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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}
useGreenRiemannformulertodetermined:I=DxydxdyD={(x,y)R2x0;y;x+y1}
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=01(01yxydx)dy=01(01yxdx)ydy=01((1y)22)ydy=1201y(y22y+1)dy=1201(y32y2+y)dy=12[y4423y3+y22]01=12{1423+12}=12{3423}=3813=124.
Commented by Cmr 237 last updated on 05/Sep/19
thk sir
thksir
Commented by mathmax by abdo last updated on 10/Sep/19
you are welcome.
youarewelcome.

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