∫∫(D)e^[-√(x^2+y^2)]dxdy,D为x^2+y^2≤1在第一象限部分

2024-11-08 17:58:54
推荐回答(1个)
回答(1):

∫∫ e^[-√(x²+y²)]dxdy
=∫∫ e^(-r)*r drdθ
=∫[0→π/2] dθ∫[0→1] re^(-r) dr
=(π/2)∫[0→1] re^(-r) dr
=-(π/2)∫[0→1] r d(e^(-r))
=-(π/2)re^(-r)+(π/2)∫[0→1] e^(-r) dr
=-(π/2)re^(-r)-(π/2)e^(-r) |[0→1]
=(π/2)(1-e⁻¹-e⁻¹)
=(π/2)(1-2/e)