Mathematical Methods for Physics and Engineering : A Comprehensive Guide

LINE, SURFACE AND VOLUME INTEGRALS

11.7 Integral forms forgrad,divandcurl

In the previous chapter we defined the vector operators grad, div and curl in purely

mathematical terms, which depended on the coordinate system in which they were

expressed. An interesting application of line, surface and volume integrals is the

expression of grad, div and curl in coordinate-free, geometrical terms. Ifφis a

scalar field andais a vector field then it may be shown that at any pointP

∇φ= lim

V→ 0

(

1

V

∮

S

φdS

)

(11.14)

∇·a= lim

V→ 0

(

1

V

∮

S

a·dS

)

(11.15)

∇×a= lim

V→ 0

(

1

V

∮

S

dS×a

)

(11.16)

whereVis a small volume enclosingPandSis its bounding surface. Indeed,

we may consider these equations as the (geometrical)definitionsof grad, div and

curl. An alternative, but equivalent, geometrical definition of∇×aat a pointP,

which is often easier to use than (11.16), is given by

(∇×a)·nˆ= lim

A→ 0

(

1

A

∮

C

a·dr

)

, (11.17)

whereCis a plane contour of areaAenclosing the pointPandˆnis the unit

normal to the enclosed planar area.

It may be shown,in any coordinate system, that all the above equations are

consistent with our definitions in the previous chapter, although the difficulty of

proof depends on the chosen coordinate system. The most general coordinate

system encountered in that chapter was one with orthogonal curvilinear coordi-

natesu 1 ,u 2 ,u 3 , of which Cartesians, cylindrical polars and spherical polars are all

special cases. Although it may be shown that (11.14) leads to the usual expression

for grad in curvilinear coordinates, the proof requires complicated manipulations

of the derivatives of the basis vectors with respect to the coordinates and is not

presented here. In Cartesian coordinates, however, the proof is quite simple.

Show that the geometrical definition ofgradleads to the usual expression for∇φin

Cartesian coordinates.

Consider the surfaceSof a small rectangular volume element ∆V=∆x∆y∆zthat has its
faces parallel to thex,y,andzcoordinate surfaces; the pointP(see above) is at one corner.
We must calculate the surface integral (11.14) over each of its six faces. Remembering that
the normal to the surface points outwards from the volume on each face, the two faces
withx= constant have areas ∆S=−i∆y∆zand ∆S=i∆y∆zrespectively. Furthermore,
over each small surface element, we may takeφto be constant, so that the net contribution