设平面区域D::x^2+y^2≤1,则∫∫D(1-√(x^2+y^2))dxdy=?
问题描述：
设平面区域D::x^2+y^2≤1,则∫∫D(1-√(x^2+y^2))dxdy=?
答
取D：x² + y² ≤ 1
∫∫D [1 - √(x² + y²)] dxdy
= ∫(0,2π) dθ ∫(0,1) (1 - r)r dr
= 2π∫(0,1) (r - r²) dr
= 2π * (1/2 - 1/3)
= π/3