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328 Advanced Calculus, Fifth Edition fA vector field u (whose components have continuous derivatives) such thatcurl u = 0in a domain D, is called irrotational in D. By virtue of the above theorems, irrotationalityin a simply connected domain is equivalent to each of the properties:IcUT ds = 0 for every simple closed curve in D;UT ds is independent of the path in D; .TS3u = grad F in D..I:,4 3ii;A theory similar to the preceding holds for surface integrals. Rather than give afull discussion here, we confine our attention to the counterpart of the last statementin Theorem 111; for more details we refer the reader to the books of Brand and Kellogglisted at the end of the chapter.ITHEOREM IVcontinuous partial derivatives in a spherical domain D. If div u = 0 in D:.; ALet u = Li + Mj + Nk be a vector field whose components have3then a vector field v = Xi + Yj + Zk in D can be found such thatthat is,curl v = u in D;Remark. The theorem provides a converse to the theorem of Section 3.6:div curl u = 0,for it asserts that if div u = 0, then u = curl v for some v. However, while div umay be 0 in an arbitrary domain Dl, the theorem provides v whose curl is u onlyin each spherical domain D contained in Dl, that is, no one v serves for all of Dl.Actually, the proof to follow gives one v for all of Dl when Dl is the interior ofa cube or of an ellipsoid or of any "convex" surface. The existence of v for all ofDl for more general cases can be established (see, for example, pages 203 ff. in the'book by Lamb listed at the end of the chapter).Vector fields u satisfying the condition div u = 0 are termed solenoidal. Thetheorem amounts to the assertion that solenoidal fields are (in suitable domains)fields of the form curl v, provided that the components of u have continuous partialderivatives. The field v is not unique (Problem 5).
Chapter 5 Vector Integral Calculus 329Proof of the theorem. Let the spherical domain D have center at Po; for simplicitywe assume Po to be the origin (0, 0, 0). If Pl(xl, yl, zl) is an arbitrary point of D,we setwhere, on the right-hand side, x, y, and z are the following functions oft:As t varies from 0 to 1, the point (x, y , z) varies from Po to P1 on the line segmentPo PI; hence (x, y, z) remains in D. We have now by Leibnitz's Rule and the chainrule (Sections 4.9 and 2.8):and, similarly,Accordingly,Since div (Li + Mj + Nk) = 0, this can be written as follows:Now
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ContentsVectors and Matrices1.1 Int
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3.6 Combined Operations 183*3.7 Cur
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*6.24 Principal Value of Improper I
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Case of Two Particles 662Case of N
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Differential Calculusof Functionsof
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Vector Integral CalculusPART I.TWO-
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