Please show me the full solution, not just the answer, cuz I got the answer.
Integrate this:
J = ∫∫D [(x-y)/(x+y)^3] dxdy
D = {(x,y) | 0 ≤x≤1, 0≤y≤1}
Integrate this:
J = ∫∫D [(x-y)/(x+y)^3] dxdy
D = {(x,y) | 0 ≤x≤1, 0≤y≤1}
-
∫∫D [(x-y)/(x+y)^3] dxdy where D = {(x,y) | 0 ≤x≤1, 0≤y≤1}
= ∫(1, 0) ∫(1, 0) [(x-y)/(x+y)^3] dxdy
= ∫(1, 0) ∫(1, 0) x/(x + y)^3 - y/(x + y)^3 dxdy
The second term is quite easy to integrate, but the first term requires partial fractions, and we can decompose it into 1/(x + y)^2 - y/(x + y)^3
= ∫(1, 0) ∫(1, 0) 1/(x + y)^2 - 2y/(x + y)^3 dxdy
= ∫(1, 0) [-1/(x + y) + y/(x + y)^2] over [1, 0] dy
= ∫(1, 0) [-1/(y + 1) + y/(y + 1)^2 + 1/y - y/y^2] dy
= ∫(1, 0) [-1/(y + 1) + y/(y + 1)^2] dy
= ∫(1, 0) [-1/(y + 1) + 1/(y + 1) - 1/(y + 1)^2] dy
= ∫(1, 0) [-1/(y + 1)^2]
= [1/(y + 1)] over [1, 0]
= 1/2 - 1/1
= -1/2
I think it would be almost similar difficulty if you chose to partially integrate with respect to y first, since the numerator is a linear expression (x - y) so it shouldn't make too much difference.
= ∫(1, 0) ∫(1, 0) [(x-y)/(x+y)^3] dxdy
= ∫(1, 0) ∫(1, 0) x/(x + y)^3 - y/(x + y)^3 dxdy
The second term is quite easy to integrate, but the first term requires partial fractions, and we can decompose it into 1/(x + y)^2 - y/(x + y)^3
= ∫(1, 0) ∫(1, 0) 1/(x + y)^2 - 2y/(x + y)^3 dxdy
= ∫(1, 0) [-1/(x + y) + y/(x + y)^2] over [1, 0] dy
= ∫(1, 0) [-1/(y + 1) + y/(y + 1)^2 + 1/y - y/y^2] dy
= ∫(1, 0) [-1/(y + 1) + y/(y + 1)^2] dy
= ∫(1, 0) [-1/(y + 1) + 1/(y + 1) - 1/(y + 1)^2] dy
= ∫(1, 0) [-1/(y + 1)^2]
= [1/(y + 1)] over [1, 0]
= 1/2 - 1/1
= -1/2
I think it would be almost similar difficulty if you chose to partially integrate with respect to y first, since the numerator is a linear expression (x - y) so it shouldn't make too much difference.