∫(0→1) ∫(x→√(2x - x²)) xy/(x² + y²) dydx
= ∫(0→1) [∫(x→√(2x - x²)) xy/(x² + y²) dy] dx
= ∫(0→1) [(x/2)ln(x² + y²)] |(x→√(2x - x²)) dx
= ∫(0→1) (x/2)[ln(x² + 2x - x²) - ln(x² + x²)] dx
= ∫(0→1) (x/2)ln[(2x)/(2x²)] dx
= ∫(0→1) (x/2)ln(x⁻¹) dx
= (- 1/2)∫(0→1) xlnx dx
= (- 1/2)(xlnx - x) |(0→1)
= (- 1/2)(- 1)
= 1/2让换成极坐标形式最后答案是1/8谢谢了，麻烦再帮一下∫(0→1) ∫(x→√(2x - x²)) xy/(x² + y²) dydx= ∫(π/4→π/2) ∫(0→2cosθ) (r²cosθsinθ)/r² • rdrdθ= ∫(π/4→π/2) cosθsinθ ∫(0→2cosθ) rdrdθ= ∫(π/4→π/2) cosθsinθ • [r²/2] |(0→2cosθ) dθ= ∫(π/4→π/2) cosθsinθ • (1/2)(4cos²θ) dθ= ∫(π/4→π/2) - 2cos³θ d(cosθ)= (- 1/2)(cos⁴θ) |(π/4→π/2)= (- 1/4)(0 - 1/2)= 1/8那肯定是上面算错了。极坐标不比直角坐标准确的。2cosθ怎么找呀极坐标的总是不会找r的区间，请详细说明非常感谢你直接代入，不用找的嘿嘿y = √(2x - x²)x² + y² = 2x(rcosθ)² + (rsinθ)² = 2(rcosθ)r²(cos²θ + sin²θ) = 2rcosθr = 2cosθ