Question

求积分 ∫ ∫xydxdy,区域D由y=x,x+y=2,x=2围成。

答案是10/3,我算的是5/3。

Answer 1

y=x与x+y=2的交点的坐标(1,1)
∫(1->2)xdx∫(x-2->x)ydy
=1/2∫(1->2)xdx y^2|(x-2->x)
=1/2∫(1->2)xdx (x^2-(x-2)^2)
=1/2∫(1->2)(4x-4)xdx
=2∫(1->2)x^2dx -2∫(1->2)xdx
=2/3x^3(1->2)-x^2(1->2)
=2/3 (2^3-1^3)-(2^2-1^2)
=14/3-3
=5/3
我也算出是5/3

追问

答案坑爹啊！！！

追答

唉

Answer 2

∫ ∫xydxdy
=∫(1,2)xdx ∫(2-x,x)ydy
=(1/2)∫(1,2)x(x^2-(2-x)^2)dx
=2∫(1,2)(x^2-x)dx
=2(x^3/3-x^2/2)|(1,2)
=2(8/3-2-1/3+1/2)
=5/3

Answer 3

1<x<2
2-x<y<x