跪求高人指导计算二重积分 ∫∫(X+Y)dθ,其中D={(X,Y)⼀X^2+Y^2

2024-11-07 09:36:56
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回答(1):

积分域:
D={(x,y)|x^2+y^2≤x+y+1}={(x,y)|x^2+y^2-x-y-1≤0}={(x,y)|(x-1/2)^2+(y-1/2)^2≤3/2}
积分域为圆心为(1/2,1/2)半径为√6/2的圆的内部(包括边界)
转换成极坐标:
设x-1/2=rcosθ,即x=(1/2)+rcosθ
y-1/2=rsinθ,即y=(1/2)+rsinθ
则积分域为
0≤θ≤2π,0≤r≤√6/2
∴∫∫(x+y)dxdy =∫[∫(1+r(cosθ+sinθ))rdr]dθ
=∫[(1/2)r^2+(1/3)r^3(cosθ+sinθ)]|dθ
=∫[3/4+(√6/4)(cosθ+sinθ)]dθ
=[(3/4)θ+(√6/4)(-sinθ+cosθ)]|
=3π/2