de-CH
utf-8
math math-format
Kurvenintegral
va-03-01-gpt
multiple
90
randRangeExclude(-4,4,[0]) randRangeExclude(-4,4,[M1,-M1,0]) M1*M1 + N1*N1
M1*M1-N1*N1 2*M1*N1 randRangeExclude(-8,8,[0,1,-1]) randRangeExclude(-8,8,[0,1,-1]) randRangeExclude(-3,3,[0]) randRangeExclude(-3,3,[0])

Gegeben sei die Funktion f: \mathbb R^2 \to \mathbb R mit f(x,y) = A x + B y und das Dreieck unten mit Randkurve \gamma.

Berechnen Sie das Kurvenintegral \displaystyle \int_\gamma f ds.



graphInit({ range: [[-20,20],[-20,20]], scale: [20,20], tickStep: [2,2], gridStep: [1,1], labelStep: [2,2], gridOpacity: 0.1, axisOpacity: 0.8, tickOpacity: 0.6, labelOpacity: 0.8 }); label( [ 0, 20 ], "y", "above" ); label( [20,0 ], "x", "above" ); line( [U, V], [X1+U, V], { stroke: GRAY} ); line( [X1+U,V], [X1+U,V+Y1], { stroke: GRAY} ); line( [U,V], [X1+U,V+Y1], { stroke: GRAY} );
a \displaystyle \int_\gamma f ds= ((A * (X1+U) + B * (V+Y1)) *W1 + abs(Y1)*(B * Y1 + 2*A*U+2*Y1*B+2*A*X1) + abs(X1)*(A * X1 + 2*A*U+2*V*B))/2

Für die Berechnung zerlegen wir das Kurvenintegral \displaystyle \int_\gamma f ds in die Summe über die Integrale entlang der Dreieckskanten:

\displaystyle \int_\gamma f ds = {\color{red} \int_{\gamma_1} f ds} + {\color{blue}\int_{\gamma_2} f ds} + {\color{purple}\int_{\gamma_3} f ds}.

line( [U, V], [X1+U, V], { stroke: RED} ); line( [X1+U,V], [X1+U,V+Y1], { stroke: BLUE} ); line( [U,V], [X1+U,V+Y1], { stroke: PURPLE} );

Mögliche **Parametrisierungen der Kanten** sind:

{\color{red}\gamma_1}(t): [0,1] \to \mathbb R^2, {\color{orange}\gamma_1}(t) = \begin{pmatrix} U\\ V \end{pmatrix} + t\cdot \begin{pmatrix} X1 \\ 0 \end{pmatrix}.

{\color{blue}\gamma_2}(t): [0,1] \to \mathbb R^2, {\color{blue}\gamma_2}(t) = \begin{pmatrix} X1+U\\ V \end{pmatrix} + t\cdot \begin{pmatrix} 0 \\ Y1 \end{pmatrix}.

{\color{purple}\gamma_3}(t): [0,1] \to \mathbb R^2, {\color{purple}\gamma_3}(t) = \begin{pmatrix} X1+U\\ V+Y1 \end{pmatrix} + t\cdot \begin{pmatrix} -X1 \\ -Y1 \end{pmatrix}.

Berechnung des jeweiligen **Kurvenintegrals**

Es ist \displaystyle \int_{\gamma_i} f ds = \int_0^{1} f(\gamma_i(t)) \cdot |{\gamma_i}'(t)| \ dt.

Das Kurvenintegral entlang \gamma_1 ergibt:

\displaystyle {\color{red} \int_{\gamma_1} f ds} = \int_0^{1} (A \cdot (U + X1 t) + B \cdot negParens(V)) \cdot \sqrt{negParens(X1)^2 + 0} \ dt = \int_0^{1} (A*U+V*B + A*X1 \cdot t) \cdot abs(X1) \ dt = {\color{red} fractionReduce(abs(X1)*(A * X1 + 2*A*U+2*V*B) ,2)}.

Das Kurvenintegral entlang \gamma_2 ergibt:

\displaystyle {\color{blue} \int_{\gamma_2} f ds} = \int_0^{1} (A \cdot negParens(U+X1) + B \cdot (V + Y1 t) \cdot \sqrt{ 0+ negParens(Y1)^2} \ dt = \int_0^{1} (A*U+A*X1 +B*Y1 + B*Y1 \cdot t) \cdot abs(Y1) \ dt = {\color{red} fractionReduce(abs(Y1)*(B * Y1 + 2*A*U+2*Y1*B+2*A*X1) ,2)}.

Das Kurvenintegral entlang \gamma_3 ergibt:

\displaystyle {\color{purple} \int_{\gamma_3} f ds} = \int_0^{1} (A \cdot (X1+U + -X1 t) + B \cdot \left(V+Y1 + -Y1 t) \right) \cdot \sqrt{negParens(-X1)^2 + negParens(-Y1)^2} \ dt = \int_0^{1} \left(A*(X1+U)+B*(Y1+V) + -A*X1-B*Y1 \cdot t\right) \cdot W1 \ dt = {\color{red} fractionReduce((A * (X1+U) + B * (V+ Y1)) * W1,2)}.

Gesamtlösung:

\displaystyle \int_\gamma f ds = fractionReduce((A * (X1+U) + B * (V+Y1)) *W1 + abs(Y1)*(B * Y1 + 2*A*U+2*Y1*B+2*A*X1) + abs(X1)*(A * X1 + 2*A*U+2*V*B),2) .