verify stoke’s theorem when f = x ^ 2 i + y ^2 j + z ^ 2k ,s is the upper hemisphere
z =√( a ^2 − x ^2 − y ^2)
x=asinφ,y=acosφ,z=0,φ:0→2πf=(a2sin2φ,a2cos2φ,0)dl=(acosφ,−asinφ,0)∮fdl=∫02π(a3sin2φcosφ−a3cos2φsinφ)dφ=0rot(f)=∣ijk∂∂x∂∂y∂∂zx2y2z2∣=0∬rot(f)dS=0Stokes formula holds.x=\mathrm{a}\sin \varphi ,y=\mathrm{a}\cos \varphi ,z=0,\varphi :0\rightarrow 2\pi \\f=\left( a^2\sin ^2\varphi ,a^2\cos ^2\varphi ,0 \right) \\dl=\left( \mathrm{a}\cos \varphi ,-\mathrm{a}\sin \varphi ,0 \right) \\\oint{fdl}=\int_0^{2\pi}{\left( a^3\sin ^2\varphi \cos \varphi -a^3\cos ^2\varphi \sin \varphi \right) d\varphi}=0\\rot\left( f \right) =\left| \begin{matrix} i& j& k\\ \frac{\partial}{\partial x}& \frac{\partial}{\partial y}& \frac{\partial}{\partial z}\\ x^2& y^2& z^2\\\end{matrix} \right|=0\\\iint{rot\left( f \right) dS}=0\\Stokes\,\,formula\,\,holds.x=asinφ,y=acosφ,z=0,φ:0→2πf=(a2sin2φ,a2cos2φ,0)dl=(acosφ,−asinφ,0)∮fdl=∫02π(a3sin2φcosφ−a3cos2φsinφ)dφ=0rot(f)=∣∣i∂x∂x2j∂y∂y2k∂z∂z2∣∣=0∬rot(f)dS=0Stokesformulaholds.
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