计算二重积分：∫∫D ln(x^2+y^2)dxdy, 其中D为e^2≤x^2+y^2≤e^4

Answer 1

{ x = rcosθ、y = rsinθ
e² ≤ x² + y² ≤ e⁴ → e² ≤ r² ≤ e⁴ → e ≤ r ≤ e²
∫∫_[D] ln(x² + y²) dxdy
= ∫[0→2π] dθ ∫[e→e²] ln(r²) rdr
= 4∫[0→π/2] dθ ∫[e→e²] ln(r²) d(r²/2)
= 4∫[0→π/2] dθ [(1/2)r²ln(r²) - ∫ r dr] |[e→e²]
= 4[π/2] * {[(1/2)(e⁴)ln(e⁴) - (1/2)e⁴] - [(1/2)(e²)ln(e²) - (1/2)e²]}
= 2π * (e²/2)(3e² - 1)
= π(3e² - 1)e² ≈ 491.36