极坐标求二重积分∫∫|x2+y2-4|dxdyD:x2+y2≤16

Question

极坐标求二重积分∫∫|x2+y2-4|dxdyD:x2+y2≤16

Answer 1

化为极坐标
x=pcosa,y=psina
0≤p≤4
0≤a≤2π
∫∫|x2+y2-4|dxdy
=4∫[0,π/2]∫[0,2] (4-p^2)pdpda+4∫[0,π/2]∫[2,4] (p^2-4)pdpda
=4∫[0,π/2]da∫[0,2] (4p-p^3)dp+4∫[0,π/2]da∫[2,4] (p^3-4p)dp
=4*π/2*(2p^2-p^4/4)[0,2]+4*π/2*(p^4/4-2p^2)[2,4]
=2π*(8-4+64-32-4+8)
=80π