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

Wir betrachten das Vektorfeld

$F:{ℝ}^2⟶{ℝ}^2,(x,y)⟼(x^2-y^3,xy)$

und den Weg

$\gamma :[0,1]⟶{ℝ}^2,t⟼(t^2,t^3-5t).$

Die Ableitung von $\gamma$ ist

${\gamma }^{\prime }(t)=(2t,3t^2-5).$

Daher ist das Wegintegral zu diesem Vektorfeld längs dieser Kurve gleich

$\begin{array}{cc}{\int }_{\gamma }F& ={\int }_0^1⟨F(\gamma (t)),{\gamma }^{\prime }(t)⟩dt\\ & ={\int }_0^1({\gamma }_1(t{)}^2-{\gamma }_2(t{)}^3)\cdot {{\gamma }_1}^{\prime }(t)+{\gamma }_1(t){\gamma }_2(t)\cdot {{\gamma }_2}^{\prime }(t)dt\\ & ={\int }_0^1(t^4-(t^3-5t{)}^3)\cdot 2t+t^2(t^3-5t)\cdot (3t^2-5)dt\\ & ={\int }_0^1(t^4-t^9+15t^7-75t^5+125t^3)2t+t^2(3t^5-5t^3-15t^3+25t)dt\\ & ={\int }_0^{1}2t^5-2t^{10}+30t^8-150t^6+250t^4+3t^7-20t^5+25t^{3}dt\\ & ={\int }_0^1-2t^{10}+30t^8+3t^7-150t^6-18t^5+250t^4+25t^{3}dt\\ & =(-\frac{2}{11}{t}^{11}+\frac{10}{3}{t}^9+\frac{3}{8}{t}^8-\frac{150}{7}{t}^7-3t^6+50t^5+\frac{25}{4}{t}^4){|}_0^1\\ & =47+\frac{-336+6160+693-39600+11550}{1848}\\ & =47-\frac{21533}{1848}\\ & =\frac{65323}{1848}.\end{array}$