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 $u$: $\frac{198}{961} \int_0^2 \frac{55u}{2} \, du = \frac{198 \cdot 55}{1922} \int_0^2 u \, du = \frac{10890}{1922} \int_0^2 u \, du$