Chapter 17 - Line and Surface Integrals - 17.3 Conservative Vector Fields - Exercises - Page 946: 29

Case 1. Figure 20 (A) $\mathop \oint \limits_C^{} {\bf{F}}\cdot{\rm{d}}{\bf{r}} = 2\pi $ Case 2. Figure 20 (B) $\mathop \oint \limits_C^{} {\bf{F}}\cdot{\rm{d}}{\bf{r}} = 2\pi $ Case 3. Figure 20 (C) $\mathop \oint \limits_C^{} {\bf{F}}\cdot{\rm{d}}{\bf{r}} = 0$ Case 4. Figure 20 (D) $\mathop \oint \limits_C^{} {\bf{F}}\cdot{\rm{d}}{\bf{r}} = - 2\pi $ Case 5. Figure 20 (E) $\mathop \oint \limits_C^{} {\bf{F}}\cdot{\rm{d}}{\bf{r}} = 4\pi $

We have the vortex field: ${\bf{F}} = \left( {\dfrac{{ - y}}{{{x^2} + {y^2}}},\dfrac{x}{{{x^2} + {y^2}}}} \right)$. Notice that ${\bf{F}}$ is not defined at the origin. As is mentioned in the Conceptual Insight on page 943, the potential function for ${\bf{F}}$ is given by $f\left( {x,y} \right) = \theta = {\tan ^{ - 1}}\dfrac{y}{x}$, where ${\bf{F}} = \nabla f$. If a closed path ${\bf{r}}$ winds around the origin $n$ times, then $\mathop \smallint \limits_C^{} {\bf{F}}\cdot{\rm{d}}{\bf{r}} = 2\pi n$ Case 1. Figure 20 (A) The closed path ${\bf{r}}$ winds around the origin 1 time in counterclockwise direction, so $\mathop \oint \limits_C^{} {\bf{F}}\cdot{\rm{d}}{\bf{r}} = 2\pi $ Case 2. Figure 20 (B) The closed path ${\bf{r}}$ winds around the origin 1 time in counterclockwise direction, so $\mathop \oint \limits_C^{} {\bf{F}}\cdot{\rm{d}}{\bf{r}} = 2\pi $ Case 3. Figure 20 (C) The closed path ${\bf{r}}$ does not wind around the origin, thus the domain of definition of ${\bf{F}}$ is simply connected. Since ${\bf{F}}$ satisfies the cross-partials condition (as is verified in Example 8), therefore, by Theorem 4 the field ${\bf{F}}$ is conservative. Hence $\mathop \oint \limits_C^{} {\bf{F}}\cdot{\rm{d}}{\bf{r}} = 0$ Case 4. Figure 20 (D) The closed path ${\bf{r}}$ winds around the origin 1 times in clockwise direction, so $\mathop \oint \limits_C^{} {\bf{F}}\cdot{\rm{d}}{\bf{r}} = 2\pi \left( { - 1} \right) = - 2\pi $ Case 5. Figure 20 (E) The closed path ${\bf{r}}$ winds around the origin 2 times in counterclockwise direction, so $\mathop \oint \limits_C^{} {\bf{F}}\cdot{\rm{d}}{\bf{r}} = 2\pi \left( 2 \right) = 4\pi $