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Theorem of the month

Green's theorem

For some [math]\displaystyle{ h }[/math]-domain [math]\displaystyle{ D \subseteq {}^{(\omega)}\mathbb{R}^{2} }[/math], infinitesimal [math]\displaystyle{ h = |{\downarrow}x|= |{\downarrow}y| = |{}^\curvearrowright \gamma(t) - \gamma(t)| = \mathcal{O}({\tilde{\omega}}^{m}) }[/math], sufficiently large [math]\displaystyle{ m \in \mathbb{N}^{*}, (x, y) \in D, {D}^{-} := \{(x, y) \in D : (x + h, y + h) \in D\} }[/math], and a simply closed path [math]\displaystyle{ \gamma: [a, b[\rightarrow {\downarrow} D }[/math] followed anticlockwise, choosing [math]\displaystyle{ {}^\curvearrowright \gamma(t) = \gamma({}^\curvearrowright t) }[/math] for [math]\displaystyle{ t \in [a, b[, A \subseteq {[a, b]}^{2} }[/math], the following equation holds for sufficiently [math]\displaystyle{ \alpha }[/math]-continuous functions [math]\displaystyle{ u, v: D \rightarrow \mathbb{R} }[/math] with not necessarily continuous [math]\displaystyle{ {\downarrow} u/{\downarrow} x, {\downarrow} u/{\downarrow} y, {\downarrow} v/{\downarrow} x }[/math] and [math]\displaystyle{ {\downarrow} v/{\downarrow} y }[/math]

[math]\displaystyle{ {\uparrow}_{\gamma }{(u\,{\downarrow}x+v\,{\downarrow}y)}={\uparrow}_{(x,y)\in {{D}^{-}}}{\left( \tfrac{{\downarrow} v}{{\downarrow} x}-\tfrac{{\downarrow} u}{{\downarrow} y} \right){\downarrow}(x,y)}. }[/math]

Proof:

Only [math]\displaystyle{ D := \{(x, y) : r \le x \le s, f(x) \le y \le g(x)\}, r, s \in {}^{(\omega)}\mathbb{R}, f, g : {\downarrow} D \rightarrow {}^{(\omega)}\mathbb{R} }[/math] is proved, since the proof is analogous for each case rotated by [math]\displaystyle{ \iota }[/math]. Every [math]\displaystyle{ h }[/math]-domian is union of such sets. Simply showing

[math]\displaystyle{ {\uparrow}_{\gamma }{u\,{\downarrow}x}=-{\uparrow}_{(x,y)\in {{D}^{-}}}{\tfrac{{\downarrow} u}{{\downarrow} y}{\downarrow}(x,y)}. }[/math]

is sufficient because the other relation is given analogously. Neglecting the regions of [math]\displaystyle{ \gamma }[/math] with [math]\displaystyle{ {\downarrow}x = 0 }[/math] and [math]\displaystyle{ t := h(u(s, g(s)) - u(r, g(r))) }[/math] shows

[math]\displaystyle{ -{\uparrow}_{\gamma }{u\,{\downarrow}x}-t={\uparrow}_{r}^{s}{u(x,g(x)){\downarrow}x}-{\uparrow}_{r}^{s}{u(x,f(x)){\downarrow}x}={\uparrow}_{r}^{s}{{\uparrow}_{f(x)}^{g(x)}{\tfrac{{\downarrow} u}{{\downarrow} y}}{\downarrow}y{\downarrow}x}={\uparrow}_{(x,y)\in {{D}^{-}}}{\tfrac{{\downarrow} u}{{\downarrow} y}{\downarrow}(x,y)}.\square }[/math]

Recommended reading

Nonstandard Mathematics