Question Number 218459 by SdC355 last updated on 10/Apr/25
$$\overset{\rightarrow} {\boldsymbol{\mathrm{F}}}\left({x},{y}\right)=−\frac{\mathrm{1}}{\mathrm{2}}{y}\overset{\rightarrow} {\boldsymbol{\mathrm{e}}}_{\mathrm{1}} +\frac{\mathrm{1}}{\mathrm{2}}{x}\overset{\rightarrow} {\boldsymbol{\mathrm{e}}}_{\mathrm{2}} \\ $$$$\overset{\rightarrow} {\bigtriangledown}×\overset{\rightarrow} {\boldsymbol{\mathrm{F}}}\left({x},{y}\right)=\begin{vmatrix}{\:\:\:\:\overset{\rightarrow} {\boldsymbol{\mathrm{e}}}_{\mathrm{1}} }&{\overset{\rightarrow} {\boldsymbol{\mathrm{e}}}_{\mathrm{2}} }&{\overset{\rightarrow} {\boldsymbol{\mathrm{e}}}_{\mathrm{3}} }\\{\:\:\:\:\:\partial_{{x}} }&{\:\partial_{{y}} }&{\partial_{{z}} }\\{−\frac{\mathrm{1}}{\mathrm{2}}{y}}&{\frac{\mathrm{1}}{\mathrm{2}}{x}}&{\mathrm{0}}\end{vmatrix}=\mathrm{0}\overset{\rightarrow} {\boldsymbol{\mathrm{e}}}_{\mathrm{1}} −\mathrm{0}\overset{\rightarrow} {\boldsymbol{\mathrm{e}}}_{\mathrm{2}} +\left(\frac{\mathrm{1}}{\mathrm{2}}−\left(−\frac{\mathrm{1}}{\mathrm{2}}\right)\right)\overset{\rightarrow} {\boldsymbol{\mathrm{e}}}_{\mathrm{3}} \\ $$$$\oint_{\:{C}} \:\overset{\rightarrow} {\boldsymbol{\mathrm{F}}}\centerdot\mathrm{d}\boldsymbol{\mathrm{l}}=\oint_{\:{C}} −\frac{\mathrm{1}}{\mathrm{2}}{y}\mathrm{d}{x}+\frac{\mathrm{1}}{\mathrm{2}}{x}\mathrm{d}{x}=\int\int_{\:\mathbb{R}^{\mathrm{2}} } \overset{\rightarrow} {\boldsymbol{\mathrm{e}}}_{\mathrm{3}} \centerdot\overset{\rightarrow} {\boldsymbol{\mathrm{n}}}\:\mathrm{dS} \\ $$$$\therefore\oint_{\:{C}} −\frac{\mathrm{1}}{\mathrm{2}}{y}\mathrm{d}{x}+\frac{\mathrm{1}}{\mathrm{2}}{x}\mathrm{d}{x}=\int\int_{\:\mathbb{R}^{\mathrm{2}} } \:\mathrm{dS}…..\mathrm{is}\:\mathrm{right}…?? \\ $$
Commented by MrGaster last updated on 10/Apr/25
（Too lazy to output pictures ...）\vec{F}(x,y)=-\frac{1}{2}y\vec{e}_1+\frac{1}{2}x\vec{e}_2\nabla\times\vec{F}=\begin{vmatrix}\vec{e}_1&\vec{e}_2&\vec{e}_3\\\partial_x&\partial_y&\partial_z\\-\frac{1}{2}y&\frac{1}{2}x&0\end{vmatrix}=0\vec{e}_1-0\vec{e}_2+\left(\frac{1}{2}-\left(-\frac{1}{2}\right)\right)\vec{e}_3=\vec{e}_3\oint_C\vec{F}\cdot d\vec{l}=\oint_C-\frac{1}{2}y\,dx+\frac{1}{2}x\,dy\iint_{\mathbb{R}^2}(\nabla\times\vec{F})\cdot\vec{n}\,dS=\iint_{\mathbb{R}^2}\vec{e}_3\cdot\vec{n}\,dS\vec{n}=\vec{e}_3\Rightarrow\iint_{\mathbb{R}^2}1\,dS\oint_C-\frac{1}{2}y\,dx+\frac{1}{2}x\,dy=\iint_{\mathbb{R}^2}dS\text{Let }C\text{ be the unit circle: }x=\cos\theta,y=\sin\thetadx=-\sin\theta\,d\theta,\quad dy=\cos\theta\,d\theta\oint_C-\frac{1}{2}\sin\theta(-\sin\theta\,d\theta)+\frac{1}{2}\cos\theta(\cos\theta\,d\theta)=\oint_C\frac{1}{2}\sin^2\theta\,d\theta+\frac{1}{2}\cos^2\theta\,d\theta=\frac{1}{2}\oint_C(\sin^2\theta+\cos^2\theta)\,d\theta=\frac{1}{2}\oint_C 1\,d\theta=\frac{1}{2}\times 2\pi=\pi\iint_{\mathbb{R}^2}1\,dS=\pi\times 1^2=\pi\oint_C\vec{F}\cdot d\vec{l}=\iint_{\mathbb{R}^2}(\nabla\times\vec{F})\cdot\vec{n}\,dS\therefore\text{The application of Stokes' theorem is correct.}\mathbf{\text{Yes}}
Commented by SdC355 last updated on 10/Apr/25
$$\mathrm{thx} \\ $$