Mathematical Methods for Physics and Engineering - Matematica.NET

LINE, SURFACE AND VOLUME INTEGRALSwhich shows that we require a · dr to be an exact differential: condition (iv). From(10.27) we can write dφ = ∇φ · dr, and so we have(a −∇φ) · dr =0.Since dr is arbitrary, we find that a = ∇φ; this immediately implies ∇×a = 0,condition (iii) (see (10.37)).Alternatively, if we suppose that there exists a single-valued function of positionφ such that a = ∇φ then ∇×a = 0 follows as before. The line integral around aclosed loop then becomes∮C∮∮a · dr = ∇φ · dr =CSince we defined φ to be single-valued, this integral is zero as required.Now suppose ∇×a = 0. From Stoke’s theorem, which is discussed in section11.9, we immediately obtain ∮ Ca · dr =0;thena = ∇φ and a · dr = dφ followas above.Finally, let us suppose a · dr = dφ. Then immediately we have a = ∇φ, andtheother results follow as above.◮Evaluate the line integral I = ∫ Ba · dr, wherea A =(xy2 + z)i +(x 2 y +2)j + xk, A is thepoint (c, c, h) and B is the point (2c, c/2,h), along the different paths(i) C 1 ,givenbyx = cu, y = c/u, z = h,(ii) C 2 ,givenby2y =3c − x, z = h.Show that the vector field a is in fact conservative, and find φ such that a = ∇φ.Expanding out the integrand, we haveI =∫ (2c, c/2,h)(c, c, h)dφ.[(xy 2 + z) dx +(x 2 y +2)dy + xdz ] , (11.7)which we must evaluate along each of the paths C 1 and C 2 .(i) Along C 1 we have dx = cdu, dy = −(c/u 2 ) du, dz = 0, and on substituting in (11.7)and finding the limits on u, weobtain∫ 2I = c(h − 2 )du = c(h − 1).u 21(ii) Along C 2 we have 2 dy = −dx, dz = 0 and, on substituting in (11.7) and using thelimits on x, weobtainI =∫ 2cc( 12 x3 − 9 4 cx2 + 9 4 c2 x + h − 1 ) dx = c(h − 1).Hence the line integral has the same value along paths C 1 and C 2 . Taking the curl of a,we have∇×a =(0− 0)i +(1− 1)j +(2xy − 2xy)k = 0,so a is a conservative vector field, and the line integral between two points must be388