The value of the double integral is 16π/5.
Consider the change of variables given by Φ(u, v) = (u + v/2, u - v/2). The Jacobian of Φ is given by J(Φ) = 1/2. Let R' be the image of R under Φ. Then, we have |J(Φ)| = 1/2.
To compute the double integral using Φ, we first need to find the bounds for R'. The vertices of R are mapped to the following points under Φ:
(π, 0) → (3π/2, π/2)
(2π, π) → (5π/2, 3π/2)
(π, 2π) → (3π/2, 5π/2)
(0, π) → (π/2, 3π/2)
Therefore, R' is the rectangle with vertices (π/2, 3π/2), (5π/2, 3π/2), (5π/2, 7π/2), and (π/2, 7π/2).
We can now rewrite the double integral using Φ:
∫∫ᵣ ((x − y) sin(x + y))² dxdy = ∫∫R' ((u − v) sin(u + v))^2 |J(Φ)| du dv
= ∫∫R' ((u − v) sin(u + v))^2 / 2 du dv
Substituting the expression for Φ(u, v) into the integrand, we get:
∫∫R' (((u + v/2) - (u - v/2)) sin((u + v/2) + (u - v/2))^2 / 2 du dv
= ∫∫R' (v sin(2u))^2 / 2 du dv
To simplify the integrand, we can use the double-angle identity:
sin(2u) = 2 sin(u) cos(u)
Substituting this identity into the integrand, we get:
∫∫R' (4v sin^2(u) cos^2(u))^2 / 2 du dv
= ∫∫R' 4v sin^4(u) cos^4(u) du dv
We can now evaluate the double integral using Fubini's Theorem. The first integral is:
∫∫R' 4v sin^4(u) cos^4(u) du dv
= 4∫0π/2 ∫π/2 7π/2 sin^4(u) cos^4(u) dv du
The second integral is:
4∫0π/2 ∫π/2 7π/2 sin^4(u) cos^4(u) dv du
= 16∫0π/2 sin^4(u) cos^4(u) du
We can use the following trigonometric identity to simplify the integral:
sin^2(2u) = 1 - 2cos(4u)
Substituting this identity into the integrand, we get:
16∫0π/2 sin^4(u) cos^4(u) du
= 16∫0π/2 (1 - 2cos(4u))^2 du
Expanding the square and evaluating the integral, we get:
16∫0π/2 (1 - 4cos(4u) + 4cos^2(4u)) du
= 16π/5
Therefore, the value of the double integral is 16π/5.