Change order: x/(x²+y²)

Short answer

Reverse the order to x from 0 to a, y from 0 to x. The integral becomes ∫₀ᵃ ∫₀ˣ x dy dx/(x²+y²) = ∫₀ᵃ (π/4) dx = πa/4.

Evaluate ∫₀ᵃ ∫_y^a x dx dy/(x²+y²) by changing order of integration. [Limits as printed: outer 0 to a, inner y to a; scan unclear]

How I read the limits. As printed: I = ∫₀^a ∫_y^a x dx dy/(x² + y²), with y the outer variable (0 to a) and x the inner variable (from y to a).

0 ≤ y ≤ a and y ≤ x ≤ a. This is the triangle with vertices (0,0), (a,0), (a,a), lying between y = 0, y = x and x = a.

y
a |       *(a,a)
  |     / |
  |   /   |
  | /_____|__ x
 0        a

Reversing the order: x runs from 0 to a, and for each x, y runs from 0 to x.

I = ∫₀^a x [ ∫₀^x dy/(x² + y²) ] dx

Inner: ∫₀^x dy/(x² + y²) = (1/x) tan⁻¹(y/x) |₀^x = (1/x)(π/4).

I = ∫₀^a x · (π/4x) dx = (π/4) ∫₀^a dx = π a/4

Answer: π a/4.

✓ Answer checked by Claude Opus 5.5 (high effort)

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