Let's break down the problem step by step: ## 1. **Determine if \( \mathbf{F} \) is Conservative and Find a Potential Function** Given: \[ \mathbf{F} = \frac{1}{y^2z^2}(yz, -zx, -xy + 4y^2z^2) \] Let’s write the components: \[ F_1 = \frac{yz}{y^2z^2} = \frac{1}{yz} \] \[ F_2 = \frac{-zx}{y^2z^2} = -\frac{x}{y^2z} \] \[ F_3 = \frac{-xy + 4y^2z^2}{y^2z^2} = -\frac{x}{z^2} + 4 \] Now, check if \( \mathbf{F} \) is conservative by looking for a potential function \( f(x, y, z) \) such that: \[ \nabla f = \mathbf{F} \] So, \[ \frac{\partial f}{\partial x} = \frac{1}{yz} \] \[ \frac{\partial f}{\partial y} = -\frac{x}{y^2z} \] \[ \frac{\partial f}{\partial z} = -\frac{x}{z^2} + 4 \] ### **Step 1: Integrate \( F_1 \) w.r.t. \( x \):** \[ f(x, y, z) = \int \frac{1}{yz} dx = \frac{x}{yz} + C(y, z) \] ### **Step 2: Differentiate w.r.t. \( y \) and set equal to \( F_2 \):** \[ \frac{\partial f}{\partial y} = \frac{\partial}{\partial y}\left(\frac{x}{yz}\right) + \frac{\partial C}{\partial y} = -\frac{x}{y^2z} + \frac{\partial C}{\partial y} \] Set equal to \( F_2 \): \[ -\frac{x}{y^2z} + \frac{\partial C}{\partial y} = -\frac{x}{y^2z} \implies \frac{\partial C}{\partial y} = 0 \] So \( C \) is only a function of \( z \): \( C(z) \) ### **Step 3: Differentiate w.r.t. \( z \) and set equal to \( F_3 \):** \[ \frac{\partial f}{\partial z} = \frac{\partial}{\partial z}\left(\frac{x}{yz}\right) + \frac{dC}{dz} = -\frac{x}{yz^2} + \frac{dC}{dz} \] Set equal to \( F_3 \): \[ -\frac{x}{z^2} + 4 = -\frac{x}{yz^2} + \frac{dC}{dz} \] But \( -\frac{x}{z^2} \neq -\frac{x}{yz^2} \) in general, **unless \( y = 1 \)**, which is not always true. #### **Conclusion:** **There is no scalar potential function for \( \mathbf{F} \) in all of \( \mathbb{R}^3 \).** So, \( \mathbf{F} \) is **not conservative**. --- ## 2. **Value of the Line Integral Along the Given Path** The path is from \( (0, 0) \to (2, 1) \to (3, 3) \) in the \( xy \)-plane. But the vector field \( \mathbf{F} \) is 3-dimensional with variables \( x, y, z \). The path shown is only in the \( xy \)-plane, with no \( z \)-component specified. **Assumption:** If the path is in the \( xy \)-plane, we can assume \( z = 0 \) along the path. But, if \( z = 0 \), then \( \mathbf{F} \) is not defined (division by zero). --- ## **Final Answers:** ### **1. Is there a potential function?** **No**, there does not exist a potential function for the given \( \mathbf{F} \) because the mixed partials do not match (the field fails the test for conservativeness). ### **2. What is the value of the line integral?** **The line integral is not defined along the path shown,** because the field \( \mathbf{F} \) is not defined when \( z = 0 \), which is the case for any path lying entirely in the \( xy \)-plane. --- ## **Summary Table** | Part | Answer | |------|------------------------------------------------------------------------| | 1 | No potential function exists; \( \mathbf{F} \) is not conservative. | | 2 | The line integral is not defined along the shown path (\( z = 0 \)). | If you have further details about the path or \( z \)-values, let me know!