Final Answer:
The integral ∫∫_R (4x + 8y) dA, where R is the described parallelogram, evaluates to 0.
Step-by-step explanation:
Change of variables: We are given the transformation:
x = 1/(u + v)
y = -3u
Jacobian determinant: We need to calculate the Jacobian determinant, which represents the area element scaling factor under the transformation:
|J| = |-∂x/∂u ∂x/∂v|
|∂y/∂u ∂y/∂v|
Calculating the partial derivatives:
|J| = |-(u + v)^−2 |u|
|3 |u + v|
Integral substitution: We can substitute the original variables with the transformed ones and multiply by the Jacobian:
∫∫_R (4x + 8y) dA
= ∫∫_S (4(1/(u + v)) + 8(-3u)) |J| du dv
Simplify the integral:
= ∫∫_S (4u^2 - 20uv - 24) |u + v| du dv
Symmetric region and odd integrand: The region R is symmetric across the y-axis, and the integrand (4u^2 - 20uv - 24) is odd in both u and v. When integrating over a symmetric region with an odd function, the positive and negative contributions cancel each other out.
Therefore, the integral ∫∫_R (4x + 8y) dA evaluates to 0.