求一道二重积分的计算

Question

求∫∫(x²+y²)dxdy,其中区域D为:(x-1)²+y²<=1

Answer 1

用极坐标变换：x=rcosa，y=rsina，对应的积分区域为(rcosa-1)^2+r^2sin^2a<=1，
即r^2<=2rcosa，等价于r<=2cosa。因此必有-pi/2<=a<=pi/2。
于是∫∫(x²+y²)dxdy
=∫_(-pi/2到pi/2) da ∫_(0到2cosa) r^2*rdr
=∫_(-pi/2到pi/2) (2cosa)^4/4 da
=3pi/2。

Answer 2

令x=pcosθ，y=psinθ
x²-2x+y²=0
p²=2pcosθ
p=2cosθ
上半圆为：
{0<=p<=2cosθ
{0<=θ<=π/2
原式=2∫∫p²pdpdθ
=2∫(0,π/2)dθ∫(0,2cosθ)p³dp
=2∫(0,π/2)1/4p^4|(0,2cosθ)dθ
=1/2∫（0，π/2）16cos^4θdθ
=8∫(0,π/2)cos^4θdθ
=8×3/4×1/2×π/2
=(3/2)π

Answer 3