Zurückziehen einer Differentialform/Punktierte Ebene nach S^1/Längenform/Aufgabe/Lösung

Es ist

${}{\begin{aligned}\pi ^{*}du&=d(u\circ \pi )\\&=d{\left({\frac {x}{\sqrt {x^{2}+y^{2}}}}\right)}\\&={\frac {\partial {\left({\frac {x}{\sqrt {x^{2}+y^{2}}}}\right)}}{\partial x}}dx+{\frac {\partial {\left({\frac {x}{\sqrt {x^{2}+y^{2}}}}\right)}}{\partial y}}dy\\&={\frac {{\sqrt {x^{2}+y^{2}}}-x^{2}(x^{2}+y^{2})^{-1/2}}{x^{2}+y^{2}}}dx-{\frac {xy}{{\sqrt {x^{2}+y^{2}}}^{3}}}dy\\&={\frac {x^{2}+y^{2}-x^{2}}{{\sqrt {x^{2}+y^{2}}}^{3}}}dx-{\frac {xy}{{\sqrt {x^{2}+y^{2}}}^{3}}}dy\\&={\frac {y^{2}}{{\sqrt {x^{2}+y^{2}}}^{3}}}dx-{\frac {xy}{{\sqrt {x^{2}+y^{2}}}^{3}}}dy\end{aligned}}$

und

${}{\begin{aligned}\pi ^{*}dv&=d(v\circ \pi )\\&=d{\left({\frac {y}{\sqrt {x^{2}+y^{2}}}}\right)}\\&={\frac {\partial {\left({\frac {y}{\sqrt {x^{2}+y^{2}}}}\right)}}{\partial x}}dx+{\frac {\partial {\left({\frac {y}{\sqrt {x^{2}+y^{2}}}}\right)}}{\partial y}}dy\\&={\frac {-xy}{{\sqrt {x^{2}+y^{2}}}^{3}}}dx+{\frac {{\sqrt {x^{2}+y^{2}}}-y^{2}(x^{2}+y^{2})^{-1/2}}{x^{2}+y^{2}}}dy\\&={\frac {-xy}{{\sqrt {x^{2}+y^{2}}}^{3}}}dx+{\frac {x^{2}}{{\sqrt {x^{2}+y^{2}}}^{3}}}dy.\end{aligned}}$

Somit ist

${}{\begin{aligned}\pi ^{*}(-vdu+udv)&=-(v\circ \pi )\pi ^{*}du+(u\circ \pi )\pi ^{*}dv\\&=-{\frac {y}{\sqrt {x^{2}+y^{2}}}}{\left({\frac {y^{2}}{{\sqrt {x^{2}+y^{2}}}^{3}}}dx-{\frac {xy}{{\sqrt {x^{2}+y^{2}}}^{3}}}dy\right)}+{\frac {x}{\sqrt {x^{2}+y^{2}}}}{\left({\frac {-xy}{{\sqrt {x^{2}+y^{2}}}^{3}}}dx+{\frac {x^{2}}{{\sqrt {x^{2}+y^{2}}}^{3}}}dy\right)}\\&={\frac {1}{{\left(x^{2}+y^{2}\right)}^{2}}}{\left({\left(-y^{3}-x^{2}y\right)}dx+{\left(xy^{2}+x^{3}\right)}dy\right)}\\&={\frac {1}{x^{2}+y^{2}}}{\left(-ydx+xdy\right)}.\,\end{aligned}}$

Zur gelösten Aufgabe