Question

根号下y-x^2的绝对值的二重积分

二重积分 交换次序
计算二重积分I=∫∫根号（y-x^2）dxdy 其中积分区域D是由0≤y≤2 绝对值X≤1

Answer 1

∫∫_D √(y - x²) dxdy
= ∫(-1-->1) dx ∫(0-->2) √(y - x²) dy
= ∫(-1-->1) dx ∫(0-->2) √(y - x²) d(y - x²)
= ∫(-1-->1) (2/3)(y - x²)^(3/2) |(0-->2) dx
= ∫(-1-->1) (2/3)(2 - x²)^(3/2) dx
= (4/3)∫(0-->1) (2 - x²)^(3/2) dx
令x = √2sinθ,dx = √2cosθdθ
当x = 0,θ = 0,当x = 1,θ = π/4
= (4/3)∫(0-->π/4) (2 - 2sin²θ)^(3/2) √2cosθdθ
= (4/3)(2√2)(√2)∫(0-->π/4) cos⁴θ dθ
= (16/3)∫(0-->π/4) [(1 + cos2θ)/2]² dθ
= (4/3)∫(0-->π/4) (1 + 2cos2θ + cos²2θ) dθ
= (4/3)∫(0-->π/4) (1 + 2cos2θ) dθ + (2/3)∫(0-->π/4) (1 + cos4θ) dθ
= (4/3)(θ + sin2θ) |(0-->π/4) + (2/3)(θ + 1/4 · sin4θ)|(0-->π/4)
= (4/3)(π/4 + 1) + (2/3)(π/4)
= π/2 + 4/3