Comments
Description
Transcript
Hints and answers
LINE, SURFACE AND VOLUME INTEGRALS 11.11 Hints and answers 11.1 11.3 11.5 11.7 11.9 11.11 11.13 11.15 11.17 11.19 11.21 11.23 11.25 11.27 Show that ∇ × F = 0. The potential φF (r) = x2 z + y 2 z 2 − z. (a) c3 ln 2 i + 2 j + (3c/2)k; (b) (−3c4 /8)i − c j − (c2 ln 2)k; (c) c4 ln 2 − c. For P , x = y = ab/(a2 + b2 )1/2 . The relevant limits are 0 ≤ θ1 ≤ tan−1 (b/a) and tan−1 (a/b) ≤ θ2 ≤ π/2. The total common area is 4ab tan−1 (b/a). 2 3 Show that, in the notation of section 11.3, ∂Q/∂x − ∂P /∂y = 2x ; I = πa b/2. M = I C r × (dr × B). Show that the horizontal sides in the first term and the whole of the second term contribute nothing to the couple. Note that, if n̂ is the outward normal to the surface, n̂z · n̂ dl is equal to −dρ. (b) φ = c + z/r. (a) Yes, F0 (x − y) exp(−r 2 /a2 ); (b) yes, −F0 [(x2 + y 2 )/(2a)] exp(−r2 /a2 ); (c) no, ∇ × F = 0. A spiral of radius c with its axis parallel to the z-direction and passing through (a, b). The pitch of the spiral is 2πc2 . No, because (i) γ is not a closed loop and (ii) the line integral must be zero for every closed loop, not just for a particular one. In fact ∇ × f = −2k = 0 shows that f is not conservative. (a) dS = (2a3 cos θ sin2 θ cos φ i + 2a3 cos θ sin2 θ sin φ j + a2 cos θ sin θ k) dθ dφ. (b) ∇ · r = 3; over the plane z = 0, r · dS = 0. The necessarily common value is 3πa4 /2. Write r as ∇( 12 r2 ). √ The answer is 3 3πα/2 in each case. Identify the expression for ∇ · (E × B) and use the divergence theorem. (a) The successive contributions to the integral are: 1 − 2e−1 , 0, 2 + 12 e, − 73 , −1 + 2e−1 , − 21 . (b) ∇ × F = 2xyz 2 i − y 2 z 2 j + yex k. Show that the contour is equivalent to the sum of two plane square contours in the planes z = 0 and x = 1, the latter being traversed in the negative sense. Integral = 16 (3e − 5). 414