Timeline for Using Stokes' theorem to define "area" enclosed by a curve

10 events

when toggle format	what		by	license	comment
Jan 23, 2019 at 2:55	answer	added	Piotr Hajlasz		timeline score: 5
Dec 10, 2014 at 15:57	vote	accept	Izhar Oppenheim
Dec 10, 2014 at 15:57	answer	added	Izhar Oppenheim		timeline score: 6
Dec 10, 2014 at 15:07	history	edited	Izhar Oppenheim	CC BY-SA 3.0	deleted 3 characters in body
Dec 10, 2014 at 14:58	history	edited	Izhar Oppenheim	CC BY-SA 3.0	added 755 characters in body
Dec 10, 2014 at 2:41	answer	added	Will Sawin		timeline score: 15
Dec 9, 2014 at 21:50	comment	added	Liviu Nicolaescu		@BrunoLeFloch Sometimes, the projection of the curve onto a $2$-plane can look like a figure $8$. In this case, integrating $xdy-ydx$ over the figure $8$ we can get $0$.
Dec 9, 2014 at 21:10	comment	added	Bruno Le Floch		Note that all terms in the sum over $i,j$ contribute equally, as $SO_n$ acts transitively on ordered pairs $x_i,x_j$ for $n\geq 3$. For the same reason, the average over $SO_n(\mathbb{R})$ would have vanished if you had omitted the squares. As far as I can tell, your Stokes area is a quadratic average over all 2d planes $P$ of the area of the projection of the curve onto $P$. Once properly normalized, that should be less than the area of a minimal surface. Instead, averaging over $SO_n$ before integrating over $\phi(S^1)$ gives an upper bound to the area of a minimal surface.
Dec 9, 2014 at 20:29	history	edited	Izhar Oppenheim	CC BY-SA 3.0	deleted 5 characters in body
Dec 9, 2014 at 19:51	history	asked	Izhar Oppenheim	CC BY-SA 3.0