Let's carefully analyze and solve the problem step-by-step. --- ### Step 1: Understand the problem Given: - Region \( R \) is a parallelogram with vertices: \[ (-1, 3), \quad (1, -3), \quad (2, -2), \quad (0, 4) \] - The integral: \[ \iint_R 8xy \, dA \] - Transformation: \[ x = \frac{1}{4}(u + v), \quad y = \frac{1}{4}(v - 3u) \] --- ### Step 2: Find the Jacobian of the transformation Express \( x \) and \( y \) in terms of \( u \) and \( v \): \[ x = \frac{u + v}{4} \] \[ y = \frac{v - 3u}{4} \] Compute the Jacobian \( J \): \[ J = \left| \frac{\partial (x, y)}{\partial (u, v)} \right| = \begin{vmatrix} \frac{\partial x}{\partial u} & \frac{\partial x}{\partial v} \\ \frac{\partial y}{\partial u} & \frac{\partial y}{\partial v} \end{vmatrix} \] Calculate the partial derivatives: \[ \frac{\partial x}{\partial u} = \frac{1}{4} \] \[ \frac{\partial x}{\partial v} = \frac{1}{4} \] \[ \frac{\partial y}{\partial u} = -\frac{3}{4} \] \[ \frac{\partial y}{\partial v} = \frac{1}{4} \] Compute the determinant: \[ J = \left| \begin{bmatrix} \frac{1}{4} & \frac{1}{4} \\ -\frac{3}{4} & \frac{1}{4} \end{bmatrix} \right| = \left( \frac{1}{4} \times \frac{1}{4} \right) - \left( \frac{1}{4} \times -\frac{3}{4} \right) = \frac{1}{16} + \frac{3}{16} = \frac{4}{16} = \frac{1}{4} \] Since the Jacobian determinant is positive, the differential area element transforms as: \[ dA = |J| du dv = \frac{1}{4} du dv \] --- ### Step 3: Express the integrand \( 8xy \) in terms of \( u \) and \( v \) Compute \( xy \): \[ x = \frac{u + v}{4} \] \[ y = \frac{v - 3u}{4} \] Then: \[ xy = \left( \frac{u + v}{4} \right) \left( \frac{v - 3u}{4} \right) = \frac{(u + v)(v - 3u)}{16} \] Multiply out numerator: \[ (u + v)(v - 3u) = u v - 3 u^2 + v^2 - 3 u v = (u v - 3 u v) + v^2 - 3 u^2 = -2 u v + v^2 - 3 u^2 \] Thus, \[ xy = \frac{-2 u v + v^2 - 3 u^2}{16} \] The integrand: \[ 8 xy = 8 \times \frac{-2 u v + v^2 - 3 u^2}{16} = \frac{8}{16} (-2 u v + v^2 - 3 u^2) = \frac{1}{2} (-2 u v + v^2 - 3 u^2) \] Simplify: \[ 8 xy = - u v + \frac{v^2}{2} - \frac{3}{2} u^2 \] --- ### Step 4: Find the transformed region \( R' \) in \( uv \)-coordinates Find the \( (u, v) \) coordinates of the vertices: | Original vertex | \(x\) | \(y\) | Solve for \(u, v\) | |------------------|--------|--------|---------------------| | \((-1, 3)\) | \(-1\) | \(3\) | \(x = \frac{u+v}{4} = -1 \Rightarrow u+v = -4\) <br> \(y = \frac{v - 3u}{4} = 3 \Rightarrow v - 3u = 12\) | | \((1, -3)\) | 1 | -3 | \(u+v=4\) <br> \(v - 3u = -12\) | | \((2, -2)\) | 2 | -2 | \(u+v=8\) <br> \(v - 3u = -8\) | | \((0, 4)\) | 0 | 4 | \(u+v=0\) <br> \(v - 3u=16\) | Solve for each: 1. \((-1, 3)\): \[ u + v = -4 \quad (1) \] \[ v - 3u = 12 \quad (2) \] From (1): \(v = -4 - u\) Substitute into (2): \[ -4 - u - 3u = 12 \Rightarrow -4 - 4u = 12 \Rightarrow -4u = 16 \Rightarrow u = -4 \] Then: \[ v = -4 - (-4) = 0 \] Vertices: \[ (-4, 0) \] 2. \((1, -3)\): \[ u + v = 4 \quad (1) \] \[ v - 3u = -12 \quad (2) \] From (1): \(v=4 - u\) Plug into (2): \[ 4 - u - 3u = -12 \Rightarrow 4 - 4u = -12 \Rightarrow -4u = -16 \Rightarrow u=4 \] Then: \[ v=4 - 4=0 \] Vertices: \[ (4, 0) \] 3. \((2, -2)\): \[ u + v=8 \quad (1) \] \[ v - 3u = -8 \quad (2) \] From (1): \(v=8 - u\) Plug into (2): \[ 8 - u - 3u = -8 \Rightarrow 8 - 4u = -8 \Rightarrow -4u = -16 \Rightarrow u=4 \] Then: \[ v=8 - 4=4 \] Vertices: \[ (4, 4) \] 4. \((0, 4)\): \[ u + v=0 \quad (1) \] \[ v - 3u=16 \quad (2) \] From (1): \(v=-u\) Plug into (2): \[ -u - 3u=16 \Rightarrow -4u=16 \Rightarrow u=-4 \] \[ v=-(-4)=4 \] Vertices: \[ (-4, 4) \] --- ### **Summary of transformed vertices:** \[ (-4, 0), \quad (4, 0), \quad (4, 4), \quad (-4, 4) \] This is a rectangle in \( uv \)-plane with: \[ u \in [-4, 4], \quad v \in [0, 4] \] --- ### **Step 5: Write the integral in \( u, v \) variables** The double integral: \[ \iint_R 8xy\, dA \] becomes: \[ \iint_{R'} \left( - u v + \frac{v^2}{2} - \frac{3}{2} u^2 \right) \times |J| du dv \] Recall \( |J| = \frac{1}{4} \), so: \[ \boxed{ \iint_{R'} \left( - u v + \frac{v^2}{2} - \frac{3}{2} u^2 \right) \times \frac{1}{4} du dv } \] with \( u \in [-4,4] \), \( v \in [0,4] \). --- ### **Final integral:** \[ \boxed{ \int_{u=-4}^{4} \int_{v=0}^{4} \left( - u v + \frac{v^2}{2} - \frac{3}{2} u^2 \right) \times \frac{1}{4} dv du } \] --- ### **Step 6: Evaluate the integral** Rewrite as: \[ \frac{1}{4} \int_{-4}^{4} \int_{0}^{4} \left( - u v + \frac{v^2}{2} - \frac{3}{2} u^2 \right) dv du \] --- ### Step 7: Integrate with respect to \( v \) Inner integral: \[ I(u) = \int_0^4 \left( - u v + \frac{v^2}{2} - \frac{3}{2} u^2 \right) dv \] Break into parts: \[ I(u) = - u \int_0^4 v dv + \frac{1}{2} \int_0^4 v^2 dv - \frac{3}{2} u^2 \int_0^4 dv \] Compute each: - \(\int_0^4 v dv = \frac{v^2}{2} \big|_0^4 = \frac{16}{2} = 8\) - \(\int_0^4 v^2 dv = \frac{v^3}{3} \big|_0^4 = \frac{64}{3}\) - \(\int_0^4 dv=4\) Thus: \[ I(u) = - u \times 8 + \frac{1}{2} \times \frac{64}{3} - \frac{3}{2} u^2 \times 4 \] Simplify: \[ I(u) = -8 u + \frac{64}{6} - 6 u^2 = -8 u + \frac{32}{3} - 6 u^2 \] --- ### Step 8: Integrate with respect to \( u \) Now, the total integral: \[ \frac{1}{4} \int_{-4}^{4} \left( -8 u + \frac{32}{3} - 6 u^2 \right) du \] Distribute \( \frac{1}{4} \): \[ \frac{1}{4} \left[ \int_{-4}^{4} -8 u du + \int_{-4}^{4} \frac{32}{3} du - \int_{-4}^{4} 6 u^2 du \right] \] Calculate each: 1. \(\int_{-4}^{4} -8 u du\): Since \(-8 u\) is an odd function over symmetric limits, the integral is zero: \[ = 0 \] 2. \(\int_{-4}^{4} \frac{32}{3} du = \frac{32}{3} \times (4 - (-4)) = \frac{32}{3} \times 8 = \frac{256}{3}\) 3. \(\int_{-4}^{4} 6 u^2 du\): Since \(u^2\) is even: \[ = 6 \times 2 \int_0^4 u^2 du = 12 \times \frac{u^3}{3} \big|_0^4 = 12 \times \frac{64}{3} = 12 \times \frac{64}{3} = \frac{768}{3} = 256 \] Putting all together: \[ \frac{1}{4} \left[ 0 + \frac{256}{3} - 256 \right] = \frac{1}{4} \left( \frac{256}{3} - 256 \right) \] Express \(256\) as \(\frac{768}{3}\): \[ = \frac{1}{4} \left( \frac{256}{3} - \frac{768}{3} \right) = \frac{1}{4} \times \left( - \frac{512}{3} \right) = - \frac{512}{3} \times \frac{1}{4} = - \frac{512}{12} = - \frac{128}{3} \] --- ### **Final answer:** \[ \boxed{ \iint_R 8xy \, dA = - \frac{128}{3} } \] --- **Summary:** - The integral over \( R \) transforms into a rectangular region in \( uv \)-space. - The integral evaluates to \(\boxed{- \frac{128}{3}}\).