What region is described by 0 <= y <= x, 0 <= x <= a?

The triangle with vertices (0,0), (a,0), (a,a)

The triangle with vertices (0,0), (0,a), (a,a)

The triangle with vertices (0,0), (a,0), (0,a)

After changing the order, what are the limits?

y from 0 to a; for each y, x from y to a

x from 0 to a; for each x, y from x to a

y from 0 to a; for each y, x from 0 to y

y from 0 to a; for each y, x from y to 0

I = 2 sin a, because sqrt(a - y) cancels leaving 2 cos y

I = sin a, because the limits become constants

I = 2 cos a, because sqrt(a - y) cancels leaving 2 sin y

I = 2 sin a, because cos y integrates to a constant

Short answer

Changing the order: y from 0 to a, x from y to a. The inner integral becomes 2√(a-y), cancelling the square root, leaving 2∫₀ᵃ cos y dy = 2 sin a.

Change the order of integration and hence evaluate the same: ∫₀ᵃ ∫₀ˣ cos y dy dx/√((a−x)(a−y)).

I = ∫₀^a ∫₀^x cos y dy dx / √((a − x)(a − y))

For the given order, y runs from 0 to x and x from 0 to a: the triangle 0 ≤ y ≤ x ≤ a (vertices (0,0), (a,0), (a,a)).

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

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

I = ∫₀^a [ cos y / √(a − y) ] [ ∫_y^a dx / √(a − x) ] dy

Inner integral: ∫_y^a (a − x)^(−1/2) dx = [ −2√(a − x) ]_y^a = 2√(a − y).

I = ∫₀^a [ cos y / √(a − y) ] · 2√(a − y) dy
  = 2 ∫₀^a cos y dy
  = 2 sin a

Changing the order removes the awkward square roots, which is why it helps.

Answer: I = 2 sin a.

✓ Answer checked by Claude Opus 5.5 (high effort)