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Question Number 31084 by abdo imad last updated on 02/Mar/18

find ∫∫_D ((dxdy)/((x+y)^4 )) with D={(x,y)∈R^2 /x≥1,y≥1,x+y≤4}

find∫∫Ddxdy(x+y)4withD={(x,y)∈R2/x⩾1,y⩾1,x+y⩽4}

Commented by abdo imad last updated on 11/Mar/18

x+y ≤4 and y≥1 ⇒−y≤−1 but x ≤4−y ⇒1≤x≤3 and 1≤y≤4−x I =∫_1 ^3 (∫_1 ^(4−x) (dy/((x+y)^4 )))dx but ∫_1 ^(4−x) (x+y)^(−4) dy=[ (1/(−3))(x+y)^(−3) ]_(y=1) ^(y=4−x) =(1/3)(x+1)^(−3) −(1/3)(4)^(−3) =(1/3)(x+1)^(−3) −(1/(3.4^3 )) I=∫_1 ^3 ((1/3)(x+1)^(−3) −(1/(192)))dx=((−2)/(192)) +(1/3)[(1/(−2))(x+1)^(−2) ]_1 ^3 =−(1/(96)) −(1/6)(4^(−2) −2^(−2) )=−(1/(96)) −(1/6)((1/(16)) −(1/4)) =−(1/6) ((−3)/(16)) −(1/(96)) = (1/(32)) −(1/(96)) .

x+y⩽4andy⩾1⇒−y⩽−1butx⩽4−y⇒1⩽x⩽3and1⩽y⩽4−xI=∫13(∫14−xdy(x+y)4)dxbut∫14−x(x+y)−4dy=[1−3(x+y)−3]y=1y=4−x=13(x+1)−3−13(4)−3=13(x+1)−3−13.43I=∫13(13(x+1)−3−1192)dx=−2192+13[1−2(x+1)−2]13=−196−16(4−2−2−2)=−196−16(116−14)=−16−316−196=132−196.

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