Wegintegral/Ebene/x^2-y^3,xy/t^2,t^3-5t/0 bis 1/Beispiel

Wir betrachten das Vektorfeld

$F\colon \mathbb {R} ^{2}\longrightarrow \mathbb {R} ^{2},\,(x,y)\longmapsto (x^{2}-y^{3},xy)$

und den Weg

$\gamma \colon [0,1]\longrightarrow \mathbb {R} ^{2},\,t\longmapsto (t^{2},t^{3}-5t).$

Die Ableitung von ${}\gamma$ ist

${}\gamma '(t)=(2t,3t^{2}-5)\,.$

Daher ist das Wegintegral zu diesem Vektorfeld längs dieser Kurve gleich

${}{\begin{aligned}\int _{\gamma }F&=\int _{0}^{1}\left\langle F(\gamma (t)),\gamma '(t)\right\rangle dt\\&=\int _{0}^{1}{\left(\gamma _{1}(t)^{2}-\gamma _{2}(t)^{3}\right)}\cdot \gamma _{1}'(t)+\gamma _{1}(t)\gamma _{2}(t)\cdot \gamma _{2}'(t)dt\\&=\int _{0}^{1}{\left(t^{4}-(t^{3}-5t)^{3}\right)}\cdot 2t+t^{2}(t^{3}-5t)\cdot (3t^{2}-5)dt\\&=\int _{0}^{1}{\left(t^{4}-t^{9}+15t^{7}-75t^{5}+125t^{3}\right)}2t+t^{2}{\left(3t^{5}-5t^{3}-15t^{3}+25t\right)}dt\\&=\int _{0}^{1}2t^{5}-2t^{10}+30t^{8}-150t^{6}+250t^{4}+3t^{7}-20t^{5}+25t^{3}dt\\&=\int _{0}^{1}-2t^{10}+30t^{8}+3t^{7}-150t^{6}-18t^{5}+250t^{4}+25t^{3}dt\\&={\left(-{\frac {2}{11}}t^{11}+{\frac {10}{3}}t^{9}+{\frac {3}{8}}t^{8}-{\frac {150}{7}}t^{7}-3t^{6}+50t^{5}+{\frac {25}{4}}t^{4}\right)}|_{0}^{1}\\&=47+{\frac {-336+6160+693-39600+11550}{1848}}\\&=47-{\frac {21533}{1848}}\\&={\frac {65323}{1848}}.\end{aligned}}$