Parametrisation of Double Integral

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int2-03-01

radio

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

Given the triangle \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 } )

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

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

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

A paramatrisation of {\orange{D}} \subset \mathbb R^2 could be given by {\orange{D}} = \left\{ (x,y) \, | \, 0 \leq x \leq X, \; fractionReduce(Y,X) x - Y \leq y \leq -fractionReduce(Y,X) x +Y\right\}.

Using and applying the defintion of the integral yield \displaystyle \int\int_{\orange{D}} f(x,y) dA = \int_{0}^{X} \int_{fractionReduce(Y,X,small=true) x - Y}^{-fractionReduce(Y,X,small=true) x +Y} f(x,y) \, dy dx