Suppose \(P(x,y) = \frac{-y}{x^2+y^2}\text{,}\) \(Q(x,y) = \frac{x}{x^2+y^2}\text{.}\) If we think of \((P,Q)\) as a vector, so that we have a so-called vector field, \((P,Q)\) is called the vortex vector field, as it gives the velocity of particles traveling in a vortex around the origin. Variations on this vector field come up often in applications. Suppose that \(\gamma\) is a path that goes counterclockwise around a rectangle whose interior contains the origin. We claim
\begin{equation*}
\int_{\gamma} \frac{-y}{x^2+y^2} \, dx + \frac{x}{x^2+y^2} \, dy = 2 \pi .
\end{equation*}
First we draw a circle
\(C\) of radius
\(r > 0\) centered at the origin such that the entire circle is within
\(\gamma\) and oriented clockwise. Consider
\(U\) to be the domain between
\(\gamma\) and
\(C\text{.}\) See
Figure 10.16. The integral around
\(\partial U\) is the integral around
\(\gamma\) plus the integral around
\(C\text{.}\) Now
\(U\) is not a domain of type III, so we cannot just apply the version of Green’s theorem we actually proved. However, if we cut the box along the axis as shown in the figure with dashed lines, the four resulting domains, let us call them
\(U_1,U_2,U_3,U_4\text{,}\) are of type III. The dashed lines are oriented in opposite directions for the two
\(U_j\) that share them, and so when we integrate along both, the integrals cancel. That is,
\begin{multline*}
\int_{\partial U} P\, dx + Q \, dy
=
\\
\int_{\partial U_1} P\, dx + Q \, dy
+
\int_{\partial U_2} P\, dx + Q \, dy
+
\int_{\partial U_3} P\, dx + Q \, dy
+
\int_{\partial U_4} P\, dx + Q \, dy .
\end{multline*}
Now we are allowed to apply Green’s theorem to every \(U_j\text{.}\) We leave it to the reader to verify that outside of the origin, \(\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} = 0\text{.}\) So for each \(j\text{,}\) we find
\begin{equation*}
\int_{\partial U_j} P\, dx + Q \, dy
=
\int_{U_j} \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial
y} \right)
=
\int_{U_j} 0 = 0 ,
\end{equation*}
and so \(\int_{\partial U} P\, dx + Q \, dy = 0\text{.}\) As \(\partial U\) is \(C\) together with \(\gamma\text{,}\) we find
\begin{equation*}
\int_C P \, dx + Q \, dy
+
\int_{\gamma} P \, dx + Q \, dy
=
\int_{\partial U} P \, dx + Q \, dy = 0 .
\end{equation*}
So the integral around \(C\) is minus the integral around \(\gamma\text{.}\) The integral around \(C\) is easy to compute as on \(C\) we have \(x^2+y^2 = r^2\text{,}\) so \(P(x,y) = \frac{-y}{r^2}\) and \(Q(x,y) = \frac{x}{r^2}\text{.}\) We leave it to the reader to compute
\begin{equation*}
\int_C P\, dx + Q \, dy
=
\int_C \frac{-y}{r^2}\, dx + \frac{x}{r^2} \, dy
=
- 2 \pi .
\end{equation*}
The claim follows.