Question Number 220660 by SdC355 last updated on 17/May/25
$$\mathrm{Hmmm}…… \\ $$$$\mathrm{can}\:\mathrm{you}\:\mathrm{guys}\:\mathrm{explain}?? \\ $$$$\mathrm{Let}\:\boldsymbol{\omega}={u}\:\mathrm{d}{x}+{v}\:\mathrm{d}{y}\:\mathrm{be}\:\mathrm{a}\:\mathrm{1}-\mathrm{form}\:\mathrm{defined}\:\mathrm{over}\:\mathbb{R}^{\mathrm{2}} \\ $$$$\mathrm{By}\:\mathrm{applying}\:\mathrm{the}\:\mathrm{above}\:\mathrm{formula}\:\mathrm{to}\:\mathrm{each}\:\mathrm{terms} \\ $$$$\mathrm{consider}\:{x}^{\mathrm{1}} ={u}\:,\:{x}^{\mathrm{2}} ={v} \\ $$$$\mathrm{d}\boldsymbol{\omega}=\left(\Sigma\:\frac{\partial{u}}{\partial{x}^{{i}} }\:\mathrm{d}{x}^{{i}} \wedge\mathrm{d}{x}\right)+\left(\Sigma\:\frac{\partial{u}}{\partial{x}^{{i}} }\:\mathrm{d}{x}^{{i}} \wedge\mathrm{d}{y}\right) \\ $$$$=\frac{\partial{u}}{\partial{x}}\:\mathrm{d}{x}\wedge\mathrm{d}{x}+\frac{\partial{u}}{\partial{y}}\:\mathrm{d}{y}\wedge\mathrm{d}{x}+\frac{\partial{u}}{\partial{x}}\:\mathrm{d}{x}\wedge\mathrm{d}{y}+\frac{\partial{u}}{\partial{y}}\:\mathrm{d}{y}\wedge\mathrm{d}{y} \\ $$$$\mathrm{0}−\frac{\partial{u}}{\partial{y}}\:\mathrm{d}{x}\wedge\mathrm{d}{y}+\frac{\partial{u}}{\partial{x}}\:\mathrm{d}{x}\wedge\mathrm{d}{y}+\mathrm{0} \\ $$$$\left(\frac{\partial{u}}{\partial{x}}−\frac{\partial{u}}{\partial{y}}\right)\:\mathrm{d}{x}\wedge\mathrm{d}{y} \\ $$$$\mathrm{d}{x}\wedge\mathrm{d}{x}=\mathrm{0} \\ $$$$\mathrm{d}{x}\wedge\mathrm{d}{y}=−\mathrm{d}{y}\wedge\mathrm{d}{x} \\ $$$$\mathrm{d}{y}\wedge\mathrm{d}{y}=\mathrm{0} \\ $$$$\mathrm{i}\:\mathrm{can}'\mathrm{t}\:\mathrm{understand}\:\mathrm{why}\:\mathrm{Exterior}\:\mathrm{derivate} \\ $$$$''\mathrm{d}\boldsymbol{\omega}=\left(\Sigma\:\frac{\partial{u}}{\partial{x}^{{i}} }\:\mathrm{d}{x}^{\boldsymbol{{i}}} \wedge\mathrm{d}{x}\right)+\left(\Sigma\:\frac{\partial{u}}{\partial{x}^{{i}} }\:\mathrm{d}{x}^{{i}} \wedge\mathrm{d}{y}\right)'' \\ $$$$\mathrm{from}\:\mathrm{the}\:\mathrm{Stoke}'\mathrm{s}\:\mathrm{Theorem} \\ $$$$\int_{\:\partial{S}} \:\boldsymbol{\omega}=\int_{\:{S}} \:\mathrm{d}\boldsymbol{\omega} \\ $$$$\int_{\:\partial{S}} \:\boldsymbol{\omega}=\int\int_{\:{S}} \:\left(\frac{\partial{u}}{\partial{x}}−\frac{\partial{u}}{\partial{y}}\right)\:\mathrm{d}{x}\wedge\mathrm{d}{y} \\ $$
Answered by MrGaster last updated on 18/May/25
$$\boldsymbol{\omega}={u}\:{du}+{v}\:{dy} \\ $$$${d}\boldsymbol{\omega}={du}\wedge{dx}+{dv}\wedge{dy} \\ $$$$=\left(\frac{\partial{y}}{\partial{x}}{dx}+\frac{\partial{u}}{\partial{y}}\partial{y}\right)\wedge{dx}+\left(\frac{\partial{u}}{\partial{x}}{dx}+\frac{\partial{v}}{\partial{y}}\partial{y}\right)\wedge{dy} \\ $$$$=\frac{\partial{u}}{\partial{x}}\underset{\mathrm{0}} {\underbrace{\mathrm{d}{x}\wedge\mathrm{d}{x}}}\:+\frac{\partial{u}}{\partial{y}}{dy}\wedge{dx}+\frac{\partial{v}}{\partial{x}}\mathrm{d}{x}+{dy}+\frac{\partial{v}}{\partial{y}}\underset{\mathrm{0}} {\underbrace{{dy}\wedge{dy}}} \\ $$$$=−\frac{\partial{u}}{\partial{y}}{dx}\wedge{dy}+\frac{\partial{v}}{{dx}}{dx}\wedge{dy} \\ $$$$=\left(\frac{\partial{v}}{\partial{x}}−\frac{\partial{u}}{\partial{y}}\right){dx}\wedge{dy} \\ $$$$\oint_{\partial\mathcal{S}} \boldsymbol{\omega}=\int\int_{{S}} {d}\boldsymbol{\omega}=\int\int_{{S}} \left(\frac{\partial{v}}{\partial{x}}−\frac{\partial{u}}{\partial{y}}\right){dx}\wedge{dy} \\ $$