Advanced Calculus fi..

Chapter 5 Vector Integral Calculus 353.- 1Figure 5.41 Distances and angles for Eq. (5.168).Figure 5.42 Potential and force magnitude for Eq. (5.168).one finds:%Ha2 - R2 - 2a2 logal, R 5 aIT ba2 log R,R>a(see Problem 1 following Section 5.18). At R = a the two expressions give U =nba2 log 1 /a so that U is continuous for all (x, y). Also(-nb(xi + yj), R 5 a= nbu2--jir(xi + yj), R > a,(5.170)from which we see that F is continuous everywhere. The force is radial everywhere,directed toward the origin. The force magnitude depends only on R; this magnitudeand U are graphed as functions of R in Fig. 5.42.U as a harmonic function. For fixed (6, q) the function log l/r is harmonic in(x, y) for (x, y) not at (t, rl). Forand the sum is 0. Leibnitz's Rule permits a similar differentiation under the integralsign for the potentials (5.163), (5.166), and (5.167). Hence we conclude that awayfrom the masses U(x, y) is harmonic. Equivalently,div F = div grad U = 0.