Question

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

追问

∫(-1-->1) (2/3)(2 - x²)^(3/2) dx
这里把2带进去 0不带进去算吗

追答

其实跳了几步的。
∫(-1-->1) (2/3)(2 - x²)^(3/2) dx
= 2∫(0-->1) (2/3)(2 - x²)^(3/2) dx
= (4/3)∫(0-->1) (2 - x²)^(3/2) dx
这个一定要用第二换元法(三角函数)代换做了。因为可消除根号
令x = √2sinθ，(2 - x²)^(3/2) = (2 - 2sin²θ)^(3/2) = (2cos²θ)^(3/2) = 2^(3/2) · cos³θ
在还未计算出被积函数的原函数之前，不能直接带入上下限。