calculate ∫∫_D (x^2 −y^2 )dxdy with D = [−1,1]^2


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Question Number 41343 by maxmathsup by imad last updated on 05/Aug/18

$c a l c u l a t e \int \int_{D} (x^{2} - y^{2}) d x d y w i t h$ $D = {[- 1, 1]}^{2}$
Commented by maxmathsup by imad last updated on 08/Aug/18
$I = ∫_(−1) ^1 ( ∫_(−1) ^1 (x^2 −y^2 )dx)dy but ∫_(−1) ^1 (x^2 −y^2 )dx =2 ∫_0 ^1 (x^2 −y^2 )dx = 2[(x^3 /3) −y^2 x]_0 ^1 =2 { (1/3) −y^2 } =(2/3) −2y^2 ⇒ I = ∫_(−1) ^1 ((2/3)−2y^2 )dy =(4/3) −2 ∫_(−1) ^1 y^2 dy =(4/3) −4 ∫_0 ^1 y^2 dy =(4/3) −4 [(y^3 /3)]_0 ^1 =(4/3) −(4/3) ⇒ I =0$
$I = \int_{- 1}^{1} (\int_{- 1}^{1} (x^{2} - y^{2}) d x) d y b u t$ $\int_{- 1}^{1} (x^{2} - y^{2}) d x = 2 \int_{0}^{1} (x^{2} - y^{2}) d x = 2 {[\frac{x^{3}}{3} - y^{2} x]}_{0}^{1}$ $= 2 {\frac{1}{3} - y^{2}} = \frac{2}{3} - 2 y^{2} \Rightarrow I = \int_{- 1}^{1} (\frac{2}{3} - 2 y^{2}) d y$ $= \frac{4}{3} - 2 \int_{- 1}^{1} y^{2} d y = \frac{4}{3} - 4 \int_{0}^{1} y^{2} d y = \frac{4}{3} - 4 {[\frac{y^{3}}{3}]}_{0}^{1} = \frac{4}{3} - \frac{4}{3} \Rightarrow$ $I = 0$
	Answered by alex041103 last updated on 08/Aug/18

	$\int \int_{D} (x^{2} - y^{2}) d x d y = \int_{- 1}^{1} \int_{- 1}^{1} (x^{2} - y^{2}) d x d y =$ $= \int_{- 1}^{1} {[\frac{x^{3}}{3} - y^{2} x]}_{x = - 1}^{x = 1} d y =$ $= \int_{- 1}^{1} (\frac{2}{3} - 2 y^{2}) d y = {[\frac{2}{3} y - \frac{2 y^{3}}{3}]}_{y = - 1}^{y = 1} =$ $= \frac{4}{3} - \frac{4}{3} = 0$ $\int \int_{D} (x^{2} - y^{2}) d x d y = 0$
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