Answer
See proof
Work Step by Step
Step 1: We are given the integral: \[ \int 2xz \, dx + yz \, dy + x^2 \, dz \] and we want to check if this integral is independent of the path it is taken over between the two points \(-\). If this is true, we say that the integral is path-independent.
Step 2: In the previous problem, we've found out that the vector field \(\mathbf{F}(x, y, z) = f(x, y, z) \mathbf{i} + g(x, y, z) \mathbf{j} + h(x, y, z) \mathbf{k}\) is conservative if the components of that vector field \(f, g\), and \(h\) satisfy these conditions: \[ \begin{align*} \frac{\partial f}{\partial x} &= \frac{\partial g}{\partial y} \\ \frac{\partial f}{\partial y} &= \frac{\partial h}{\partial x} \\ \frac{\partial g}{\partial z} &= \frac{\partial h}{\partial y} \end{align*} \] The work integral, \(\int \mathbf{F} \cdot d\mathbf{r}\), is path-independent if \(\mathbf{F}\) is conservative. Also, this work integral can be represented as: \[ \int \mathbf{F} \cdot d\mathbf{r} = \int f(x, y, z) \, dx + g(x, y, z) \, dy + h(x, y, z) \, dz \] Step 3: By comparing the form of the work integral and the given integral, we can see that the given integral is also a work integral of some unknown vector field. We can recognize the components of the given work integral as: \[ \begin{align*} f(x, y, z) &= 2xz \\ g(x, y, z) &= yz \\ h(x, y, z) &= x^2 \end{align*} \] This enables us to check if these components satisfy the conditions for the vector field to be conservative and, by extension, if the given integral is path-independent. Step 4: Let's check the first of the three conditions for the components (4) and (5): \[ \begin{align*} \frac{\partial f}{\partial x} &= 0 \\ \frac{\partial g}{\partial y} &= 0 \end{align*} \] So, the first condition, \(\frac{\partial f}{\partial x} = \frac{\partial g}{\partial y} = 0\), is satisfied. Step 5: Let's check if the second condition is satisfied for the components (4) and (6): \[ \begin{align*} \frac{\partial f}{\partial y} &= 2x \\ \frac{\partial h}{\partial x} &= 2x \end{align*} \] So, the second condition is satisfied as well. Step 6: Finally, we check the condition \(\frac{\partial f}{\partial x} = \frac{\partial h}{\partial y}\) for the components (6) and (7): \[ \begin{align*} \frac{\partial g}{\partial z} &= y \\ \frac{\partial h}{\partial y} &= 0 \end{align*} \] In this case, we see that the third condition is not satisfied: \(\frac{\partial g}{\partial z} \neq \frac{\partial h}{\partial y}\). Since all three conditions must be satisfied for the integral to be path-independent, this means that the given integral is not path-independent. Step 7: Let's recap on what we've just done: We have compared the given integral to the work integral of some unknown vector field. This enabled us to use the conditions to check if the unknown vector field is conservative, meaning that the corresponding work integral is path-independent. We checked all three conditions for the components of the given work integral and concluded that the integral is not path-independent.
