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  • 積分區(qū)域D是由y=x,y=2x-x^2所圍,二重積分∫∫y^(1/2)dxdy=?

    積分區(qū)域D是由y=x,y=2x-x^2所圍,二重積分∫∫y^(1/2)dxdy=?
    數(shù)學人氣:921 ℃時間:2020-06-21 18:14:22
    優(yōu)質(zhì)解答
    兩線交點為(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
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