Question Number 224789 by fkwow344 last updated on 04/Oct/25
$$\mathrm{prove}\:\mathrm{Sphere}\:\mathcal{S};\mathbb{R}^{\mathrm{3}} \rightarrow\mathbb{R} \\ $$$${x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{z}^{\mathrm{2}} ={R}^{\mathrm{2}} \:,\:\mathrm{Euler}\:\mathrm{characteristic}\:\boldsymbol{\chi}=\mathrm{2} \\ $$$$\mathrm{by}\:\mathrm{gauss}-\mathrm{Bonnet}\:\mathrm{theorem} \\ $$$$\mathrm{2}\pi\boldsymbol{\chi}\left(\boldsymbol{\Omega}\right)=\int_{\:\boldsymbol{\Omega}} \:\mathrm{d}{A}\:{K} \\ $$$$\mathrm{Gauss}\:\mathrm{curvature}\:\mathrm{defined}\:\mathrm{as}\:{K}=\frac{\mathrm{det}\:\Pi}{\mathrm{det}\:\mathrm{I}}=\frac{{LN}−{M}^{\mathrm{2}} }{{EG}−{F}^{\mathrm{2}} } \\ $$$$\mathrm{such}\:\mathrm{that}\: \\ $$$$\mathrm{I}=\left(\mathrm{d}\zeta^{\mathrm{1}} \:\mathrm{d}\zeta^{\mathrm{2}} \right)\begin{pmatrix}{{E}}&{{F}}\\{{F}}&{{G}}\end{pmatrix}\begin{pmatrix}{\mathrm{d}\zeta^{\mathrm{1}} }\\{\mathrm{d}\zeta^{\mathrm{2}} }\end{pmatrix}=\underset{{jk}} {\sum}\:\frac{\partial\boldsymbol{\mathrm{f}}}{\partial\zeta^{{j}} }\centerdot\frac{\partial\boldsymbol{\mathrm{f}}}{\partial\zeta^{{k}} }\:\mathrm{d}\zeta^{{j}} \mathrm{d}\zeta^{{k}} \\ $$$$\mathrm{E}={x}_{{u}} \ast{x}_{{u}} \:,\mathrm{F}={x}_{{u}} \ast{x}_{{v}} \:,\:{G}={x}_{{v}} \ast{x}_{{v}} \: \\ $$$$\Pi=\left(\mathrm{d}\zeta^{\mathrm{1}} \:\mathrm{d}\zeta^{\mathrm{2}} \right)\begin{pmatrix}{{L}}&{{M}}\\{{M}}&{{N}}\end{pmatrix}\begin{pmatrix}{\mathrm{d}\zeta^{\mathrm{1}} }\\{\mathrm{d}\zeta^{\mathrm{2}} }\end{pmatrix}=\underset{\imath\jmath} {\sum}\:\hat {\boldsymbol{\mathrm{n}}}\ast\frac{\partial^{\mathrm{2}} \boldsymbol{\mathrm{f}}}{\partial\zeta^{\imath} \partial\zeta^{\jmath} }\:\mathrm{d}\zeta^{\imath} \mathrm{d}\zeta^{\jmath} \\ $$$$\hat {\boldsymbol{\mathrm{n}}}=\frac{{x}_{{u}} ×{x}_{{v}} }{\mid\mid{x}_{{u}} ×{x}_{{v}} \mid\mid} \\ $$$${L}={x}_{{uu}} \ast\hat {\boldsymbol{\mathrm{n}}},\:{M}={x}_{{uv}} \ast\hat {\boldsymbol{\mathrm{n}}}\:,\:{N}={x}_{{vv}} \ast\hat {\boldsymbol{\mathrm{n}}} \\ $$