To evaluate the double integral

$\iint_R -6x + 2y x - 4y \, dA$

over the region $R$, which is the parallelogram enclosed by the lines:

$-6x + 2y = 0, \quad -6x + 2y = 2, \quad x - 4y = 1, \quad x - 4y = 6,$

we’ll simplify the region $R$ by using a transformation of variables. Let’s proceed with these steps.

Step 1: Set up the transformation

Observe the given lines:

1. $-6x + 2y = 0$ and $-6x + 2y = 2$
2. $x - 4y = 1$ and $x - 4y = 6$

These pairs of lines are parallel, so they form a parallelogram. Let’s introduce new variables $u$ and $v$ to simplify the boundary.

Define:

$u = -6x + 2y$ $v = x - 4y$

Then the region $R$ becomes a rectangle in the $u$-$v$ plane, as it’s bounded by constant values of $u$ and $v$:

• $u = 0$ to $u = 2$
• $v = 1$ to $v = 6$

Step 2: Solve for $x$ and $y$ in terms of $u$ and $v$

We have two equations:

1. $u = -6x + 2y$
2. $v = x - 4y$

To solve for $x$ and $y$, we can use these equations as a system:

1. From the first equation: $u = -6x + 2y$
2. From the second equation: $v = x - 4y$

Solve for $y$ and $x$ in terms of $u$ and $v$.

Step 3: Find the Jacobian determinant