Esercizio – calcolo dell’integrale curvilineo di un campo su una curva

Calcolare l’integrale curvilineo del campo \(F=\left( \frac{{{x}^{2}}y+{{y}^{3}}+y}{{{x}^{2}}+{{y}^{2}}},\frac{x{{y}^{2}}+{{x}^{3}}-x}{{{x}^{2}}+{{y}^{2}}} \right)\) lungo il bordo del quadrato \(Q=\left[ -1,1 \right]\times \left[ -1,1 \right]\) percorso in senso antiorario e dire se il campo è conservativo.

 Soluzione

Una parametrizzazione del quadrato è data da:

\({{\gamma }_{1}}=\left( 1,t \right)\,\,\,,\,\,t\in \left[ -1,1 \right]\,\,\,\,\,\,{{\gamma }_{1}}^{\prime }=\left( 0,1 \right)\)

\({{\gamma }_{2}}=\left( t,1 \right)\,\,\,,\,\,t\in \left[ 1,-1 \right]\,\,\,\,\,\,{{\gamma }_{2}}^{\prime }=\left( 1,0 \right)\)

\({{\gamma }_{3}}=\left( -1,t \right)\,\,\,,\,\,t\in \left[ 1,-1 \right]\,\,\,\,\,\,{{\gamma }_{3}}^{\prime }=\left( 0,1 \right)\)

\({{\gamma }_{4}}=\left( t,-1 \right)\,\,\,,\,\,t\in \left[ -1,1 \right]\,\,\,\,\,\,{{\gamma }_{3}}^{\prime }=\left( 1,0 \right)\)

\(\gamma ={{\gamma }_{1}}\oplus {{\gamma }_{2}}\oplus {{\gamma }_{3}}\oplus {{\gamma }_{4}}\)

\(d\mathbf{l}={\gamma }’dt\)

\(\int\limits_{\gamma }{\mathbf{F}\,\centerdot \,\,d\mathbf{l}}=\int\limits_{{{\gamma }_{1}}}{\mathbf{F}}\centerdot d\mathbf{l}+\int\limits_{{{\gamma }_{2}}}{\mathbf{F}\centerdot d\mathbf{l}}+\int\limits_{{{\gamma }_{3}}}{\mathbf{F}\centerdot d\mathbf{l}}+\int\limits_{{{\gamma }_{4}}}{\mathbf{F}\centerdot d\mathbf{l}}\)

\(\int\limits_{-1}^{1}{\frac{{{t}^{2}}+1-1}{{{t}^{2}}+1}dt}+\int\limits_{1}^{-1}{\frac{{{t}^{2}}+1+1}{{{t}^{2}}+1}dt+}\int\limits_{1}^{-1}{\frac{{{t}^{2}}-1+1}{{{t}^{2}}+1}dt+}\int\limits_{-1}^{1}{\frac{{{t}^{2}}-1-1}{{{t}^{2}}+1}dt}\)

\(\int\limits_{-1}^{1}{1-\frac{1}{{{t}^{2}}+1}dt}-\int\limits_{-1}^{1}{1+\frac{1}{{{t}^{2}}+1}dt-}\int\limits_{-1}^{1}{1-\frac{1}{{{t}^{2}}+1}dt+}\int\limits_{-1}^{1}{1-\frac{3}{{{t}^{2}}+1}dt}\)

\(-\int\limits_{-1}^{1}{\frac{1}{{{t}^{2}}+1}dt}-\int\limits_{-1}^{1}{\frac{1}{{{t}^{2}}+1}dt+}\int\limits_{-1}^{1}{\frac{1}{{{t}^{2}}+1}dt}-\int\limits_{-1}^{1}{\frac{3}{{{t}^{2}}+1}dt}=-4\int\limits_{-1}^{1}{\frac{1}{{{t}^{2}}+1}dt}=\left[ -4\arctan t \right]_{-1}^{1}=-2\pi \)

Il campo non è conservativo perché la circuitazione su un percorso chiuso è diversa da zero.

Lezioni di Analisi Matematica 2