The proof that the condition \(\nabla \times \mathbf{F}=0\) guarantees the path
independence of the work \(\int_{1}^{2} \mathbf{F} \cdot d \mathbf{r}\) done by
\(\mathbf{F}\) is unfortunately too lengthy to be included here. However, the
following three exercises capture the main points: \(^{16}\) (a) Show that the
path independence of \(\int_{1}^{2} \mathbf{F} \cdot d \mathbf{r}\) is
equivalent to the statement that the integral \(\oint_{\mathrm{T}} \mathbf{F}
\cdot d \mathbf{r}\) around any closed path \(\Gamma\) is zero. (By tradition,
the symbol \(\oint\) is used for integrals around a closed path \(-\) a path that
starts and stops at the same point.) [Hint: For any two points 1 and 2 and any
two paths from 1 to 2 , consider the work done by \(\mathbf{F}\) going from 1 to
2 along the first path and then back to 1 along the second in the reverse
direction. \((\) b) Stokes's theorem asserts that \(\oint_{\mathrm{T}} \mathbf{F}
\cdot d \mathbf{r}=\int(\nabla \times \mathbf{F}) \cdot \hat{\mathbf{n}} d A,\)
where the integral on the right is a surface integral over a surface for which
the path \(\Gamma\) is the boundary, and \(\hat{\mathbf{n}}\) and \(d A\) are a unit
normal to the surface and an element of area. Show that Stokes's theorem
implies that if \(\nabla \times \mathbf{F}=0\) everywhere, then
\(\oint_{\mathrm{T}} \mathbf{F} \cdot d \mathbf{r}=0 .\) (c) While the general
proof of Stokes's theorem is beyond our scope here, the following special case
is quite easy to prove (and is an important step toward the general proof):
Let \(\Gamma\) denote a rectangular closed path lying in a plane perpendicular
to the \(z\) direction and bounded by the lines \(x=B, x=B+b, y=C\) and \(y=C+c .\)
For this simple path (traced counterclockwise as seen from above), prove
Strokes's theorem that \(\oint_{\Gamma} \mathbf{F} \cdot d
\mathbf{r}=\int(\mathbf{\nabla} \times \mathbf{F}) \cdot \hat{\mathbf{n}} d A\)
where \(\hat{\mathbf{n}}=\hat{\mathbf{z}}\) and the integral on the right runs
over the flat, rectangular area inside \(\Gamma\). [Hint: The integral on the
left contains four terms, two of which are integrals over \(x\) and two over
\(y\). If you pair them in this way, you can combine each pair into a single
integral with an integrand of the form \(F_{x}(x, C+c, z)-F_{x}(x, C, z)\) (or a
similar term with the roles of \(x\) and \(y\) exchanged). You can rewrite this
integrand as an integral over \(y\) of \(\partial F_{x}(x, y, z) / \partial y\)
(and similarly with the other term), and you're home.]
