D is limited by x^2+y^2>=1
∫∫ 1/(x² + y²)^a dx dy
over the unit circle ......... I assume you mean x² + y² ≤ 1
let x = r cos(θ) and y = r sin(θ) ; dx dy = r dr dθ
= ∫∫ r^(1-2a) dr dθ [0,1] [0,2π]
= π/(1-a)
provided a < 1, otherwise I'm not sure the integral converges.
If x² + y² ≥ was correct, then
= π/(a-1) ; provided a > 1, otherwise I'm not sure the integral converges.
Answer: π/(1-a)
â«â« 1/(x² + y²)^a dx dy
Covert to polar
= â«â« r^(1-2a) dr dt [1,inf] [0,2Ï], this integral depends on value of a for converge
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∫∫ 1/(x² + y²)^a dx dy
over the unit circle ......... I assume you mean x² + y² ≤ 1
let x = r cos(θ) and y = r sin(θ) ; dx dy = r dr dθ
= ∫∫ r^(1-2a) dr dθ [0,1] [0,2π]
= π/(1-a)
provided a < 1, otherwise I'm not sure the integral converges.
If x² + y² ≥ was correct, then
= π/(a-1) ; provided a > 1, otherwise I'm not sure the integral converges.
Answer: π/(1-a)
â«â« 1/(x² + y²)^a dx dy
Covert to polar
= â«â« r^(1-2a) dr dt [1,inf] [0,2Ï], this integral depends on value of a for converge