∫∫√∣y-x^2∣dxdy,D:-1≤x≤1,0≤y≤2请给出详细过程谢谢

Question

∫∫√∣y-x^2∣dxdy,D:-1≤x≤1,0≤y≤2
请给出详细过程谢谢

Answer 1

∫∫ √|y - x²| dxdy
= ∫(- 1→1) dx ∫(0→2) √|y - x²| dy
= ∫(- 1→1) dx [ ∫(0→x²) √|y - x²| dy + ∫(x²→2) √|y - x²| dy ]
= ∫(- 1→1) dx [ ∫(0→x²) √(x² - y) dy + ∫(x²→2) √(y - x²) dy ]
= ∫(- 1→1) [ (2/3)|x|³ + (2/3)(2 - x²)^(3/2) ] dx
= (4/3)∫(0→1) x³ dx + (4/3)∫(0→1) (2 - x²)^(3/2) dx、可令x = √2sinθ计算
= (4/3)(1/4)[ x⁴ ]：(0→1) + (4/3)[ (3/2)arcsin(x/√2) - (x/4)(x² - 5)√(2 - x²) ]：(0→1)
= 1/3 + (4/3)[ (3/2)(π/4) - (1/4)(- 4) ]
= 1/3 + 4/3 + π/2
= 5/3 + π/2