Mathematical Methods for Physics and Engineering - Matematica.NET

11.1 LINE INTEGRALSA 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 thesecond 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 thepath C along which the line integrals are evaluated has A and B as itsend-points then∫ BAa · dr = −∫ ABa · dr.This implies that if the path C is a loop then integrating around the loopin the opposite direction changes the sign of the integral.(ii) If the path of integration is subdivided into smaller segments then the sumof the separate line integrals along each segment is equal to the line integralalong the whole path. So, if P is any point on the path of integration thatlies between the path’s end-points A and B then∫ BAa · dr =∫ PAa · dr +∫ BPa · dr.◮Evaluate the line integral I = ∫ a · dr, wherea =(x + y)i +(y − x)j, along each of theCpaths in the xy-plane shown in figure 11.1, namely(i) the parabola y 2 = x from (1, 1) to (4, 2),(ii) the curve x =2u 2 + u +1, y =1+u 2 from (1, 1) to (4, 2),(iii) the line y = 1 from (1, 1) to (4, 1), followed by the line x = 4 from (4, 1)to (4, 2).Since each of the paths lies entirely in the xy-plane, we have dr = dx i + dy j. Wecantherefore write the line integral as∫ ∫I = a · dr = [(x + y) dx +(y − x) dy]. (11.3)CCWe must now evaluate this line integral along each of the prescribed paths.Case (i). Along the parabola y 2 = x we have 2ydy = dx. Substituting for x in (11.3)and using just the limits on y, weobtainI =∫ (4,2)(1,1)[(x + y) dx +(y − x) dy] =∫ 21[(y 2 + y)2y +(y − y 2 )] dy =11 1 3 .Note that we could just as easily have substituted for y and obtained an integral in x,which would have given the same result.Case (ii). The second path is given in terms of a parameter u. We could eliminate ubetween the two equations to obtain a relationship between x and y directly and proceedas above, but it is usually quicker to write the line integral in terms of the parameter u.Along the curve x =2u 2 + u +1, y =1+u 2 we have dx =(4u +1)du and dy =2udu.379