cos(x+y)绝对值的二重积分。0≦x≦π/2。0≦y≦π/2

Question

如题所述

Answer 1

简单计算一下即可，答案如图所示

Answer 2

设直线x+y=π/2，将区域D分为D1，D2,
其中D1={(x,y)|0<=x<=π/2,0<=y<=π/2-x},
D2 = {(x,y)|0<=x<=π/2,π/2-x<=y<=π/2},
则∫∫|cos(x+y)|dxdy = (D1)∫∫cos(x+y)dxdy-(D2)∫∫cos(x+y)dxdy,
(D1)∫∫cos(x+y)dxdy = ∫(0至π/2)dx∫(0至π/2-x)cos(x+y)dy = ∫(0至π/2)(1-sinx)dx = π/2-1
(D2)∫∫cos(x+y)dxdy=∫(0至π/2)dx∫(π/2-x至π/2)cos(x+y)dy = ∫(0至π/2)(1-sinx)dx = 1-π/2,
所以∫∫|cos(x+y)|dxdy = π/2-1-(1-π/2) = π-2。

Answer 3

最后的结果是0，先将绝对值去掉，第一次积分时将x当成常数

Answer 4

本回答被网友采纳

Answer 5

先去绝对值符号，由题中给的D可以知道0≤x+y≤π，所以用x+y=二分之π将积分D分成两部分，去绝对值之后I=∫∫D1-∫∫D2=∫0→π/2dx∫0→π/2-x cos(x+y)dy-∫0→π/2dx∫π/2-x→π/2 cos(x+y)dy=（π/2-1）-（1-π/2）=π-2