I=∫∫D（x^2+4y^2+9）dδ,其中D={(x,y)|x^2+y^2

Question

I=∫∫D（x^2+4y^2+9）dδ,其中D={(x,y)|x^2+y^2

Answer 1

x² + y² ≤ 4 → r² ≤ 4 → 0 ≤ r ≤ 2、0 ≤ θ ≤ 2π
I = ∫∫_D (x² + 4y² + 9) dσ
= ∫(0→2π) dθ ∫(0→2) (r²cos²θ + 4r²sin²θ + 9) * rdr
= ∫(0→2π) dθ ∫(0→2) (r³cos²θ + 4r³sin²θ + 9r) dr
= ∫(0→2π) [(1/4)r⁴cos²θ + r⁴sin²θ + (9/2)r²] |(0→2)
= ∫(0→2π) (4cos²θ + 16sin²θ + 18) dθ
= ∫(0→2π) (28 - 6cos2θ) dθ
= [ 28θ - 3sin2θ ] |(0→2π)
= 28(2π)
= 56π