Advanced Calculus fi..

298 Advanced Calculus, Fifth Editionin Fig. 5.21. If one integrates in a positive direction around the boundary of eachsubregion and then adds the results, one finds that the integrals along the auxiliaryarcs cancel out, leaving just the integral around C1 in the positive direction plus theintegrals around CZ, C3, . . . , in the negative direction. On the other hand, the lineintegral around the boundary of each subregion can be expressed as a double integralover the subregion by Green's theorem. Hence the sum of the line integrals is equalto the double integral over R. This gives (5.55); Eq. (5.55') then follows as a specialcase.If one denotes by BR the directed boundary of R, that is, the curves C, , CZ, . . . , Cnwith the given directions, then Eqs. (5.55) and (5.55') can be written in the conciseform:It is to be noted that the correct direction of integration always keeps the region tothe left; that is, the normal pointing away from the region is 90" behind the tangent.As an application of the theorem just proved, let us consider the case of a doublyconnected region D with a hole which, for the sake of simplicity, we take to be apoint A, as in Fig. 5.22. Let aP/ay = dQ/ax in D. What are the possible values ofan integral'about a simple closed path in D? We assert that there are only two possible values,namely, the value 0 when C does not enclose the hole A (the curve Co of Fig. 5.22)and a value k, which is the same for all curves C enclosing the hole A. That the valueis 0 for a curve such as Co follows from the fact that Co lies in a simply connectedFigure 5.22Doubly connected domain (point A excluded).