Mathematical Methods for Physics and Engineering : A Comprehensive Guide

11.1 LINE INTEGRALS

A similar procedure may be followed for the third type of line integral in (11.1),

which involves a cross product.

Line integrals have properties that are analogous to those of ordinary integrals.

In particular, the following are useful properties (which we illustrate using the

second form of line integral in (11.1) but which are valid for all three types).

(i) Reversing the path of integration changes the sign of the integral. If the

pathCalong which the line integrals are evaluated hasAandBas its

end-points then

∫B

A

a·dr=−

∫A

B

a·dr.

This implies that if the pathCis a loop then integrating around the loop

in the opposite direction changes the sign of the integral.

(ii) If the path of integration is subdivided into smaller segments then the sum

of the separate line integrals along each segment is equal to the line integral

along the whole path. So, ifPis any point on the path of integration that

lies between the path’s end-pointsAandBthen

∫B

A

a·dr=

∫P

A

a·dr+

∫B

P

a·dr.

Evaluate the line integralI=

∫

Ca·dr,wherea=(x+y)i+(y−x)j, along each of the

paths in thexy-plane shown in figure 11.1, namely

(i)the parabolay^2 =xfrom(1,1)to(4,2),

(ii)the curvex=2u^2 +u+1,y=1+u^2 from(1,1)to(4,2),

(iii)the liney =1from(1,1)to(4,1), followed by the linex =4from(4,1)

to(4,2).

Since each of the paths lies entirely in thexy-plane, we havedr=dxi+dyj.Wecan
therefore write the line integral as

I=

∫

C

a·dr=

∫

C

[(x+y)dx+(y−x)dy]. (11.3)

We must now evaluate this line integral along each of the prescribed paths.
Case (i). Along the parabolay^2 =xwe have 2ydy=dx. Substituting forxin (11.3)
and using just the limits ony,weobtain

I=

∫(4,2)

(1,1)

[(x+y)dx+(y−x)dy]=

∫ 2

1

[(y^2 +y)2y+(y−y^2 )]dy=11^13.

Note that we could just as easily have substituted foryand obtained an integral inx,
which would have given the same result.
Case (ii). The second path is given in terms of a parameteru. We could eliminateu
between the two equations to obtain a relationship betweenxandydirectly and proceed
as above, but it is usually quicker to write the line integral in terms of the parameteru.
Along the curvex=2u^2 +u+1,y=1+u^2 we havedx=(4u+1)duanddy=2udu.