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Mathematical Methods for Physics and Engineering - Matematica.NET

Mathematical Methods for Physics and Engineering - Matematica.NET

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LINE, SURFACE AND VOLUME INTEGRALS11.11 Hints <strong>and</strong> answers11.1 Show that ∇×F = 0. The potential φ F (r) =x 2 z + y 2 z 2 − z.11.3 (a) c 3 ln 2 i +2j +(3c/2)k; (b)(−3c 4 /8)i − c j − (c 2 ln 2)k; (c)c 4 ln 2 − c.11.5 For P , x = y = ab/(a 2 + b 2 ) 1/2 . The relevant limits are 0 ≤ θ 1 ≤ tan −1 (b/a) <strong>and</strong>tan −1 (a/b) ≤ θ 2 ≤ π/2. The total common area is 4ab tan −1 (b/a).11.7 Show that, in the notation of section 11.3, ∂Q/∂x − ∂P/∂y =2x 2 ; I = πa 3 b/2.11.9 M = I ∫ r × (dr × B). Show that the horizontal sides in the first term <strong>and</strong> theCwhole of the second term contribute nothing to the couple.11.11 Note that, if ˆn is the outward normal to the surface, ˆn z · ˆn dl is equal to −dρ.11.13 (b) φ = c + z/r.11.15 (a) Yes, F 0 (x − y)exp(−r 2 /a 2 ); (b) yes, −F 0 [(x 2 + y 2 )/(2a)] exp(−r 2 /a 2 );(c) no, ∇×F ≠ 0.11.17 A spiral of radius c with its axis parallel to the z-direction <strong>and</strong> passing through(a, b). The pitch of the spiral is 2πc 2 . No, because (i) γ is not a closed loop <strong>and</strong>(ii) the line integral must be zero <strong>for</strong> every closed loop, not just <strong>for</strong> a particularone. In fact ∇×f = −2k ≠ 0 shows that f is not conservative.11.19 (a) dS =(2a 3 cos θ sin 2 θ cos φ i +2a 3 cos θ sin 2 θ sin φ j + a 2 cos θ sin θ k) dθ dφ.(b) ∇ · r = 3; over the plane z =0,r · dS =0.The necessarily common value is 3πa 4 /2.11.21 Write r as ∇( 1 2 r2 ).11.23 The answer is 3 √ 3πα/2 in each case.11.25 Identify the expression <strong>for</strong> ∇ · (E × B) <strong>and</strong> use the divergence theorem.11.27 (a) The successive contributions to the integral are:1 − 2e −1 , 0, 2+ 1 2 e, − 7 3 , −1+2e−1 , − 1 2 .(b) ∇×F =2xyz 2 i − y 2 z 2 j + ye x k. Show that the contour is equivalent to thesum of two plane square contours in the planes z =0<strong>and</strong>x = 1, the latter beingtraversed in the negative sense. Integral = 1 (3e − 5).6414

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