Mathematical Methods for Physics and Engineering : A Comprehensive Guide

LINE, SURFACE AND VOLUME INTEGRALS

Each of the line integrals in (11.1) is evaluated over some curveCthat may be

either open (AandBbeing distinct points) or closed (the curveCforms a loop,

so thatAandBare coincident). In the case whereCis closed, the line integral

is written

∮
Cto indicate this. The curve may be given either parametrically by
r(u)=x(u)i+y(u)j+z(u)kor by means of simultaneous equations relatingx, y, z

for the given path (in Cartesian coordinates). A full discussion of the different

representations of space curves was given in section 10.3.

In general, the value of the line integral depends not only on the end-points

AandBbut also on the pathCjoining them. For a closed curve we must also

specify the direction around the loop in which the integral is taken. It is usually

taken to be such that a person walking around the loopCin this direction

always has the regionRon his/her left; this is equivalent to traversingCin the

anticlockwise direction (as viewed from above).

11.1.1 Evaluating line integrals

The method of evaluating a line integral is to reduce it to a set of scalar integrals.

It is usual to work in Cartesian coordinates, in which casedr=dxi+dyj+dzk.

The first type of line integral in (11.1) then becomes simply
∫

C

φdr=i

∫

C

φ(x, y, z)dx+j

∫

C

φ(x, y, z)dy+k

∫

C

φ(x, y, z)dz.

The three integrals on the RHS are ordinary scalar integrals that can be evaluated

in the usual way once the path of integrationChas been specified. Note that in

the above we have used relations of the form
∫
φidx=i

∫

φdx,

which is allowable since the Cartesian unit vectors are of constant magnitude

and direction and hence may be taken out of the integral. If we had been using

a different coordinate system, such as spherical polars, then, as we saw in the

previous chapter, the unit basis vectors would not be constant. In that case the

basis vectors could not be factorised out of the integral.

The second and third line integrals in (11.1) can also be reduced to a set of

scalar integrals by writing the vector fieldain terms of its Cartesian components

asa=axi+ayj+azk,whereax,ay,azare each (in general) functions ofx, y, z.

The second line integral in (11.1), for example, can then be written as
∫

C

a·dr=

∫

C

(axi+ayj+azk)·(dxi+dyj+dzk)

=

∫

C

(axdx+aydy+azdz)

=

∫

C

axdx+

∫

C

aydy+

∫

C

azdz. (11.2)