Evaluate the given integral by making an appropriate change of variables.
 
6 

*x* − 8*y*

4*x* − *y*

 *dA*,

*R*

 where *R* is the parallelogram enclosed by the lines 

*x* − 8*y* = 0,
 
*x* − 8*y* = 2,
 
4*x* − *y* = 3,

 and 

4*x* − *y* = 8

To evaluate the integral

\[
\iint_R \left(6(x - 8y)(4x - y)\right) \, dA
\]

where \( R \) is the region enclosed by the lines \( x - 8y = 0 \), \( x - 8y = 2 \), \( 4x - y = 3 \), and \( 4x - y = 8 \), we’ll start by performing a change of variables based on the given lines.

### Step 1: Define New Variables
Let:
\[
u = x - 8y
\]
\[
v = 4x - y
\]

### Step 2: Find the Jacobian Determinant
To perform this change of variables, we need to express \( x \) and \( y \) in terms of \( u \) and \( v \) and then compute the Jacobian determinant, \( \frac{\partial(x, y)}{\partial(u, v)} \).

1. Solve the equations for \( x \) and \( y \):
    \[
    x = u + 8y
    \]
    Substitute into the second equation:
    \[
    v = 4(u + 8y) - y = 4u + 32y - y = 4u + 31y
    \]
    Solving for \( y \):
    \[
    y = \frac{v - 4u}{31}
    \]
    Substitute \( y \) back into \( x = u + 8y \):
    \[
    x = u + 8\left(\frac{v - 4u}{31}\right) = \frac{31u + 8v - 32u}{31} = \frac{u + 8v}{31}
    \]

2. Now we compute the partial derivatives for the Jacobian matrix:
    \[
    \frac{\partial x}{\partial u} = \frac{1}{31}, \quad \frac{\partial x}{\partial v} = \frac{8}{31}
    \]
    \[
    \frac{\partial y}{\partial u} = -\frac{4}{31}, \quad \frac{\partial y}{\partial v} = \frac{1}{31}
    \]

3. Calculate the Jacobian determinant:
    \[
    J = \det\begin{pmatrix} \frac{1}{31} & \frac{8}{31} \\ -\frac{4}{31} & \frac{1}{31} \end{pmatrix} = \frac{1 \cdot 1 - (8)(-4)}{31 \cdot 31} = \frac{1 + 32}{31 \cdot 31} = \frac{33}{961}
    \]

Thus, the area element \( dA \) becomes \( dA = \frac{33}{961} \, du \, dv \).

### Step 3: Set Up the Integral
The region \( R \) in the \( uv \)-plane is now described by:
\[
0 \leq u \leq 2, \quad 3 \leq v \leq 8
\]

The integrand, \( 6(x - 8y)(4x - y) \), in terms of \( u \) and \( v \) is simply \( 6uv \).

The integral in terms of \( u \) and \( v \) becomes:
\[
\iint_R 6uv \, dA = \int_0^2 \int_3^8 6uv \cdot \frac{33}{961} \, dv \, du
\]

### Step 4: Evaluate the Integral
Separate constants and integrate with respect to \( v \) and \( u \):
\[
= \frac{198}{961} \int_0^2 \int_3^8 uv \, dv \, du
\]

1. **Inner integral in terms of \( v \):**
   \[
   \int_3^8 uv \, dv = u \cdot \frac{v^2}{2} \Big|_3^8 = u \cdot \frac{64 - 9}{2} = \frac{55u}{2}
   \]

2. **Outer integral in terms of \( u \):**
   \[
   \frac{198}{961} \int_0^2 \frac{55u}{2} \, du = \frac{198 \cdot 55}{1922} \int_0^2 u \, du = \frac{10890}{1922} \int_0^2 u \, du
   \]
   


Evaluate the given integral by making an appropriate change of variables.

6 (x − 8y)(4x − y) dA, where R is the parallelogram enclosed by the lines x − 8y = 0, x − 8y = 2, 4x − y = 3, and 4x − y = 8.

Learn how to evaluate the double integral 6(x − 8y)(4x − y) over a parallelogram region using a change of variables. This solution covers key multivariable calculus concepts such as Jacobian determinants and double integrals. Ideal for university-level calculus students.

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