Section 13.3 Application: Conservative Vector Fields in Space
In Section 8.2 we stated without a proof that a closed vector field, \(\vect f\text{,}\) on a simply connected domain \(D\subset\mathbb R^3\) is conservative. We now want to apply the theorem of Stokes to sketch a proof. Simply connected means that every closed curve in \(D\) can be continuously deformed into a point in \(D\) without ever leaving \(D\) as shown in Figure 8.9. It turns out that this is equivalent to the following. Suppose that \(C_0\) and \(C_1\) are smooth curves in \(D\) connecting \(\vect x_0\in D\) to \(\vect x_1\in D\text{.}\) Then it is possible to deform \(C_0\) smoothly into \(C_1\) without leaving \(D\text{.}\) When doing this the “trace” left by the moving curve defines a surface, \(S\text{,}\) with boundary \(\partial S=C_0\cup C_1\) as shown in Figure 13.5.
Choosing consistent orientations for \(S\) and \(\partial S\) we get from Stokes' Theorem Theorem 13.2
\begin{align*}
\int_{C_0}\vect f\cdot d\vect x-\int_{C_1}\vect f\cdot d\vect x
\amp=\int_{\partial S}\vect f\cdot d\vect x\\
\amp=\int_S(\curl\vect f)\cdot\vect n\,dS\text{.}
\end{align*}
We saw in Section 8.2 that a vector field defined on a subset \(D\) of \(\mathbb R^3\) is closed if and only if \(\curl\vect f=0\) on \(D\text{.}\) Hence, if \(\vect f\) is closed, then the above surface integral is zero, and therefore
\begin{equation*}
\int_{C_0}\vect f\cdot d\vect x
=\int_{C_1}\vect f\cdot d\vect x\text{.}
\end{equation*}
This shows that line integrals are path independent and thus \(\vect f\) is conservative.
