The proof that the condition V x F = 0 guarantees the path

QUESTION:

The proof that the condition V x F = 0 guarantees the path independence of the work fi2 F dr done by 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 fi2 F dr is equivalent to the statement that the integral fr, F dr around any closed path F is zero. (By tradition, the symbol f 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 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 A, F dr = f (V x F) dA, where the integral on the right is a surface integral over a surface for which the path F is the boundary, and ii. and dA are a unit normal to the surface and an element of area. Show that Stokes's theorem implies that if V x F = 0 everywhere, then A, F dr = 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 F 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 Stokes's theorem that F dr = f (V x F) ill d A where it = z and the integral on the right runs over the flat, rectangular area inside F. [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 Fx(x, C c, z) Fx(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 a Fx(x, y, z)/ay (and similarly with the other term), and you're home.]