Gebietsintegral übersetzen

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int2-03-01

radio

randRange(1,8) randRangeExclude(1,8,[X])

Gegeben sei das Dreieck \color{orange}D.

graphInit({ range: [[-2, 9],[-9, 9]], scale: [20,20], tickStep: [1,1], gridStep: [1,1], labelStep: [2,2], gridOpacity: 0.1, axisOpacity: 0.8, tickOpacity: 0.6, labelOpacity: 0.8 }); label( [ 0, 9.5 ], "y", "above" ); label( [9.5,0 ], "x", "right" ); line( [0, -Y], [0, Y], { stroke: ORANGE } ); line( [0,-Y], [X, 0], { stroke: ORANGE } ); line( [X,0], [0, Y], { stroke: ORANGE } )

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

\displaystyle 
				\int_{-Y}^{0}  \int_0^{fractionReduce(X,Y,small=true) y + X} f(x,y) \, dx dy
			+ 
			\int_0^{Y}  \int_0^{-fractionReduce(X,Y,small=true) y + X} f(x,y) \, dx dy

\displaystyle \int_{-Y}^{0} \int_0^{fractionReduce(X,Y,small=true) y + X} f(x,y) \, dy dx + \int_0^{Y} \int_0^{-fractionReduce(X,Y,small=true) y + X} f(x,y) \, dy dx
\displaystyle \int_{-X}^{0} \int_0^{fractionReduce(X,Y,small=true) y + X} f(x,y) \, dy dx + \int_0^{X} \int_0^{-fractionReduce(X,Y,small=true) y + X} f(x,y) \, dy dx
\displaystyle 2 \int_0^{X} \int_0^{-fractionReduce(X,Y,small=true) y + X} f(x,y) \, dx dy
\displaystyle \int_{-Y}^{Y} \int_{fractionReduce(X,Y,small=true) y - X}^{-fractionReduce(X,Y,small=true) y + X} f(x,y) \, dx dy
Keines dieser.

Es ist (z.B.) {\orange{D}} \subset \mathbb R^2 gegeben durch {\orange{D}} = \left\{ (x,y) \, | \, -Y \leq y \leq Y, \; 0 \leq x \leq -fractionReduce(X,Y) |y| + X\right\}.

Damit schreibt sich dann \displaystyle \int\int_{\orange{D}} f(x,y) dA = \int_{-Y}^{Y} \int_0^{f_O(y)} f(x,y) \, dx dy mit f_O(y) = -fractionReduce(X,Y) |y| + X. (Cave: Reihenfolge)

Für die innere Integration (zuerst nach x) zerlegt sich das Integral in y \geq 0 und y < 0:

\displaystyle \int_{-Y}^{Y} \int_0^{f_O(y)} f(x,y) \, dx dy = \int_{-Y}^{0} \int_0^{fractionReduce(X,Y,small=true) y + X} f(x,y) \, dx dy + \int_0^{Y} \int_0^{-fractionReduce(X,Y,small=true) y + X} f(x,y) \, dx dy.