题目

积分区域D是由y=x,y=2x-x^2所围,二重积分∫∫y^(1/2)dxdy=?

提问时间：2021-01-13

答案

两线交点为(0,0),(1,1)
∫∫ √y dxdy
= ∫(0,1) dx ∫(x,2x - x²) √y dy
= ∫(0,1) (2/3)y^(3/2) |(x,2x - x²) dx
= (2/3)∫(0,1) [(2x - x²)^(3/2) - x^(3/2)] dx
= (2/3)∫(0,1) [1 - (x - 1)²]^(3/2) dx - (2/3)∫(0,1) x^(3/2) dx
x - 1 = sinθ,dx = cosθ dθ
= (2/3)∫(- π/2,0) cos⁴θ dθ - (2/3) * (2/5)x^(5/2) |(0,1)
= (2/3)(3/8)(π/2) - (2/3)(2/5)
= π/8 - 4/15
= (15π - 32)/120