Mathematical Methods for Physics and Engineering : A Comprehensive Guide

LINE, SURFACE AND VOLUME INTEGRALS

y

x

(i)

(ii)

(1,1) (iii)

(4,2)

Figure 11.1 Different possible paths between the points (1, 1) and (4, 2).

Substituting forxandyin (11.3) and writing the correct limits onu,weobtain

I=

∫(4,2)

(1,1)

[(x+y)dx+(y−x)dy]

=

∫ 1

0

[(3u^2 +u+ 2)(4u+1)−(u^2 +u)2u]du=10^23.

Case (iii). For the third path the line integral must be evaluated along the two line
segments separately and the results added together. First, along the liney= 1 we have
dy= 0. Substituting this into (11.3) and using just the limits onxforthissegment,we
obtain
∫(4,1)

(1,1)

[(x+y)dx+(y−x)dy]=

∫ 4

1

(x+1)dx=10^12.

Next, along the linex= 4 we havedx= 0. Substituting this into (11.3) and using just the
limits onyfor this segment, we obtain
∫(4,2)

(4,1)

[(x+y)dx+(y−x)dy]=

∫ 2

1

(y−4)dy=− 212.

The value of the line integral along the whole path is just the sum of the values of the line
integrals along each segment, and is given byI=10^12 − 212 =8.

When calculating a line integral along some curveC, which is given in terms

ofx,yandz, we are sometimes faced with the problem that the curveCis such

thatx,yandzare not single-valued functions of one another over the entire

length of the curve. This is a particular problem for closed loops in thexy-plane

(and also for some open curves). In such cases the path may be subdivided into

shorter line segments along which one coordinate is a single-valued function of

the other two. The sum of the line integrals along these segments is then equal

to the line integral along the entire curveC. A better solution, however, is to

represent the curve in a parametric formr(u) that is valid for its entire length.