The value of the integral is 2√2 + 2/9.
To evaluate the double integral ∫∫ᵣ (x² - xy + y²) dA, where R is the region bounded by the ellipse x² - xy + y² = 2, we can use the given transformation x = √2 u - √2/3 v and y = √2 u + √2/3 v.
First, we need to find the bounds of integration for u and v. The ellipse x² - xy + y² = 2 can be rewritten as (x - y/2)² = 3(y/2)².
This is the equation of an ellipse centered at (1/2, 0) with semi-axes of length √3 and 1, respectively.
Therefore, the bounds of integration are 0 ≤ u ≤ √3 and -1 ≤ v ≤ 1.
Next, we need to substitute the transformation into the integral. We have:
x² - xy + y² = (√2 u - √2/3 v)² - √2 u(√2 u + √2/3 v) + (√2 u + √2/3 v)²
= 3u² - 2uv + v²
Therefore, the integral becomes:
∫∫ᵣ (3u² - 2uv + v²) |J(u, v)| dA
where J(u, v) is the Jacobian of the transformation.
The Jacobian of the transformation is given by:
J(u, v) = |√2/3 - √2/3| = √6/3
Therefore, the integral becomes:
∫∫ᵣ (3u² - 2uv + v²) (√6/3) dA
= √6/3 ∫∫ᵣ (3u² - 2uv + v²) dA
We can now evaluate the integral using double integration.
∫∫ᵣ (3u² - 2uv + v²) dA
= √6/3 ∫₀^√³ ∫_{-1}^¹ (3u² - 2uv + v²) du dv
= √6/3 [u³ - u²v + v³/3]₀^√³ ∫_{-1}^¹ dv
= √6/3 (9√3 - 3√3 + 1/3) ∫_{-1}^¹ dv
= √6/3 (6√3 + 1/3) ∫_{-1}^¹ dv
= √18 + 2/9 [-v]_{-1}^¹
= 2√2 + 2/9