Tensors for Physics

120 8 Integration of Fields

Fig. 8.7 Spherical polar
coordinates with the vectors
eφ,eθander

8.2.3 Surface Integrals as Integrals Over Two Parameters

Consider a finite surfaceAwhich has a rim and which is simply connected, i.e.Ais
without any holes. Furthermore, the surface should everywhere have a well oriented
normal direction. A counter example is theMoebius strip.
The surfaceAis described by a parameter representationr=r(p,q)where the
parameterspandqvary between the valuesp 1 ,p 2 andq 1 ,q 2. Thus the rim of the
surface inR^3 corresponds to the rim of a rectangle in thep–q-plane, cf. Fig.8.8.
Now letf= f(r)be a function ofrwhich depends on the parametersqandq
viar=r(p,q). The surface integral of this function over the surfaceAis defined by

Sμ=

∫

A

f(r)dsμ. (8.22)

Thesurface element dsμis the cross product of the tangential vectors (8.13), see
also (8.14). The axial vector dsμis perpendicular to the surface. More specifically,
one has

dsμ=εμνλ

∂rν

∂p

∂rλ

∂q

dpdq=̂sμ(p,q)d^2 s. (8.23)

Herêsisaunitvectororthogonaltothesurface,andd^2 sisthemagnitudeofthesurface
element. It quantifies the change ofrwith the change ofpandq. The exponent 2 in the
symbol “d^2 s” used here indicates that the surface integration is “two dimensional”.
It is clearly distinguished from the arc length element dsoccurring sometimes in
“one dimensional” line integrals. Notice that the surface element ds, being defined
in (8.23) as the cross product of two tangential vectors, is an axial vector. The same
applies to the unit vector̂s.
Due to (8.23), the surface integral (8.22) can be computed as a double integral
overpandq:

Sμ=

∫p 2

p 1

dp

∫q 2

q 1

dqf(r)εμνλ

∂rν

∂p

∂rλ

∂q

. (8.24)