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Use the map Φ(u, v) = u + v/2 , u − v/2 to compute ∫∫ᵣ ((x − y) sin(x + y))² dxdy where R is the square with vertices (π, 0), (2π, π), (π, 2π), and (0, π)

User Seanpj
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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.

User Samir Rahimy
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