计算I=∫∫(1-sin²(x+y))½dxdy,其中0≤x≤π/2 ,0≤y≤π/2

计算I=∫∫(1-sin²(x+y))½dxdy,其中0≤x≤π/2 ,0≤y≤π/2
计算I=∫∫(1-sin²(x+y))½dxdy,其中0≤x≤π/2 ,0≤y≤π/2

计算I=∫∫(1-sin²(x+y))½dxdy,其中0≤x≤π/2 ,0≤y≤π/2
积分域用x + y = π/2划分如下,对于函数cos(x + y),区域1为正值,区域2为负值.
∫∫ √[1 - sin²(x + y)] dxdy
= ∫∫ √[cos²(x + y)] dxdy
= ∫∫ |cos(x + y)| dxdy
= ∫(0,π/2) dx [∫(0,π/2 - x) cos(x + y) dx + ∫(π/2 - x,π/2) (- )cos(x + y) dx ]
= ∫(0,π/2) [sin(x + y)：(0,π/2 - x)] dx - ∫(0,π/2) [sin(x + y)：(π/2 - x,π/2)] dx
= ∫(0,π/2) [sin(x + π/2 - x) - sinx] dx - ∫(0,π/2) [sin(x + π/2) - sin(x + π/2 - x)] dx
= ∫(0,π/2) (1 - sinx) dx - ∫(0,π/2) (cosx - 1) dx
= (x + cosx - sinx + x)：(0,π/2)
= (π/2 + 0 - 1 + π/2) - (0 + 1 - 0 + 0)
= π - 2