计算二重积分:I=∫∫Df(x,y)dxdy,其中D为x2+y2≤4,f(x,y)=
{1,x2+y2≤1,
xy2,1<x2+y2≤4.
【正确答案】:利用二重积分的可加性,∫∫Df(x,y)dxdy=∫∫D1f(x,y)dxdy+
∫∫D2f(x,y)dxdy,其中D1和 D2分别是由x2+y2≤1和1<x2+y2≤4围成的区域,而∫∫D1f(x,y)dxdy=∫∫D1dxdy=D1的面积=π.
∫∫Df(x,y)dxdy=π+∫02πdθ∫12rcosθ•r2
sin2θ•rdr=π+∫02πcosθ•sin2θ•(1/5)r5∣12
=π+(31/5)•∫02πsin2θd(sinθ)=π+(31/5)•(1/3)sin3θ∣02π=π
计算二重积分:I=∫∫Df(x,y)dxdy,其中D为x2+y2≤4,f(x,y)= {1,x2+y2≤1, xy2,1<x2+
- 2024-11-07 09:09:44
- 高等数学(经管类)(13125)