Answer :
To solve the problem using the Fundamental Theorem of Line Integrals, we need to proceed through the following steps:
1. Verify if the vector field [tex]\(\vec{F}\)[/tex] is conservative.
2. Find the potential function [tex]\(f(x, y, z)\)[/tex] if the vector field is conservative.
3. Evaluate the potential function at the endpoints [tex]\((1, 0, 2)\)[/tex] and [tex]\((1, 1, 1)\)[/tex].
4. Calculate the line integral using the fundamental theorem for the conservative vector field.
---
### Step 1: Verify if the Vector Field is Conservative
A vector field [tex]\(\vec{F}(x, y, z) = \langle P, Q, R \rangle\)[/tex] is conservative if there exists a potential function [tex]\(f(x, y, z)\)[/tex] such that:
[tex]\[ \vec{F} = \nabla f = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z} \right) \][/tex]
One way to check if [tex]\(\vec{F}\)[/tex] is conservative is to verify that the curl of [tex]\(\vec{F}\)[/tex] is zero:
[tex]\[ \nabla \times \vec{F} = \vec{0} \][/tex]
Given:
[tex]\[ \vec{F}(x, y, z) = \langle 4x - 2y + 2z, -2x + 2yz + z^2, 2x + 2yz + y^2 \rangle \][/tex]
We compute the curl [tex]\(\nabla \times \vec{F}\)[/tex]:
[tex]\[ \nabla \times \vec{F} = \left( \frac{\partial R}{\partial y} - \frac{\partial Q}{\partial z}, \frac{\partial P}{\partial z} - \frac{\partial R}{\partial x}, \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right) \][/tex]
Calculate each component:
1. [tex]\(\frac{\partial R}{\partial y} - \frac{\partial Q}{\partial z}\)[/tex]:
[tex]\[ \frac{\partial}{\partial y} (2x + 2yz + y^2) = 2z + 2y \][/tex]
[tex]\[ \frac{\partial}{\partial z} (-2x + 2yz + z^2) = 2y + 2z \][/tex]
[tex]\[ \frac{\partial R}{\partial y} - \frac{\partial Q}{\partial z} = (2z + 2y) - (2y + 2z) = 0 \][/tex]
2. [tex]\(\frac{\partial P}{\partial z} - \frac{\partial R}{\partial x}\)[/tex]:
[tex]\[ \frac{\partial}{\partial z} (4x - 2y + 2z) = 2 \][/tex]
[tex]\[ \frac{\partial}{\partial x} (2x + 2yz + y^2) = 2 \][/tex]
[tex]\[ \frac{\partial P}{\partial z} - \frac{\partial R}{\partial x} = 2 - 2 = 0 \][/tex]
3. [tex]\(\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}\)[/tex]:
[tex]\[ \frac{\partial}{\partial x} (-2x + 2yz + z^2) = -2 \][/tex]
[tex]\[ \frac{\partial}{\partial y} (4x - 2y + 2z) = -2 \][/tex]
[tex]\[ \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} = -2 - (-2) = 0 \][/tex]
Since all components of the curl are zero, [tex]\(\nabla \times \vec{F} = \langle 0, 0, 0 \rangle\)[/tex]. Thus, [tex]\(\vec{F}\)[/tex] is conservative.
---
### Step 2: Find the Potential Function [tex]\(f(x, y, z)\)[/tex]
Since [tex]\(\vec{F}\)[/tex] is conservative, there exists a potential function [tex]\(f(x, y, z)\)[/tex] such that:
[tex]\[ \vec{F} = \nabla f \][/tex]
We need to find [tex]\(f\)[/tex] such that:
[tex]\[ \nabla f = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z} \right) = \left( 4x - 2y + 2z, -2x + 2yz + z^2, 2x + 2yz + y^2 \right) \][/tex]
Integrate the first component with respect to [tex]\(x\)[/tex]:
[tex]\[ \frac{\partial f}{\partial x} = 4x - 2y + 2z \implies f(x, y, z) = 2x^2 - 2xy + 2xz + g(y, z) \][/tex]
Here, [tex]\(g(y, z)\)[/tex] is a function of [tex]\(y\)[/tex] and [tex]\(z\)[/tex] because the derivative with respect to [tex]\(x\)[/tex] should not change [tex]\(g(y, z)\)[/tex].
Integrate the second component with respect to [tex]\(y\)[/tex]:
[tex]\[ \frac{\partial f}{\partial y} = -2x + 2yz + z^2 \implies f(x, y, z) = -2xy + y^2z + z^2y + h(x, z) \][/tex]
Combining and resolving the terms, we assume consistent forms:
[tex]\[ f(x, y, z) = 2x^2 - 2xy + 2xz + y^2z + z^2y \][/tex]
Thus, potential function [tex]\( f \)[/tex] is:
[tex]\[ f(x, y, z) = 2x^2 - 2xy + 2xz + y^2z + z^2y \][/tex]
---
### Step 3: Evaluate the Potential Function at the Endpoints
Evaluate [tex]\( f \)[/tex] at [tex]\((1, 0, 2)\)[/tex]:
[tex]\[ f(1, 0, 2) = 2(1)^2 - 2(1)(0) + 2(1)(2) + (0)^2(2) + (2)^2(0) = 2 + 4= 6 \][/tex]
Evaluate [tex]\( f \)[/tex] at [tex]\((1, 1, 1)\)[/tex]:
[tex]\[ f(1, 1, 1) = 2(1)^2 - 2(1)(1) + 2(1)(1) + (1)^2(1) + (1)^2(1) = 2-2+2 +1+1 = 4 \][/tex]
---
### Step 4: Calculate the Line Integral
Using the Fundamental Theorem of Line Integrals:
[tex]\[ \int_C \vec{F} \cdot d\vec{r} = f(1, 1, 1) - f(1, 0, 2) \][/tex]
Thus,
[tex]\[ \int_C \vec{F} \cdot d\vec{r} = 4 - 6 = -2 \][/tex]
Final Answer:
[tex]\[ \boxed{-2} \][/tex]
1. Verify if the vector field [tex]\(\vec{F}\)[/tex] is conservative.
2. Find the potential function [tex]\(f(x, y, z)\)[/tex] if the vector field is conservative.
3. Evaluate the potential function at the endpoints [tex]\((1, 0, 2)\)[/tex] and [tex]\((1, 1, 1)\)[/tex].
4. Calculate the line integral using the fundamental theorem for the conservative vector field.
---
### Step 1: Verify if the Vector Field is Conservative
A vector field [tex]\(\vec{F}(x, y, z) = \langle P, Q, R \rangle\)[/tex] is conservative if there exists a potential function [tex]\(f(x, y, z)\)[/tex] such that:
[tex]\[ \vec{F} = \nabla f = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z} \right) \][/tex]
One way to check if [tex]\(\vec{F}\)[/tex] is conservative is to verify that the curl of [tex]\(\vec{F}\)[/tex] is zero:
[tex]\[ \nabla \times \vec{F} = \vec{0} \][/tex]
Given:
[tex]\[ \vec{F}(x, y, z) = \langle 4x - 2y + 2z, -2x + 2yz + z^2, 2x + 2yz + y^2 \rangle \][/tex]
We compute the curl [tex]\(\nabla \times \vec{F}\)[/tex]:
[tex]\[ \nabla \times \vec{F} = \left( \frac{\partial R}{\partial y} - \frac{\partial Q}{\partial z}, \frac{\partial P}{\partial z} - \frac{\partial R}{\partial x}, \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right) \][/tex]
Calculate each component:
1. [tex]\(\frac{\partial R}{\partial y} - \frac{\partial Q}{\partial z}\)[/tex]:
[tex]\[ \frac{\partial}{\partial y} (2x + 2yz + y^2) = 2z + 2y \][/tex]
[tex]\[ \frac{\partial}{\partial z} (-2x + 2yz + z^2) = 2y + 2z \][/tex]
[tex]\[ \frac{\partial R}{\partial y} - \frac{\partial Q}{\partial z} = (2z + 2y) - (2y + 2z) = 0 \][/tex]
2. [tex]\(\frac{\partial P}{\partial z} - \frac{\partial R}{\partial x}\)[/tex]:
[tex]\[ \frac{\partial}{\partial z} (4x - 2y + 2z) = 2 \][/tex]
[tex]\[ \frac{\partial}{\partial x} (2x + 2yz + y^2) = 2 \][/tex]
[tex]\[ \frac{\partial P}{\partial z} - \frac{\partial R}{\partial x} = 2 - 2 = 0 \][/tex]
3. [tex]\(\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}\)[/tex]:
[tex]\[ \frac{\partial}{\partial x} (-2x + 2yz + z^2) = -2 \][/tex]
[tex]\[ \frac{\partial}{\partial y} (4x - 2y + 2z) = -2 \][/tex]
[tex]\[ \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} = -2 - (-2) = 0 \][/tex]
Since all components of the curl are zero, [tex]\(\nabla \times \vec{F} = \langle 0, 0, 0 \rangle\)[/tex]. Thus, [tex]\(\vec{F}\)[/tex] is conservative.
---
### Step 2: Find the Potential Function [tex]\(f(x, y, z)\)[/tex]
Since [tex]\(\vec{F}\)[/tex] is conservative, there exists a potential function [tex]\(f(x, y, z)\)[/tex] such that:
[tex]\[ \vec{F} = \nabla f \][/tex]
We need to find [tex]\(f\)[/tex] such that:
[tex]\[ \nabla f = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z} \right) = \left( 4x - 2y + 2z, -2x + 2yz + z^2, 2x + 2yz + y^2 \right) \][/tex]
Integrate the first component with respect to [tex]\(x\)[/tex]:
[tex]\[ \frac{\partial f}{\partial x} = 4x - 2y + 2z \implies f(x, y, z) = 2x^2 - 2xy + 2xz + g(y, z) \][/tex]
Here, [tex]\(g(y, z)\)[/tex] is a function of [tex]\(y\)[/tex] and [tex]\(z\)[/tex] because the derivative with respect to [tex]\(x\)[/tex] should not change [tex]\(g(y, z)\)[/tex].
Integrate the second component with respect to [tex]\(y\)[/tex]:
[tex]\[ \frac{\partial f}{\partial y} = -2x + 2yz + z^2 \implies f(x, y, z) = -2xy + y^2z + z^2y + h(x, z) \][/tex]
Combining and resolving the terms, we assume consistent forms:
[tex]\[ f(x, y, z) = 2x^2 - 2xy + 2xz + y^2z + z^2y \][/tex]
Thus, potential function [tex]\( f \)[/tex] is:
[tex]\[ f(x, y, z) = 2x^2 - 2xy + 2xz + y^2z + z^2y \][/tex]
---
### Step 3: Evaluate the Potential Function at the Endpoints
Evaluate [tex]\( f \)[/tex] at [tex]\((1, 0, 2)\)[/tex]:
[tex]\[ f(1, 0, 2) = 2(1)^2 - 2(1)(0) + 2(1)(2) + (0)^2(2) + (2)^2(0) = 2 + 4= 6 \][/tex]
Evaluate [tex]\( f \)[/tex] at [tex]\((1, 1, 1)\)[/tex]:
[tex]\[ f(1, 1, 1) = 2(1)^2 - 2(1)(1) + 2(1)(1) + (1)^2(1) + (1)^2(1) = 2-2+2 +1+1 = 4 \][/tex]
---
### Step 4: Calculate the Line Integral
Using the Fundamental Theorem of Line Integrals:
[tex]\[ \int_C \vec{F} \cdot d\vec{r} = f(1, 1, 1) - f(1, 0, 2) \][/tex]
Thus,
[tex]\[ \int_C \vec{F} \cdot d\vec{r} = 4 - 6 = -2 \][/tex]
Final Answer:
[tex]\[ \boxed{-2} \][/tex]
