use Green−Riemann formuler to determined: I=∫∫_D xydxdy D={(x,y)∈R^2 ∣x≥0;y≥;x+y≤1}


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Question Number 67946 by Cmr 237 last updated on 02/Sep/19
$use Green−Riemann formuler to determined: I=∫∫_D xydxdy D={(x,y)∈R^2 ∣x≥0;y≥;x+y≤1}$
$use Green - Riemann formuler$ $to determined :$ $I = \int \int_{D} xy dxdy$ $D = {(x, y) \in R^{2} ∣ x ⩾ 0; y ⩾; x + y ⩽ 1}$
Commented by mathmax by abdo last updated on 02/Sep/19
$I =∫_0 ^1 (∫_0 ^(1−y) xydx)dy =∫_0 ^1 (∫_0 ^(1−y) xdx)ydy =∫_0 ^1 ((((1−y)^2 )/2))ydy =(1/2) ∫_0 ^1 y(y^2 −2y+1)dy =(1/2)∫_0 ^1 (y^3 −2y^2 +y)dy =(1/2)[(y^4 /4)−(2/3)y^3 +(y^2 /2)]_0 ^1 =(1/2){(1/4)−(2/3)+(1/2)} =(1/2){(3/4)−(2/3)} =(3/8)−(1/3) =(1/(24)) .$
$I = \int_{0}^{1} (\int_{0}^{1 - y} x y d x) d y = \int_{0}^{1} (\int_{0}^{1 - y} x d x) y d y$ $= \int_{0}^{1} (\frac{{(1 - y)}^{2}}{2}) y d y = \frac{1}{2} \int_{0}^{1} y (y^{2} - 2 y + 1) d y$ $= \frac{1}{2} \int_{0}^{1} (y^{3} - 2 y^{2} + y) d y = \frac{1}{2} {[\frac{y^{4}}{4} - \frac{2}{3} y^{3} + \frac{y^{2}}{2}]}_{0}^{1}$ $= \frac{1}{2} {\frac{1}{4} - \frac{2}{3} + \frac{1}{2}} = \frac{1}{2} {\frac{3}{4} - \frac{2}{3}} = \frac{3}{8} - \frac{1}{3} = \frac{1}{24} .$
Commented by Cmr 237 last updated on 05/Sep/19

$thk sir$
Commented by mathmax by abdo last updated on 10/Sep/19

$y o u a r e w e l c o m e .$
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