Answer
See proof
Work Step by Step
Step 1: We will discuss some of the ways that are mentioned throughout the chapters and the previous problems that can tell us if some vector field \(\vec{F}\) is a conservative vector field. In order to do this, we should first define what a conservative vector field is: A vector field is a conservative vector field if it can be expressed in terms of its potential function: \[ \vec{F} = \nabla \phi \] Where \(\phi\) is the potential function of a conservative vector field \(\vec{F}\).
Step 2: Let's say that we have some open and simply connected region \(\mathcal{D}\) where the conservative vector field \(\vec{F}\) is defined. One of the theorems we proved in the previous problems states that the work integral, \[ \int_{\vec{C}} \vec{F} \cdot d\vec{r} \] of a conservative vector field \(\vec{F}\) over a path \(\vec{C}\) that is also located in \(\mathcal{D}\), between the points \(\vec{P}\) and \(\vec{Q}\) is determined only by the values of \(\phi\) at \(\vec{P}\) and \(\vec{Q}\): \[ \int_{\vec{C}} \vec{F} \cdot d\vec{r} = \phi(\vec{Q}) - \phi(\vec{P}) \] This means that if we can find two paths \(\vec{C_1}\) and \(\vec{C_2}\) in \(\mathcal{D}\) that have the same start and end point \(\vec{P}\) and \(\vec{Q}\), but give a different value for the work integral of \(\vec{F}\) along them: \[ \int_{\vec{C_1}} \vec{F} \cdot d\vec{r} \neq \int_{\vec{C_2}} \vec{F} \cdot d\vec{r} \] the vector field \(\vec{F}\) cannot be conservative.
Step 3: Also, another way we can tell if the vector field \(\vec{F}\) is conservative is by determining its line integral over a closed smooth curve \(\vec{C}\), also defined over region \(\mathcal{D}\). If this line integral gives any other value except \(0\), this means that \(\vec{F}\) is not conservative. This result is closely related to what we've discussed above, since it can be concluded from (2) by taking that, \(\vec{C} = \vec{P} = \vec{Q}\), which implies that the path of integration is a closed curve: \[ \int_{\vec{C}} \vec{F} \cdot d\vec{r} = \phi(\vec{Q}) - \phi(\vec{P}) = \phi(\vec{Q}) - \phi(\vec{Q}) = 0 \]
Step 4: Note that both of the previous methods are applicable only for the vector field \(\vec{F}\) and the path of integration \(\vec{C}\) that are defined in an open and simply connected region \(\mathcal{D}\). Open means that the boundary points of the region are not a part of the region itself and simply connected means that there is no smooth curve in \(\mathcal{D}\) that encloses a region inside \(\mathcal{D}\) with elements that do not belong in \(\mathcal{D}\) (i.e., there are no holes in the region). In the next steps, we will focus on some methods that can tell us if the field is conservative for the particular cases of two and three-dimensional vector fields.
Step 5: If we are given a two-dimensional vector field: \[ \vec{F}(x, y) = f(x, y) \mathbf{i} + g(x, y) \mathbf{j} \] that is defined with its derivatives in the region \(\mathcal{D}\), there is a theorem, that was discussed in the previous problems, that tells us that the vector field (3) is conservative if the condition: \[ \frac{\partial g}{\partial x} = \frac{\partial f}{\partial y} \] is satisfied. So, if we were given a two-dimensional vector field \(\vec{F}\) and we find that (4) does not hold in the region where \(\vec{F}\) is defined with its derivatives, then the vector field is not conservative.
Step 6: The theorem above can be generalized to three dimensions, i.e. we are given a vector field: \[ \vec{F}(x, y, z) = f(x, y, z) \mathbf{i} + g(x, y, z) \mathbf{j} + h(x, y, z) \mathbf{k} \] The analogous condition (4) for three dimensions is given by equations: \[ \frac{\partial h}{\partial x} = \frac{\partial g}{\partial y}, \quad \frac{\partial h}{\partial y} = \frac{\partial f}{\partial z}, \quad \frac{\partial g}{\partial x} = \frac{\partial h}{\partial z} \] So, like in the two-dimensional case, if (6) is not true in the region where (5) is defined with its derivatives, then the vector field (5) cannot be conservative.
Step 7: Note that even if (6) or (4) are satisfied, but there is a "hole" in the region \(\mathcal{D}\), meaning it is not simply connected, then these two theorems don't hold and they can't tell us if the field is conservative. Generally, if we find any way to prove that the vector field cannot be expressed in terms of its potential function as (1), we can say that the vector field is not conservative.
Step 8: Let's recap the methods we discussed: 1. If we can find two paths with the same start and endpoints that give a different value for the work integral of a vector field \(\vec{F}\), then the vector field \(\vec{F}\) is not conservative. 2. If the work integral of a vector field \(\vec{F}\) around a closed smooth curve is anything except \(0\), then the vector field is not conservative. 3. If \(\frac{\partial g}{\partial x} \neq \frac{\partial f}{\partial y}\) anywhere in the region where \(\vec{F}(x, y)\) is defined with its derivatives, then \(\vec{F}\) is not conservative. 4. If any of the equations: \(\frac{\partial h}{\partial x} = \frac{\partial g}{\partial y}, \frac{\partial h}{\partial y} = \frac{\partial f}{\partial z}, \frac{\partial g}{\partial x} = \frac{\partial h}{\partial z}\) are not satisfied in the region where \(\vec{F}(x, y, z)\) is defined with its derivatives, then \(\vec{F}\) is not conservative. 5. If the region where two previous conditions hold is not open and simply connected, then the two previous conclusions don't hold. 6. If we can prove that \(\vec{F}\) cannot be expressed in terms of any scalar function \(\phi\) as \(\vec{F} = \nabla \phi\), then the field is not conservative.
