Let's solve the given problem step by step: ### **Step 1: Change of Variables** Given the lines: - \( x - 7y = 0 \) - \( x - 7y = 5 \) - \( 6x - y = 4 \) - \( 6x - y = 6 \) Let’s make the change of variables: - \( u = x - 7y \) - \( v = 6x - y \) ### **Step 2: Express the Region \( R \) in Terms of \( u \) and \( v \)** - \( 0 \leq u \leq 5 \) - \( 4 \leq v \leq 6 \) So, the region \( R \) in the \( (u, v) \) plane is a rectangle. ### **Step 3: Express \( x \) and \( y \) in Terms of \( u \) and \( v \)** Solve: \[ \begin{cases} u = x - 7y \\ v = 6x - y \end{cases} \] Multiply the first by 6: \( 6u = 6x - 42y \) Subtract from the second: \[ v - 6u = (6x - y) - (6x - 42y) = 41y \implies y = \frac{v - 6u}{41} \] Now substitute \( y \) in the first equation: \[ u = x - 7y \implies x = u + 7y = u + 7\left(\frac{v - 6u}{41}\right) = \frac{41u + 7v - 42u}{41} = \frac{-u + 7v}{41} \] So, \[ x = \frac{-u + 7v}{41}, \quad y = \frac{v - 6u}{41} \] ### **Step 4: Compute the Jacobian** \[ J = \left|\frac{\partial(x, y)}{\partial(u, v)}\right| \] \[ \frac{\partial x}{\partial u} = -\frac{1}{41}, \quad \frac{\partial x}{\partial v} = \frac{7}{41} \] \[ \frac{\partial y}{\partial u} = -\frac{6}{41}, \quad \frac{\partial y}{\partial v} = \frac{1}{41} \] So, \[ J = \begin{vmatrix} -\frac{1}{41} & \frac{7}{41} \\ -\frac{6}{41} & \frac{1}{41} \end{vmatrix} = \left(-\frac{1}{41}\right)\left(\frac{1}{41}\right) - \left(\frac{7}{41}\right)\left(-\frac{6}{41}\right) \] \[ = -\frac{1}{1681} + \frac{42}{1681} = \frac{41}{1681} \] ### **Step 5: Change the Integral** The integrand: \[ 6 \frac{x - 7y}{6x - y} = 6 \frac{u}{v} \] So the integral becomes: \[ \int_{u=0}^{5} \int_{v=4}^{6} 6 \frac{u}{v} \cdot \frac{41}{1681}\ dv\ du \] \[ = \frac{246}{1681} \int_{u=0}^5 \int_{v=4}^6 \frac{u}{v}\ dv\ du \] ### **Step 6: Evaluate the Integral** First, integrate with respect to \( v \): \[ \int_{v=4}^6 \frac{u}{v}\ dv = u \int_{v=4}^6 \frac{1}{v}\ dv = u [\ln v]_{4}^{6} = u (\ln 6 - \ln 4) = u \ln\left(\frac{6}{4}\right) = u \ln\left(\frac{3}{2}\right) \] Now, integrate with respect to \( u \): \[ \int_{u=0}^5 u \ln\left(\frac{3}{2}\right)\ du = \ln\left(\frac{3}{2}\right) \int_{u=0}^5 u\ du = \ln\left(\frac{3}{2}\right) \left[\frac{u^2}{2}\right]_{u=0}^{5} = \ln\left(\frac{3}{2}\right) \cdot \frac{25}{2} \] ### **Step 7: Combine Everything** \[ \text{Final answer:} \quad \frac{246}{1681} \cdot \frac{25}{2} \cdot \ln\left(\frac{3}{2}\right) \] \[ = \frac{246 \times 25}{2 \times 1681} \ln\left(\frac{3}{2}\right) = \frac{6150}{3362} \ln\left(\frac{3}{2}\right) \] --- ## **Final Answer** \[ \boxed{ \iint_R 6 \frac{x-7y}{6x-y}\ dA = \frac{6150}{3362} \ln\left(\frac{3}{2}\right) } \]

Let's carefully go through the problem step by step, providing detailed calculations, explanations, and verification. --- ### **Problem Restatement:** Evaluate the double integral over the region \( R \): \[ \iint_R 6 \frac{x - 7y}{6x - y} \, dA \] where \( R \) is the parallelogram enclosed by the lines: - \( x - 7y = 0 \) - \( x - 7y = 5 \) - \( 6x - y = 4 \) - \( 6x - y = 6 \) --- ### **Step 1: Recognize the Region \( R \)** The region is bounded by two pairs of lines: - \( x - 7y = 0 \) and \( x - 7y = 5 \), which are parallel lines. - \( 6x - y = 4 \) and \( 6x - y = 6 \), also parallel lines. These form a parallelogram. To evaluate the integral efficiently, **make a change of variables** that maps these lines to simple coordinate axes. --- ### **Step 2: Choosing the Change of Variables** Given the equations: - \( u = x - 7y \) - \( v = 6x - y \) In the \( (u, v) \)-plane: - \( u \) ranges from 0 to 5, - \( v \) ranges from 4 to 6. Thus, the region in \( (u, v) \) coordinates is a rectangle: \[ 0 \leq u \leq 5, \quad 4 \leq v \leq 6 \] --- ### **Step 3: Express \( x \) and \( y \) in terms of \( u \) and \( v \)** From: \[ u = x - 7y \quad (1) \] \[ v = 6x - y \quad (2) \] **Goal:** Solve for \( x \), \( y \) in terms of \( u \), \( v \). --- ### **Step 4: Solving for \( x \) and \( y \)** Multiply (1) by 6: \[ 6u = 6x - 42y \] Subtract this from (2): \[ v - 6u = (6x - y) - (6x - 42y) = (6x - y) - 6x + 42y = 41 y \] \[ \Rightarrow y = \frac{v - 6u}{41} \] Now, from (1): \[ x = u + 7 y = u + 7 \times \frac{v - 6u}{41} = \frac{41 u + 7 v - 42 u}{41} = \frac{-u + 7 v}{41} \] **Summary:** \[ \boxed{ x = \frac{-u + 7 v}{41}, \quad y = \frac{v - 6 u}{41} } \] --- ### **Step 5: Compute the Jacobian \( |\partial(x, y)/\partial(u, v)| \)** Partial derivatives: \[ \frac{\partial x}{\partial u} = -\frac{1}{41}, \quad \frac{\partial x}{\partial v} = \frac{7}{41} \] \[ \frac{\partial y}{\partial u} = -\frac{6}{41}, \quad \frac{\partial y}{\partial v} = \frac{1}{41} \] Jacobian determinant: \[ J = \left| \begin{bmatrix} -\frac{1}{41} & \frac{7}{41} \\ -\frac{6}{41} & \frac{1}{41} \end{bmatrix} \right| = \left(-\frac{1}{41}\right)\left(\frac{1}{41}\right) - \left(\frac{7}{41}\right)\left(-\frac{6}{41}\right) \] Calculate: \[ J = -\frac{1}{1681} + \frac{42}{1681} = \frac{41}{1681} \] **Note:** The absolute value of the Jacobian (for area transformation) is: \[ |J| = \frac{41}{1681} \] --- ### **Step 6: Rewrite the integrand in terms of \( u \) and \( v \)** Original integrand: \[ 6 \times \frac{x - 7 y}{6 x - y} = 6 \times \frac{u}{v} \] --- ### **Step 7: Set up the transformed integral** The double integral becomes: \[ \iint_R 6 \frac{x - 7 y}{6 x - y} \, dA = \iint_{u=0}^{5} \iint_{v=4}^{6} 6 \frac{u}{v} \times |J| \, dv \, du \] Substitute \( |J| = \frac{41}{1681} \): \[ = \int_{u=0}^{5} \int_{v=4}^{6} 6 \frac{u}{v} \times \frac{41}{1681} \, dv \, du \] \[ = \frac{246}{1681} \int_{u=0}^{5} \left( u \int_{v=4}^{6} \frac{1}{v} \, dv \right) du \] --- ### **Step 8: Evaluate the inner integral over \( v \)** \[ \int_{v=4}^{6} \frac{1}{v} \, dv = \left[\ln v \right]_4^6 = \ln 6 - \ln 4 = \ln \left(\frac{6}{4}\right) = \ln \left(\frac{3}{2}\right) \] --- ### **Step 9: Evaluate the outer integral over \( u \)** \[ \int_{0}^{5} u \, du = \left[\frac{u^2}{2}\right]_0^{5} = \frac{25}{2} \] Putting it all together: \[ \boxed{ \text{Integral} = \frac{246}{1681} \times \frac{25}{2} \times \ln \left(\frac{3}{2}\right) } \] --- ### **Step 10: Simplify the final answer** Calculate the coefficient: \[ \frac{246 \times 25}{2 \times 1681} \] Compute numerator: \[ 246 \times 25 = (200 + 46) \times 25 = 200 \times 25 + 46 \times 25 = 5000 + 1150 = 6150 \] Compute denominator: \[ 2 \times 1681 = 3362 \] So, the final answer: \[ \boxed{ \frac{6150}{3362} \ln \left(\frac{3}{2}\right) } \] --- ### **### **Final Verification:** - The change of variables mapped the parallelogram to a rectangle in \( (u, v) \). - The Jacobian determinant was correctly computed. - The integrand was correctly expressed as \( 6 \frac{u}{v} \). - The integrals over \( v \) and \( u \) were straightforward and correctly evaluated. **Optional:** The fraction \( \frac{6150}{3362} \) can be simplified: Divide numerator and denominator by 2: \[ \frac{6150/2}{3362/2} = \frac{3075}{1681} \] since 1681 is \( 41^2 \), and 3075 factors as \( 3 \times 5^2 \times 41 \), but 1681 is \( 41^2 \). No common factors other than 1. So, the simplest form: \[ \boxed{ \frac{6150}{3362} \ln \left(\frac{3}{2}\right) = \frac{3075}{1681} \ln \left(\frac{3}{2}\right) } \] --- ### **### **Final Answer:** \[ \boxed{ \boxed{ \frac{3075}{1681} \ln \left(\frac{3}{2}\right) } } \] This completes the detailed step-by-step solution with verification.