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Zurückziehen einer Differentialform/Punktierte Ebene nach S^1/Längenform/Aufgabe/Lösung
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Zurückziehen einer Differentialform/Punktierte Ebene nach S^1/Längenform/Aufgabe
Es ist
π
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{\displaystyle {}{\begin{aligned}\pi ^{*}du&=d(u\circ \pi )\\&=d{\left({\frac {x}{\sqrt {x^{2}+y^{2}}}}\right)}\\&={\frac {\partial {\left({\frac {x}{\sqrt {x^{2}+y^{2}}}}\right)}}{\partial x}}dx+{\frac {\partial {\left({\frac {x}{\sqrt {x^{2}+y^{2}}}}\right)}}{\partial y}}dy\\&={\frac {{\sqrt {x^{2}+y^{2}}}-x^{2}(x^{2}+y^{2})^{-1/2}}{x^{2}+y^{2}}}dx-{\frac {xy}{{\sqrt {x^{2}+y^{2}}}^{3}}}dy\\&={\frac {x^{2}+y^{2}-x^{2}}{{\sqrt {x^{2}+y^{2}}}^{3}}}dx-{\frac {xy}{{\sqrt {x^{2}+y^{2}}}^{3}}}dy\\&={\frac {y^{2}}{{\sqrt {x^{2}+y^{2}}}^{3}}}dx-{\frac {xy}{{\sqrt {x^{2}+y^{2}}}^{3}}}dy\end{aligned}}}
und
π
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{\displaystyle {}{\begin{aligned}\pi ^{*}dv&=d(v\circ \pi )\\&=d{\left({\frac {y}{\sqrt {x^{2}+y^{2}}}}\right)}\\&={\frac {\partial {\left({\frac {y}{\sqrt {x^{2}+y^{2}}}}\right)}}{\partial x}}dx+{\frac {\partial {\left({\frac {y}{\sqrt {x^{2}+y^{2}}}}\right)}}{\partial y}}dy\\&={\frac {-xy}{{\sqrt {x^{2}+y^{2}}}^{3}}}dx+{\frac {{\sqrt {x^{2}+y^{2}}}-y^{2}(x^{2}+y^{2})^{-1/2}}{x^{2}+y^{2}}}dy\\&={\frac {-xy}{{\sqrt {x^{2}+y^{2}}}^{3}}}dx+{\frac {x^{2}}{{\sqrt {x^{2}+y^{2}}}^{3}}}dy.\end{aligned}}}
Somit ist
π
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{\displaystyle {}{\begin{aligned}\pi ^{*}(-vdu+udv)&=-(v\circ \pi )\pi ^{*}du+(u\circ \pi )\pi ^{*}dv\\&=-{\frac {y}{\sqrt {x^{2}+y^{2}}}}{\left({\frac {y^{2}}{{\sqrt {x^{2}+y^{2}}}^{3}}}dx-{\frac {xy}{{\sqrt {x^{2}+y^{2}}}^{3}}}dy\right)}+{\frac {x}{\sqrt {x^{2}+y^{2}}}}{\left({\frac {-xy}{{\sqrt {x^{2}+y^{2}}}^{3}}}dx+{\frac {x^{2}}{{\sqrt {x^{2}+y^{2}}}^{3}}}dy\right)}\\&={\frac {1}{{\left(x^{2}+y^{2}\right)}^{2}}}{\left({\left(-y^{3}-x^{2}y\right)}dx+{\left(xy^{2}+x^{3}\right)}dy\right)}\\&={\frac {1}{x^{2}+y^{2}}}{\left(-ydx+xdy\right)}.\,\end{aligned}}}
Zur gelösten Aufgabe
