Question Number 220226 by SdC355 last updated on 09/May/25
Commented by SdC355 last updated on 09/May/25
$$\mathrm{From}\:\mathrm{Generalized}\:\mathrm{Stoke}'\mathrm{s}\:\mathrm{theorem} \\ $$$$\oint_{\partial\boldsymbol{\Sigma}} \:\alpha=\int\int_{\:\boldsymbol{\Sigma}} \:\mathrm{d}\alpha \\ $$$$\oint_{\partial\boldsymbol{\Sigma}} \overset{\rightarrow} {\boldsymbol{\mathrm{F}}}\centerdot\mathrm{d}\boldsymbol{\mathrm{l}}=\int\int_{\:\boldsymbol{\Sigma}} \:\overset{\rightarrow} {\bigtriangledown}×\overset{\rightarrow} {\boldsymbol{\mathrm{F}}}\centerdot\mathrm{d}\overset{\rightarrow} {\boldsymbol{\mathrm{S}}}.. \\ $$$$\mathrm{i}\:\mathrm{don}'\mathrm{t}\:\mathrm{understand}\:\mathrm{why} \\ $$$$\oint_{\:\partial\boldsymbol{\Sigma}} {P}\:\mathrm{d}{x}+{Q}\:\mathrm{d}{y}+{R}\:\mathrm{d}{z}=\int\int_{\:\boldsymbol{\Sigma}} \left(\frac{\partial{R}}{\partial{y}}−\frac{\partial{Q}}{\partial{z}}\right)\mathrm{d}{y}\mathrm{d}{z}+\left(\frac{\partial{P}}{\partial{z}}−\frac{\partial{R}}{\partial{x}}\right)\mathrm{d}{z}\mathrm{d}{x}+\left(\frac{\partial{Q}}{\partial{x}}−\frac{\partial{P}}{\partial{y}}\right)\mathrm{d}{x}\mathrm{d}{y} \\ $$$$\overset{\rightarrow} {\bigtriangledown}×\begin{pmatrix}{{P}}\\{{Q}}\\{{R}}\end{pmatrix}\centerdot\mathrm{d}\overset{\rightarrow} {\boldsymbol{\mathrm{S}}}=\begin{pmatrix}{{R}_{{y}} −{Q}_{{z}} }\\{{P}_{{z}} −{R}_{{x}} }\\{{Q}_{{x}} −{P}_{{y}} }\end{pmatrix}\centerdot\mathrm{d}\overset{\rightarrow} {\boldsymbol{\mathrm{S}}},\: \\ $$$$\mathrm{i}.\mathrm{e}. \\ $$$$\mathrm{d}\left({P}\:\mathrm{d}{x}+{Q}\:\mathrm{d}{y}+{R}\:\mathrm{d}{z}\right)=\left({R}_{{x}} −{Q}_{{y}} \right)\mathrm{d}{y}\wedge\mathrm{d}{x}+\left({P}_{{z}} −{R}_{{x}} \right)\mathrm{d}{z}\wedge\mathrm{d}{x}+\left({Q}_{{x}} −{P}_{{y}} \right)\mathrm{d}{x}\wedge\mathrm{d}{y} \\ $$