Question

计算∫∫D|cos(x+y)|dxdy,D:0<=x<=π/2,0<=y<=π/2

是带绝对值的！在线等！！

Answer 1

记O(0, 0), A(π/2, 0), B(π/2, π/2), C(0, π/2). 则积分域 D: 为正方形 OABC，
连接AC， 则 在D1: △OAC内， x+y<=π/2，
在D2: △ABC内， π/2<=x+y<=π。于是
I = ∫ ∫<D> |cos(x+y)|dxdy = ∫ ∫<D1> cos(x+y)dxdy +∫ ∫<D2> -cos(x+y)dxdy
= ∫<0,π/2>dx∫<0,π/2-x> cos(x+y)dy - ∫<0,π/2>dx∫<π/2-x, π/2> cos(x+y)dy
= ∫<0,π/2>(1-sinx)dx - ∫<0,π/2>(cosx-1)dx
= ∫<0,π/2>(2-sinx-cosx)dx = [2x+cosx-sinx]<0, π/2> = π-2.

Answer 2

可以使用变量替换法.
u=x+y, v=x-y
结果等于 π-2