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Hmmm-can-you-guys-explain-Let-u-dx-v-dy-be-a-1-form-defined-over-R-2-By-applying-the-above-formula-to-each-terms-consider-x-1-u-x-2-v-d-u-x-i-dx-i-dx-u-x-i-dx-i-dy

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

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