Answer
See proof
Work Step by Step
In b): \[ \int_{\vec{C}} \vec{r} \cdot \vec{r}_i = 0 \quad \text{for every piecewise smooth closed curve } \vec{C} \text{ in } D \] \[ \int_C \mathbf{F} \cdot d\mathbf{r} = 0 \quad \text{for every piecewise smooth closed curve } C \text{ in } D \] In c): \[ \int_{\vec{C}} \vec{r} \cdot \vec{r}_i \text{ is independent of the path from any point } P \text{ in } D \text{ to any point } Q \text{ in } D \text{ for every piecewise smooth curve } \] \[ \int_{C} \mathbf{F} \cdot d\mathbf{r} \text{ is independent of the path from any point } P \text{ in } D \text{ to any point } Q \text{ in } D \text{ for every piecewise smooth curve } \] Let $\vec{P}, \vec{Q}$ be any two points in $D$ and $\vec{C}$ any smooth closed curve in $D$ such that $\vec{P}, \vec{Q}$ are in the curve. Then, by b): \[ \int_{\vec{C}} \vec{r} \cdot \vec{r}_i = 0 \quad \text{and} \quad \int_{C} \mathbf{F} \cdot d\mathbf{r} = 0 \] Now, we recall $\vec{C}_1, -\vec{C}_2$ such that $\vec{C}_1$ goes from $\vec{P}$ to $\vec{Q}$ and $-\vec{C}_2$ goes from $\vec{Q}$ to $\vec{P}$. Using the result in exercise 31 (solution available): \[ \int_{\vec{C}} \vec{r} \cdot \vec{r}_i = \int_{\vec{C}_1} \vec{r} \cdot \vec{r}_i + \int_{-\vec{C}_2} \vec{r} \cdot \vec{r}_i = 0 \] \[ \int_{C} \mathbf{F} \cdot d\mathbf{r} = \int_{C_1} \mathbf{F} \cdot d\mathbf{r} + \int_{C_2} \mathbf{F} \cdot d\mathbf{r} = 0 \] But \[ \int_{\vec{C}_1} \vec{r} \cdot \vec{r}_i - \int_{-\vec{C}_2} \vec{r} \cdot \vec{r}_i = 0 \equiv \int_{\vec{C}_1} \vec{r} \cdot \vec{r}_i = \int_{\vec{C}_2} \vec{r} \cdot \vec{r}_i \] \[ \int_{C_1} \mathbf{F} \cdot d\mathbf{r} - \int_{C_2} \mathbf{F} \cdot d\mathbf{r} = 0 \equiv \int_{C_1} \mathbf{F} \cdot d\mathbf{r} = \int_{C_2} \mathbf{F} \cdot d\mathbf{r} \] For two arbitrary curves $\vec{C}_1, \vec{C}_2$ in $D$ that go from any point $\vec{P}$ to any point $\vec{Q}$ in $D$, there exists a closed curve $\vec{C} = \vec{C}_1 - \vec{C}_2$ in $D$ such that \[ \int_{\vec{C}} \vec{r} \cdot \vec{r}_i = 0 \equiv \int_{\vec{C}_1} \vec{r} \cdot \vec{r}_i = \int_{\vec{C}_2} \vec{r} \cdot \vec{r}_i \] \[ \int_{C} \mathbf{F} \cdot d\mathbf{r} = 0 \equiv \int_{C_1} \mathbf{F} \cdot d\mathbf{r} = \int_{C_2} \mathbf{F} \cdot d\mathbf{r} \]
