利用极坐标计算二重积分∫∫(x^2+y^2)^(-1⼀2)dxdy,D:y=x与y=x^2所围成。

2025-01-20 19:17:07
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回答1:

极坐标方法:
{ x = rcosθ
{ y = rsinθ
1/√(x² + y²) = 1/√(r²cos²θ + r²sin²θ) = 1/r
y = x ==> θ = π/4
y = x² ==> rsinθ = r²cos²θ ==> sinθ = rcos²θ ==> r = secθtanθ
∫∫_D 1/√(x² + y²) dxdy
= ∫(0→π/4) dθ ∫(0→secθtanθ) 1/r · r dr
= ∫(0→π/4) dθ · r:(0→secθtanθ)
= ∫(0→π/4) secθtanθ dθ
= secθ:(0→π/4)
= sec(π/4) - sec(0)
= √2 - 1