Mathematical Methods for Physics and Engineering - Matematica.NET

11.9 STOKES’ THEOREM AND RELATED THEOREMSis the circle x 2 + y 2 = a 2 in the xy-plane. This is given by∮ ∮a · dr = (y i − x j + z k) · (dx i + dy j + dz k)CC∮= (ydx− xdy).CUsing plane polar coordinates, on C we have x = a cos φ, y = a sin φ so that dx =−a sin φdφ, dy = a cos φdφ, and the line integral becomes∮∫ 2π∫ 2π(ydx− xdy)=−a 2 (sin 2 φ +cos 2 φ) dφ = −a 2 dφ = −2πa 2 .C0Since the surface and line integrals have the same value, we have verified Stokes’ theoremin this case. ◭The two-dimensional version of Stokes’ theorem also yields Green’s theorem ina plane. Consider the region R in the xy-plane shown in figure 11.11, in which avector field a is defined. Since a = a x i+a y j, we have ∇×a =(∂a y /∂x−∂a x /∂y) k,and Stokes’ theorem becomes∫∫R( ∂ay∂x − ∂a x∂y)∮dx dy = (a x dx + a y dy).CLetting P = a x and Q = a y we recover Green’s theorem in a plane, (11.4).011.9.1 Related integral theoremsAs for the divergence theorem, there exist two other integral theorems that areclosely related to Stokes’ theorem. If φ is a scalar field and b is a vector field,and both φ and b satisfy our usual differentiability conditions on some two-sidedopen surface S bounded by a closed perimeter curve C, then∫∮dS ×∇φ = φdr, (11.24)SC∫∮(dS ×∇) × b = dr × b. (11.25)SC◮Use Stokes’ theorem to prove (11.24).In Stokes’ theorem, (11.23), let a = φc, wherec is a constant vector. We then have∫∮[∇×(φc)] · dS = φc · dr. (11.26)SExpanding out the integrand on the LHS we have∇×(φc) =∇φ × c + φ∇×c = ∇φ × c,since c is constant, and the scalar triple product on the LHS of (11.26) can therefore bewritten[∇×(φc)] · dS =(∇φ × c) · dS = c · (dS ×∇φ).407C