Consider the vector field \(\vF =
\displaystyle\frac{-y}{{x^2+y^2}}\vi +
\frac{x}{{x^2+y^2}}\vj\text{.}\) Notice that \(\vF\) is smooth everywhere in the plane other than at the point \((0,0)\text{.}\) This vector field is plotted in Figure 13.8.6, but we have not plotted the vectors close to the origin as their magnitudes get so large that they make it hard to interpret the figure. Here Green’s Theorem applies to any simple closed curve \(C\) that neither passes through \((0,0)\) nor bounds a region containing \((0,0)\text{.}\)
Suppose that \(C\) is the unit circle centered at the origin. Without doing any calculations, what can you say about \(\oint_C \vF\cdot d \vr\text{?}\) what does this tell you about if \(\vF\) is path-independent?
Is the vector field \(\vG =
\displaystyle\frac{x}{x^2+y^2}\vi +\frac{y}{x^2+y^2}\vj\text{,}\) which is shown in Figure 13.8.7, path-independent? Why or why not?
