计算二重积分∫∫(4-x-y)dxdy,D是区域x^2+y^2<=2y

2024-11-08 11:10:44
推荐回答(1个)
回答(1):

x=rcost,y=rsint,代入方程得r^2<=2rsint,于是0<=r<=2sint,故sint必须大于等于0,也就是
0<=t<=pi。
∫∫(4-x-y)dxdy
=∫
(从0到pi)dt

(从0到2sint)
(4-rcost-rsint)*rdr
=∫
(从0到pi)dt
[2r^2-(cost+sint)*r^3/3]|上限2sint下限0
=∫
(从0到pi)
[8sin^2t
-(cost+sint)*8sin^3t
/3]dt
这一步利用二倍角公式:
sin^2t=(1-cos2t)/2,sin^4t=(1-cos2t)^2/4=[1-2cos2t+(1+cos4t)/2]/4
=3/8-0.5cos2t+0.125cos4t
=3pi。