Gebietsintegral über Dreieck

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int2-01-02

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10192

randRangeExclude(-8,8,[0,1,-1]) randRangeExclude(-8,8,[0,1,A,-1]) randRange(1,8) randRangeExclude(1,8,[X])

Gegeben sei die Funktion f: \mathbb R^2 \to \mathbb R mit f(x,y) = A x + B y.

Berechnen Sie das Integral \displaystyle \int \int_{{\color{orange}D}} f(x,y) dA über dem Gebiet \color{orange}D.

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\displaystyle \int \int_{{\color{orange}D}} f(x,y) dA =

(X*X*X*(A*X*Y+B*B*X*X)*2 - 3*Y*Y*(A*Y+2*B*X)*(-X)- 6*Y*Y*B*Y*Y*(-X) + X*X*X*(-A*X*Y+B*B*X*X)*2 + 3*Y*Y*(A*Y-2*B*X)*X+ 6*Y*Y*B*Y*Y*X)/(6*Y*Y)

Es ist \displaystyle \int\int_{\orange{D}} f(x,y) dA = \int_{-X}^{X} \int_0^{f_O(x)} (Ax + By) \; dy \; dx mit f_O(x) = - fractionReduce(Y,X) |x| + Y.

Für die innere Integration zerlegen wir das Integral in x \geq 0 und x < 0:

\displaystyle \int_{-X}^{X} \int_0^{f_O(x)} (Ax + By) \; dy \; dx = \int_{-X}^{0} \int_0^{fractionReduce(Y,X,small=true) x + Y} (Ax + By) \; dy \; dx + \int_0^{X} \int_0^{-fractionReduce(Y,X,small=true) x + Y} (Ax + By) \; dy \; dx

und verwenden \displaystyle \int (Ax + By) \; dy = Axy + fractionReduce(B,2)y^2 +C.

Damit erhalten wir für die äussere Integration:

\displaystyle \int_{-X}^{0} fractionReduce(A*X*Y+B*B*X*X,Y*Y) x^2 + A*Y+2*B*X x + B*Y*Y \; dx + \int_0^{X} fractionReduce(-A*X*Y+B*B*X*X,Y*Y) x^2 + A*Y-2*B*X x + B*Y*Y \; dx = fractionReduce(X*X*X*(A*X*Y+B*B*X*X)*2 - 3*Y*Y*(A*Y+2*B*X)*(-X)- 6*Y*Y*B*Y*Y*(-X) ,6*Y*Y) + fractionReduce(X*X*X*(-A*X*Y+B*B*X*X)*2 + 3*Y*Y*(A*Y-2*B*X)*X+ 6*Y*Y*B*Y*Y*X ,6*Y*Y) = fractionReduce(X*X*X*(A*X*Y+B*B*X*X)*2 - 3*Y*Y*(A*Y+2*B*X)*(-X)- 6*Y*Y*B*Y*Y*(-X) + X*X*X*(-A*X*Y+B*B*X*X)*2 + 3*Y*Y*(A*Y-2*B*X)*X+ 6*Y*Y*B*Y*Y*X,6*Y*Y).