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.

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Let's tackle both parts of your problem step by step.

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### **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 $$\displaystyle \frac{64\, \sqrt{3}}{3}$$.

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### **Summary of Results**
1. **Line Integral:** $$ \displaystyle \int_{C} (2x - y)\, dx - (y - x)\, dy = 18 $$
2. **Surface Integral:** $$ \displaystyle \iint_{\sigma} (x + y)\, dS = \frac{64\, \sqrt{3}}{3} $$
**Part 1:** The value of the line integral is **18**. **Part 2:** The value of the surface integral is $$\frac{64\, \sqrt{3}}{3}$$.

Evaluate along the rectangle with vertices ,
and
Evaluate the surface integral is the part of the plane
that lies in the first octant.

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Evaluate along the rectangle with vertices , and Evaluate the surface integral is the part of the plane that lies in the first octant.