Wegintegral/Vektorfeld/(2t-1,t^2+1)/-1 bis 1/(xy^2-x,2xy-y^2)/Aufgabe/Lösung

Die Ableitung der Kurve ist

{}\gamma '(t)={\begin{pmatrix}2\\2t\end{pmatrix}}\,

und das Vektorfeld auf dem Weg ausgewertet ist

{}{\begin{aligned}F(\gamma (t))&={\begin{pmatrix}\gamma _{1}(t)\gamma _{2}(t)^{2}-\gamma _{1}(t)\\2\gamma _{1}(t)\gamma _{2}(t)-\gamma _{2}(t)^{2}\end{pmatrix}}\\&={\begin{pmatrix}{\left(2t-1\right)}{\left(t^{2}+1\right)}^{2}-{\left(2t-1\right)}\\2{\left(2t-1\right)}{\left(t^{2}+1\right)}-{\left(t^{2}+1\right)}^{2}\end{pmatrix}}\\&={\begin{pmatrix}{\left(2t-1\right)}{\left(t^{4}+2t^{2}+1\right)}-{\left(2t-1\right)}\\2{\left(2t-1\right)}{\left(t^{2}+1\right)}-{\left(t^{4}+2t^{2}+1\right)}\end{pmatrix}}\\&={\begin{pmatrix}2t^{5}+4t^{3}+2t-t^{4}-2t^{2}-1-2t+1\\4t^{3}+4t-2t^{2}-2-t^{4}-2t^{2}-1\end{pmatrix}}\\&={\begin{pmatrix}2t^{5}-t^{4}+4t^{3}-2t^{2}\\-t^{4}+4t^{3}-4t^{2}+4t-3\end{pmatrix}}.\end{aligned}}

Damit ist das Wegintegral gleich

{}{\begin{aligned}\int _{\gamma }F&=\int _{-1}^{1}\left\langle {\begin{pmatrix}2t^{5}-t^{4}+4t^{3}-2t^{2}\\-t^{4}+4t^{3}-4t^{2}+4t-3\end{pmatrix}},{\begin{pmatrix}2\\2t\end{pmatrix}}\right\rangle dt\\&=\int _{-1}^{1}2{\left(2t^{5}-t^{4}+4t^{3}-2t^{2}\right)}+2t{\left(-t^{4}+4t^{3}-4t^{2}+4t-3\right)}dt\\&=\int _{-1}^{1}2t^{5}+6t^{4}+4t^{2}-6tdt\\&={\left({\frac {1}{3}}t^{6}+{\frac {6}{5}}t^{5}+{\frac {4}{3}}t^{3}-3t^{2}\right)}_{-1}^{1}\\&={\frac {12}{5}}+{\frac {8}{3}}\\&={\frac {76}{15}}.\end{aligned}}