Let's solve the problem **step by step**. --- ### **1. Understanding the Problem** We need to evaluate the line integral: \[ \int_C \mathbf{F} \cdot d\mathbf{r} \] where \(\mathbf{F} = (z, -z, x^2 - y^2)\), and \(C\) bounds the plane \[ z = 8 - 4x - 2y \] in the **first octant**, oriented counterclockwise when viewed from above. --- ### **2. Find the Curve \(C\) (Boundary of the Plane in the First Octant)** Set \(z = 0\) to find the intersection with the \(xy\)-plane: \[ 0 = 8 - 4x - 2y \implies 4x + 2y = 8 \implies 2x + y = 4 \] **Intercepts:** - \(x\)-intercept: \(y = 0 \implies x = 2\) - \(y\)-intercept: \(x = 0 \implies y = 4\) **Intercepts with axes:** - \(x = 0, y = 0 \implies z = 8\) - \(x = 2, y = 0 \implies z = 0\) - \(x = 0, y = 4 \implies z = 0\) So, the triangle has vertices at \((0,0,8)\), \((2,0,0)\), and \((0,4,0)\). --- ### **3. Using Stokes' Theorem** Since the curve is the boundary of a surface, we use **Stokes' Theorem**: \[ \int_C \mathbf{F} \cdot d\mathbf{r} = \iint_S (\nabla \times \mathbf{F}) \cdot d\mathbf{S} \] --- ### **4. Compute \(\nabla \times \mathbf{F}\)** Given \(\mathbf{F} = (z, -z, x^2 - y^2)\): \[ \nabla \times \mathbf{F} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ z & -z & x^2 - y^2 \end{vmatrix} \] Compute components: **i-component:** \[ \frac{\partial}{\partial y}(x^2 - y^2) - \frac{\partial}{\partial z}(-z) = (-2y) - (-1) = -2y + 1 \] **j-component:** \[ -\left( \frac{\partial}{\partial x}(x^2 - y^2) - \frac{\partial}{\partial z}(z) \right) = -\left(2x - 1\right) = -2x + 1 \] **k-component:** \[ \frac{\partial}{\partial x}(-z) - \frac{\partial}{\partial y}(z) = 0 - 0 = 0 \] So, \[ \nabla \times \mathbf{F} = (-2y + 1,\, -2x + 1,\, 0) \] --- ### **5. The Surface \(S\):** The surface is \(z = 8 - 4x - 2y\), over the triangle in the \(xy\)-plane with vertices \((0,0), (2,0), (0,4)\). --- ### **6. Surface Element \(d\mathbf{S}\):** For \(z = f(x,y)\), \[ d\mathbf{S} = (-f_x, -f_y, 1)\,dx\,dy \] Here, \[ f(x,y) = 8 - 4x - 2y \implies f_x = -4,\, f_y = -2 \] So, \[ d\mathbf{S} = (4, 2, 1) dx\,dy \] --- ### **7. Compute Dot Product** \[ (\nabla \times \mathbf{F}) \cdot d\mathbf{S} = (-2y + 1,\, -2x + 1,\, 0) \cdot (4, 2, 1) dx\,dy \] \[ = [4(-2y + 1) + 2(-2x + 1) + 0] dx\,dy \] \[ = [-8y + 4 -4x + 2] dx\,dy \] \[ = [-8y - 4x + 6] dx\,dy \] --- ### **8. Set up the Double Integral (over the triangle)** The triangle has vertices \((0,0), (2,0), (0,4)\): - \(x\) goes from \(0\) to \(2\) - For each \(x\), \(y\) goes from \(0\) to \(4 - 2x\) So, \[ \iint_T (-8y - 4x + 6)\,dy\,dx \] --- ### **9. Integrate** First, integrate with respect to \(y\): \[ \int_{x=0}^{2} \int_{y=0}^{4-2x} (-8y - 4x + 6)\,dy\,dx \] Integrate inside: \[ \int_{y=0}^{4-2x} (-8y - 4x + 6)\,dy = \left[ -4y^2 - 4x y + 6y \right]_{y=0}^{y=4-2x} \] Plug in \(y = 4-2x\): - \(y^2 = (4-2x)^2 = 16 - 16x + 4x^2\) - \(-4x y = -4x (4-2x) = -16x + 8x^2\) - \(6y = 6(4-2x) = 24 - 12x\) So, \[ -4y^2 - 4x y + 6y = -4(16 - 16x + 4x^2) - 16x + 8x^2 + 24 - 12x = -64 + 64x - 16x^2 - 16x + 8x^2 + 24 - 12x \] \[ = (-64 + 24) + (64x - 16x - 12x) + (-16x^2 + 8x^2) = (-40) + (36x) + (-8x^2) \] Now subtract value at \(y=0\) (which is zero): So, the inner integral simplifies to: \[ -40 + 36x - 8x^2 \] --- Now integrate with respect to \(x\) from \(0\) to \(2\): \[ \int_{x=0}^2 (-40 + 36x - 8x^2) dx \] Integrate term by term: - \(\int -40 dx = -40x\) - \(\int 36x dx = 18x^2\) - \(\int -8x^2 dx = -\frac{8}{3} x^3\) So, \[ [-40x + 18x^2 - \frac{8}{3}x^3]_{x=0}^{x=2} \] Compute at \(x=2\): - \(-40 \times 2 = -80\) - \(18 \times 4 = 72\) - \(-\frac{8}{3} \times 8 = -\frac{64}{3}\) So, \[ -80 + 72 - \frac{64}{3} = (-80 + 72) - \frac{64}{3} = -8 - \frac{64}{3} = -\frac{24}{3} - \frac{64}{3} = -\frac{88}{3} \] At \(x=0\), everything is 0. So, the answer is: \[ \boxed{-\frac{88}{3}} \] --- ## **Final Answer** \[ \boxed{ \int_C \mathbf{F} \cdot d\mathbf{r} = -\frac{88}{3} } \] **This is the value of the line integral.**