Let's solve the problem step by step. We are given the parallelogram \( R \) bounded by the lines: 1. \( -x + 3y = 0 \) 2. \( -x + 3y = 2 \) 3. \( -4x - 6y = 1 \) 4. \( -4x - 6y = 9 \) We are asked to evaluate: \[ \iint_R \frac{-x + 3y}{-4x - 6y} \, dA \] --- ### Step 1: Change of Variables Let: \[ u = -x + 3y \\ v = -4x - 6y \] The region \( R \) in the \(xy\)-plane is transformed to a rectangle in the \(uv\)-plane: - \( u = 0 \) to \( u = 2 \) - \( v = 1 \) to \( v = 9 \) --- ### Step 2: Jacobian Calculation Express \(x\) and \(y\) in terms of \(u\) and \(v\): \[ u = -x + 3y \quad (1) \\ v = -4x - 6y \quad (2) \] Multiply (1) by 2: \[ 2u = -2x + 6y \] Add to (2): \[ 2u + v = -2x + 6y -4x - 6y = -6x \implies x = -\frac{2u + v}{6} \] Now, solve for \(y\): Multiply (1) by 4: \[ 4u = -4x + 12y \] Add to (2): \[ 4u + v = -4x + 12y -4x - 6y = 6y \implies y = \frac{4u + v}{6} \] Now, compute the Jacobian: \[ J = \begin{vmatrix} \frac{\partial x}{\partial u} & \frac{\partial x}{\partial v} \\ \frac{\partial y}{\partial u} & \frac{\partial y}{\partial v} \end{vmatrix} \] From above: \[ x = -\frac{2u + v}{6} \implies \frac{\partial x}{\partial u} = -\frac{2}{6} = -\frac{1}{3},\quad \frac{\partial x}{\partial v} = -\frac{1}{6} \] \[ y = \frac{4u + v}{6} \implies \frac{\partial y}{\partial u} = \frac{4}{6} = \frac{2}{3},\quad \frac{\partial y}{\partial v} = \frac{1}{6} \] Therefore, \[ J = \begin{vmatrix} -\frac{1}{3} & -\frac{1}{6} \\ \frac{2}{3} & \frac{1}{6} \end{vmatrix} = \left(-\frac{1}{3}\right)\left(\frac{1}{6}\right) - \left(-\frac{1}{6}\right)\left(\frac{2}{3}\right) = -\frac{1}{18} + \frac{2}{18} = \frac{1}{18} \] --- ### Step 3: Transform the Integral \[ \iint_R \frac{-x + 3y}{-4x - 6y} \, dA = \iint_{S} \frac{u}{v} \left| J \right| du\,dv = \iint_{u=0}^{2} \int_{v=1}^{9} \frac{u}{v} \cdot \frac{1}{18} dv\,du \] --- ### Step 4: Compute the Integral \[ \frac{1}{18} \int_{u=0}^{2} \left[ \int_{v=1}^{9} \frac{u}{v} dv \right] du \] \[ = \frac{1}{18} \int_{u=0}^{2} u \left[ \int_{v=1}^{9} \frac{1}{v} dv \right] du \] \[ = \frac{1}{18} \int_{u=0}^{2} u [\ln v]_{v=1}^{v=9} du \] \[ = \frac{1}{18} \int_{u=0}^{2} u (\ln 9 - \ln 1) du \] \[ \ln 1 = 0, \quad \ln 9 = 2\ln 3 \] \[ = \frac{1}{18} \int_{u=0}^{2} u \cdot 2\ln 3 \, du \] \[ = \frac{2\ln 3}{18} \int_{u=0}^{2} u \, du \] \[ = \frac{\ln 3}{9} \left[ \frac{u^2}{2} \right]_{0}^{2} \] \[ = \frac{\ln 3}{9} \cdot \frac{4}{2} = \frac{\ln 3}{9} \cdot 2 = \frac{2\ln 3}{9} \] --- ### Final Answer \[ \boxed{\frac{2 \ln 3}{9}} \]