3. Evaluate \( \int_{c}(2 x-y) d x-(y-x) d y \) along the rectangle with vertices \( (0 \), \( 0),(0,3),(3,3) \) and \( (3,0) \) Evaluate the surface integral \( \iint_{\sigma}(x+y) d s ; \sigma \) is the part of the plane \( x+y+ \) \( z=4 \) that lies in the first octant.

Let's tackle both parts of your problem step by step. --- ### **Part 1: Evaluate the Line Integral** \[ \int_{C} (2x - y)\, dx - (y - x)\, dy \] **Along the rectangle with vertices \((0, 0)\), \((0, 3)\), \((3, 3)\), and \((3, 0)\).** #### **Step 1: Analyze the Vector Field** The given line integral can be expressed in terms of a vector field \(\mathbf{F}\): \[ \mathbf{F} = (P, Q) = (2x - y, -(y - x)) = (2x - y, x - y) \] #### **Step 2: Check if the Vector Field is Conservative** A vector field \(\mathbf{F} = (P, Q)\) is **conservative** if there exists a scalar potential function \( \phi \) such that: \[ \frac{\partial P}{\partial y} = \frac{\partial Q}{\partial x} \] Compute the partial derivatives: \[ \frac{\partial P}{\partial y} = \frac{\partial (2x - y)}{\partial y} = -1 \] \[ \frac{\partial Q}{\partial x} = \frac{\partial (x - y)}{\partial x} = 1 \] Since \(-1 \neq 1\), the vector field is **not conservative**. #### **Step 3: Apply Green's Theorem** For a non-conservative vector field, **Green's Theorem** is applicable. Green's Theorem relates the line integral around a simple closed curve \( C \) to a double integral over the region \( D \) enclosed by \( C \): \[ \oint_{C} P\, dx + Q\, dy = \iint_{D} \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right) dx\, dy \] Compute the expression inside the double integral: \[ \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} = 1 - (-1) = 2 \] Thus, the line integral becomes: \[ \oint_{C} (2x - y)\, dx - (y - x)\, dy = \iint_{D} 2\, dx\, dy = 2 \times \text{Area}(D) \] #### **Step 4: Compute the Area of \( D \)** The region \( D \) is a rectangle with vertices \((0, 0)\), \((0, 3)\), \((3, 3)\), and \((3, 0)\). Its area is: \[ \text{Area}(D) = \text{length} \times \text{width} = 3 \times 3 = 9 \] #### **Step 5: Final Calculation** \[ \oint_{C} (2x - y)\, dx - (y - x)\, dy = 2 \times 9 = 18 \] **Answer for Part 1:** The value of the line integral is **18**. --- ### **Part 2: Evaluate the Surface Integral** \[ \iint_{\sigma} (x + y)\, dS \] **Where \(\sigma\) is the part of the plane \(x + y + z = 4\) that lies in the first octant.** #### **Step 1: Describe the Surface \(\sigma\)** The plane \(x + y + z = 4\) intersects the axes at: - \(x\)-axis: \((4, 0, 0)\) - \(y\)-axis: \((0, 4, 0)\) - \(z\)-axis: \((0, 0, 4)\) In the **first octant** (\(x, y, z \geq 0\)), the region \(\sigma\) is a triangular region bounded by these intercepts. #### **Step 2: Express \(z\) in Terms of \(x\) and \(y\)** \[ z = 4 - x - y \] This allows us to parameterize the surface in terms of \(x\) and \(y\). #### **Step 3: Determine the Differential Surface Element \(dS\)** For a surface described by \(z = f(x, y)\), the differential surface element is: \[ dS = \sqrt{1 + \left( \frac{\partial z}{\partial x} \right)^2 + \left( \frac{\partial z}{\partial y} \right)^2} \, dx\, dy \] Compute the partial derivatives: \[ \frac{\partial z}{\partial x} = -1, \quad \frac{\partial z}{\partial y} = -1 \] \[ dS = \sqrt{1 + (-1)^2 + (-1)^2} \, dx\, dy = \sqrt{3} \, dx\, dy \] #### **Step 4: Set Up the Double Integral** The integral becomes: \[ \iint_{\sigma} (x + y)\, dS = \sqrt{3} \iint_{D} (x + y)\, dx\, dy \] Where \( D \) is the projection of \(\sigma\) onto the \(xy\)-plane, defined by: \[ 0 \leq x \leq 4, \quad 0 \leq y \leq 4 - x \] #### **Step 5: Compute the Integral** \[ \iint_{D} (x + y)\, dx\, dy = \int_{0}^{4} \int_{0}^{4 - x} (x + y)\, dy\, dx \] First, integrate with respect to \( y \): \[ \int_{0}^{4 - x} (x + y)\, dy = \left[ xy + \frac{1}{2}y^2 \right]_{0}^{4 - x} = x(4 - x) + \frac{1}{2}(4 - x)^2 \] Simplify: \[ x(4 - x) + \frac{1}{2}(16 - 8x + x^2) = 4x - x^2 + 8 - 4x + \frac{1}{2}x^2 = 8 - \frac{1}{2}x^2 \] Now, integrate with respect to \( x \): \[ \int_{0}^{4} \left(8 - \frac{1}{2}x^2\right) dx = \left[8x - \frac{1}{6}x^3 \right]_{0}^{4} = 8(4) - \frac{1}{6}(64) = 32 - \frac{64}{6} = 32 - \frac{32}{3} = \frac{96 - 32}{3} = \frac{64}{3} \] Finally, multiply by \(\sqrt{3}\): \[ \iint_{\sigma} (x + y)\, dS = \sqrt{3} \times \frac{64}{3} = \frac{64\sqrt{3}}{3} \] **Answer for Part 2:** The value of the surface integral is \(\displaystyle \frac{64\, \sqrt{3}}{3}\). --- ### **Summary of Results** 1. **Line Integral:** \( \displaystyle \int_{C} (2x - y)\, dx - (y - x)\, dy = 18 \) 2. **Surface Integral:** \( \displaystyle \iint_{\sigma} (x + y)\, dS = \frac{64\, \sqrt{3}}{3} \)