Question

求∫∫D㏑﹙1＋x²＋y²﹚dxddy，D：﹛﹙x，y﹚|x²＋y²≦1﹜，要有过程

Answer 1

Let { x = rcosθ {y = rsinθ
x² + y² = 1 ==> r = 1，where 0 ≤ θ ≤ 2π
∫∫D ln[1 + x² + y²] dxdy
= ∫(0~2π) dθ ∫(0~1) ln[1 + r²] rdr
= 4∫(0~π/2) dθ ∫(0~1) [1/2]ln[1 + r²] d[1 + r²]
= 2∫(0~π/2) dθ * [(1 + r²)ln(1 + r²) |(0~1) - ∫(0~1) (1 + r²)/(1 + r²) * 2r dr]
= 2∫(0~π/2) dθ * [2ln(2) - ∫(0~1) 2r dr]
= 2∫(0~π/2) dθ * [2ln(2) - (r²) |(0~1)]
= 2[π/2 - 0] * [2ln(2) - 1]
= π[2ln(2) - 1]