Example 8.12. A closed but not conservative vector field.
Let \(\vect f\) be the vector field given by
\begin{equation*}
\vect f(x,y)=\Bigl(\frac{-y}{x^2+y^2},\frac{x}{x^2+y^2}\Bigr)
\end{equation*}
for \((x,y)\neq(0,0)\text{.}\) It is shown in Figure 8.13.
Note that \(\|\vect f\|=\|(x,y)\|^{-1}\text{,}\) so the length of the vectors go to infinity as \((x,y)\) approaches \((0,0)\text{.}\) We say that \(\vect f\) has a singularity at \((0,0)\text{.}\) We check whether \(\vect f\) is closed
Solution.
We have that
\begin{align*}
\frac{\partial}{\partial y}\frac{-y}{x^2+y^2}
\amp=\frac{-(x^2+y^2)+2y^2}{(x^2+y^2)^2}
=\frac{y^2-x^2}{(x^2+y^2)^2}\\
\frac{\partial}{\partial x}\frac{x}{x^2+y^2}
\amp=\frac{x^2+y^2-2y^2}{(x^2+y^2)^2}
=\frac{y^2-x^2}{(x^2+y^2)^2}\text{.}
\end{align*}
Hence \(\vect f\) is indeed a closed vector field on \(\mathbb R^2\setminus\{\vect 0\}\text{.}\) We now want to compute the integral along the circle of radius \(R\) centred at \((0,0)\text{.}\) Assume that the circle is oriented counterclockwise. We parametrise it by
\begin{align*}
\vect\gamma(t)\amp=(R\cos t,R\sin t),\amp\amp t\in[0, 2\pi]\text{.}
\end{align*}
Then
\begin{equation*}
\vect\gamma'(t)=(-R\sin t,R\cos t)\text{,}
\end{equation*}
and using the definition of the line integral we see that
\begin{align*}
\int_C\vect f\cdot\,d\vect x
\amp=\int_0^{2\pi}\frac{R\sin t}{R^2\sin^2+R^2\cos^t}R\sin t
+\int_0^{2\pi}\frac{R\cos t}{R^2\sin^2+R^2\cos^t}R\cos t\,dt\\
\amp=\int_0^{2\pi}\frac{R^2\cos t+R^2\sin^2}{R^2\sin^2+R^2\cos^t}\,dt
=\int_0^{2\pi}1\,dt=2\pi\text{.}
\end{align*}
The circle is a closed curve. Hence if \(\vect f\) were conservative the integral would be zero. As it is not zero, \(\vect f\) cannot be conservative, even though it is closed. Note that \(\mathbb R^2\setminus\{\vect 0\}\) is not simply connected, so Theorem 8.11 does not apply!
