Find ∫∫R-6x+2yx-4ydA, where R is the parallelogram enclosed by the lines -6x+2y=0, -6x+2y=2, x-4y=1, x-4y=6


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



Find ∫∫R -6x + 2y x - 4y dA, where R is the parallelogram enclosed by the lines -6x + 2y = 0, -6x + 2y = 2, x - 4y = 1, x - 4y = 6

Step-by-step solution for evaluating the double integral ∫∫R -6x + 2y x - 4y dA over a parallelogram region defined by lines -6x + 2y = 0, -6x + 2y = 2, x - 4y = 1, and x - 4y = 6. This involves using a coordinate transformation and calculating the Jacobian determinant. Ideal for students studying double integrals and variable transformations in undergraduate calculus.

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