Mathematical Methods for Physics and Engineering - Matematica.NET

11.7 INTEGRAL FORMS FOR grad, div AND curlto the surface integral from these two faces is then([(φ +∆φ) − φ]∆y ∆z i = φ + ∂φ )∂x ∆x − φ ∆y ∆z i= ∂φ ∆x ∆y ∆z i.∂xThe surface integral over the pairs of faces with y =constantandz = constant respectivelymay be found in a similar way, and we obtain∮ ( ∂φφdS =S ∂x i + ∂φ∂y j + ∂φ )∂z k ∆x ∆y ∆z.Therefore ∇φ at the point P is given by[ (1 ∂φ∇φ = lim∆x,∆y,∆z→0 ∆x ∆y ∆z ∂x i + ∂φ∂y j + ∂φ ) ]∂z k ∆x ∆y ∆z= ∂φ∂x i + ∂φ∂y j + ∂φ∂z k. ◭We now turn to (11.15) and (11.17). These geometrical definitions may beshown straightforwardly to lead to the usual expressions for div and curl inorthogonal curvilinear coordinates.◮By considering the infinitesimal volume element dV = h 1 h 2 h 3 ∆u 1 ∆u 2 ∆u 3 shown in figure11.10, show that (11.15) leads to the usual expression for ∇·a in orthogonal curvilinearcoordinates.Let us write the vector field in terms of its components with respect to the basis vectorsof the curvilinear coordinate system as a = a 1 ê 1 + a 2 ê 2 + a 3 ê 3 . We consider first thecontribution to the RHS of (11.15) from the two faces with u 1 = constant, i.e. PQRSand the face opposite it (see figure 11.10). Now, the volume element is formed from theorthogonal vectors h 1 ∆u 1 ê 1 , h 2 ∆u 2 ê 2 and h 3 ∆u 3 ê 3 at the point P andsoforPQRS wehave∆S = h 2 h 3 ∆u 2 ∆u 3 ê 3 × ê 2 = −h 2 h 3 ∆u 2 ∆u 3 ê 1 .Reasoning along the same lines as in the previous example, we conclude that the contributionto the surface integral of a · dS over PQRS and its opposite face taken together isgiven by∂(a · ∆S)∆u 1 =∂ (a 1 h 2 h 3 )∆u 1 ∆u 2 ∆u 3 .∂u 1 ∂u 1The surface integrals over the pairs of faces with u 2 = constant and u 3 = constantrespectively may be found in a similar way, and we obtain∮ [ ∂a · dS = (a 1 h 2 h 3 )+∂ (a 2 h 3 h 1 )+∂u 1 ∂u 2STherefore ∇ · a at the point P is given by[1∇ · a = lim∆u 1 ,∆u 2 ,∆u 3 →0 h 1 h 2 h 3 ∆u 1 ∆u 2 ∆u 3∮S= 1 [ ∂(a 1 h 2 h 3 )+h 1 h 2 h 3 ∂u 1∂∂u 3(a 3 h 1 h 2 )∂∂u 2(a 2 h 3 h 1 )+]a · dS]∆u 1 ∆u 2 ∆u 3 .∂ ](a 3 h 1 h 2 ) . ◭∂u 3399