\( R \iint_{R}(x+y)^{3} d x d y \) The area limited to \( x+y=4 \) and \( x+y=1, x-2 y=1 \) and \( y-2 x=-2 \) ?

To solve the double integral \( R \iint_{R}(x+y)^{3} d x d y \) over the region defined by the lines \( x+y=4 \), \( x+y=1 \), \( x-2y=1 \), and \( y-2x=-2 \), we first need to determine the vertices of the region \( R \) formed by these lines. ### Step 1: Find the intersection points of the lines 1. **Intersection of \( x+y=4 \) and \( x+y=1 \)**: - These lines are parallel and do not intersect. 2. **Intersection of \( x+y=4 \) and \( x-2y=1 \)**: - From \( x+y=4 \), we can express \( x = 4 - y \). - Substitute into \( x-2y=1 \): \[ (4 - y) - 2y = 1 \implies 4 - 3y = 1 \implies 3y = 3 \implies y = 1 \] \[ x = 4 - 1 = 3 \] - Intersection point: \( (3, 1) \). 3. **Intersection of \( x+y=4 \) and \( y-2x=-2 \)**: - From \( x+y=4 \), express \( y = 4 - x \). - Substitute into \( y-2x=-2 \): \[ (4 - x) - 2x = -2 \implies 4 - 3x = -2 \implies 3x = 6 \implies x = 2 \] \[ y = 4 - 2 = 2 \] - Intersection point: \( (2, 2) \). 4. **Intersection of \( x+y=1 \) and \( x-2y=1 \)**: - From \( x+y=1 \), express \( x = 1 - y \). - Substitute into \( x-2y=1 \): \[ (1 - y) - 2y = 1 \implies 1 - 3y = 1 \implies 3y = 0 \implies y = 0 \] \[ x = 1 - 0 = 1 \] - Intersection point: \( (1, 0) \). 5. **Intersection of \( x+y=1 \) and \( y-2x=-2 \)**: - From \( x+y=1 \), express \( y = 1 - x \). - Substitute into \( y-2x=-2 \): \[ (1 - x) - 2x = -2 \implies 1 - 3x = -2 \implies 3x = 3 \implies x = 1 \] \[ y = 1 - 1 = 0 \] - Intersection point: \( (1, 0) \) (already found). 6. **Intersection of \( x-2y=1 \) and \( y-2x=-2 \)**: - From \( x-2y=1 \), express \( x = 1 + 2y \). - Substitute into \( y-2x=-2 \): \[ y - 2(1 + 2y) = -2 \implies y - 2 - 4y = -2 \implies -3y = 0 \implies y = 0 \] \[ x = 1 + 2(0) = 1 \] - Intersection point: \( (1, 0) \) (already found). ### Step 2: Identify the vertices of the region \( R \) The vertices of the region \( R \) are: - \( (3, 1) \) - \( (2, 2) \) - \( (1, 0) \) ### Step 3: Set up the double integral The region \( R \) is bounded by the lines \( x+y=4 \) and \( x+y=1 \) and the lines \( x-2y=1 \) and \( y-2x=-2 \). We can express the integral in terms of \( u = x+y \) and \( v = x-2y \). The limits for \( u \) will be from \( 1 \) to \( 4 \), and for \( v \), we need to find the corresponding limits based on the lines. ### Step 4: Calculate the double integral We will now compute the double integral \( \iint_{R}(x+y)^{3} d x d y \). Using the transformation \( u = x+y \) and \( v = x-2y \), we can find the Jacobian and set up the integral. 1. **Jacobian**: \[ J = \begin{vmatrix} \frac{\partial x}{\partial u} & \frac{\partial x}{\partial v} \\ \frac{\partial y}{\partial u} & \frac{\partial y}{\partial v} \end{vmatrix} \] 2. **Integral**: \[ \iint_{R} u^{3} \left| J \right| du dv \] Now, let's calculate the integral. ### Step 5: Solve the integral We will compute the integral using the limits we found. Let's proceed with the calculation. First, we need to find the Jacobian determinant and then set up the integral. The transformation equations are: - \( x = \frac{u + 2v}{3} \) - \( y = \frac{u - v}{3} \) Now, we can calculate the Jacobian \( J \) and then set up the integral. Let's calculate the Jacobian and then the integral. ### Step 6: Calculate the Jacobian The Jacobian \( J \) is given by: \[ J = \begin{vmatrix} \frac{\partial x}{\partial u} & \frac{\partial x}{\partial v} \\ \frac{\partial y}{\partial u} & \frac{\partial y}{\partial v} \end{vmatrix} \] Calculating the partial derivatives: - \( \frac{\partial x}{\partial u} = \frac{1}{3} \) - \( \frac{\partial x}{\partial v} = \frac{2}{3} \) - \( \frac{\partial y}{\partial u} = \frac{1}{3} \) - \( \frac{\partial y}{\partial v} = -\frac{1}{3} \) Now, calculating the determinant: \[ J = \left