计算(x+y2)dxdy,其中D:x2+y2≤2x+2y-1.

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问题 计算(x+y2)dxdy,其中D:x2+y2≤2x+2y-1.

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答案D:x2+y2≤2x+2y-1可化为D:(x-1)2+(y-1)2≤1, 令[*] 0≤t≤2π,0≤r≤1, 则[*](x+y2)dxdy=∫0dt∫01(1+rcost+1+2rsint+r2sin2t)rdr =∫0(1+[*]sin2t)dt=2π+[*]sin2tdt=2π+[*]

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