the equation template should alleviate some of this. Mizar (talk) 11:18, 4 July 2019 (EDT)
- You could try splitting the lines up using the "ro" parameter to put the operator in. Then you have something like this:
| \(\ds \) | \(=\) | \(\ds \iint_R \Biggl(
\paren { \dfrac {\partial f_3} {\partial y} \dfrac{\partial y}{\partial s} \dfrac{\partial z}{\partial t}
- \dfrac {\partial f_3} {\partial y} \dfrac{\partial z}{\partial s} \dfrac{\partial y}{\partial t}
- \dfrac {\partial f_2} {\partial z} \dfrac{\partial y}{\partial s} \dfrac{\partial z}{\partial t}
+ \dfrac {\partial f_2} {\partial z} \dfrac{\partial z}{\partial s} \dfrac{\partial y}{\partial t}
}\)
|
||||||||||||
| \(\ds \) | \(\) | \(\, \ds + \, \) | \(\ds \paren {
\dfrac {\partial f_1} {\partial z} \dfrac{\partial z}{\partial s} \dfrac{\partial x}{\partial t} - \dfrac {\partial f_1} {\partial z} \dfrac{\partial x}{\partial s} \dfrac{\partial z}{\partial t}
- \dfrac {\partial f_3} {\partial x} \dfrac{\partial z}{\partial s} \dfrac{\partial x}{\partial t}
+ \dfrac {\partial f_3} {\partial x} \dfrac{\partial x}{\partial s} \dfrac{\partial z}{\partial t}
}\)
|
|||||||||||
| \(\ds \) | \(\) | \(\, \ds + \, \) | \(\ds \paren {
\dfrac {\partial f_2} {\partial x} \dfrac{\partial x}{\partial s} \dfrac{\partial y}{\partial t} - \dfrac {\partial f_2} {\partial x} \dfrac{\partial y}{\partial s} \dfrac{\partial x}{\partial t}
- \dfrac {\partial f_1} {\partial y} \dfrac{\partial x}{\partial s} \dfrac{\partial y}{\partial t}
+ \dfrac {\partial f_1} {\partial y} \dfrac{\partial y}{\partial s} \dfrac{\partial x}{\partial t}
} \Biggr) \rd s \rd t\) |
- ... but that means you have to handle the bracket sizes yourself as you can't split up a \paren in this context. It's messy and fiddly to set up, but it works. --prime mover (talk) 12:41, 4 July 2019 (EDT)
- That's the idea. Nice one. --prime mover (talk) 18:52, 4 July 2019 (EDT)
Doesn't this proof only apply when the boundary is planar? Otherwise, there is no $st$-plane on which to apply Green's Theorem. --CircuitCraft (talk) 15:38, 24 April 2023 (UTC)