Mathematical Methods for Physics and Engineering - Matematica.NET

LINE, SURFACE AND VOLUME INTEGRALSy(i)(4, 2)(ii)(1, 1)(iii)Figure 11.1 Different possible paths between the points (1, 1) and (4, 2).xSubstituting for x and y in (11.3) and writing the correct limits on u, weobtainI ==∫ (4,2)(1,1)∫ 10[(x + y) dx +(y − x) dy][(3u 2 + u + 2)(4u +1)− (u 2 + u)2u] du =10 2 3 .Case (iii). For the third path the line integral must be evaluated along the two linesegments separately and the results added together. First, along the line y = 1 we havedy = 0. Substituting this into (11.3) and using just the limits on x forthissegment,weobtain∫ (4,1)∫ 4[(x + y) dx +(y − x) dy] = (x +1)dx =10 1 . 2(1,1)Next, along the line x = 4 we have dx = 0. Substituting this into (11.3) and using just thelimits on y for this segment, we obtain∫ (4,2)(4,1)[(x + y) dx +(y − x) dy] =1∫ 21(y − 4) dy = −2 1 2 .The value of the line integral along the whole path is just the sum of the values of the lineintegrals along each segment, and is given by I =10 1 2 − 2 1 2 =8.◭When calculating a line integral along some curve C, which is given in termsof x, y and z, we are sometimes faced with the problem that the curve C is suchthat x, y and z are not single-valued functions of one another over the entirelength of the curve. This is a particular problem for closed loops in the xy-plane(and also for some open curves). In such cases the path may be subdivided intoshorter line segments along which one coordinate is a single-valued function ofthe other two. The sum of the line integrals along these segments is then equalto the line integral along the entire curve C. A better solution, however, is torepresent the curve in a parametric form r(u) that is valid for its entire length.380