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# h8s - 18.02 HOMEWORK#8 DUE DECEMBER 2 2010 BJORN POONEN 1...

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Unformatted text preview: 18.02 HOMEWORK #8, DUE DECEMBER 2, 2010 BJORN POONEN 1. Part A (After Nov. 18) p. 1063: 15, 27* (7 pts.) Solution. (to p. 1063, 15) By definition, ∇ f = ∂f ∂x , ∂f ∂y , ∂f ∂z ∇ 2 f = ∇ · ∂f ∂x , ∂f ∂y , ∂f ∂z = ∂ ∂x ∂f ∂x + ∂ ∂y ∂f ∂y + ∂ ∂z ∂f ∂z = ∂ 2 f ∂x 2 + ∂ 2 f ∂y 2 + ∂ 2 f ∂z 2 . Solution. (to p. 1063, 27) Let S be a cylinder of radius r and height h with central axis along the wire. The normal n to a point on either flat round end of the cylinder is parallel to the axis, and hence perpendicular to E , so the flux of E through the round ends of the cylinder is 0. On the other hand, at a point on the curved part of the cylinder, the unit normal n is pointing in the same direction as E , so E · n = | E | at such points; since | E | at such points is constant, the left hand side of Gauss’s law (14) is ZZ S E · n dS = | E | ZZ S dS = | E | (lateral surface area of S ) = | E | 2 πrh. The right hand side is Q = hq since there are h meters of wire inside the cylinder, and q coulombs per meter. Equating these yields | E | 2 πrh = hq | E | = q 2 π r . (After Nov. 18) 6C-7b* (5 pts.), 6C-9* (3 + 7 + 5 + 10 pts.) 1 Solution. (to 6C-7b) On the lateral surface of the cylinder, the normal vector n is perpen- dicular to F , so the flux of F through the lateral surface is 0. On the bottom of the cylinder, z = 0, so F = 0, so the flux of F through the bottom is 0 too. The flux of F through the top is ZZ top zy k · k dS = ZZ top y dS = 0 , by symmetry. Thus RR cylinder F · d S = 0. On the other hand, div F = ∂ ∂x 0 + ∂ ∂y 0 + ∂ ∂z zy = y, so the integral of div F over the interior of the cylinder is 0, by symmetry again. Thus both sides in the divergence theorem are 0. Solution. (to 6C-9*) a) The domain consists of all points with ρ 6 = 0; in other words, all of R 3 except the origin. b) The radially outward unit vector is h x, y, z i ρ , so F = 1 ρ 2 h x, y, z i ρ = h ρ- 3 x, ρ- 3 y, ρ- 3 z i . Differentiating ρ 2 = x 2 + y 2 + z 2 with respect to x shows that 2 ρρ x = 2 x , so ρ x = x/ρ . Therefore ∂ ∂x ( ρ- 3 x ) =- 3 ρ- 4 ρ x x + ρ- 3 =- 3 ρ- 4 x ρ x + ρ- 3 = ρ- 5 (- 3 x 2 + ρ 2 ) . Similarly, ∂ ∂y ( ρ- 3 y ) = ρ- 5 (- 3 y 2 + ρ 2 ) ∂ ∂z ( ρ- 3 z ) = ρ- 5 (- 3 z 2 + ρ 2 ) , and summing all three gives div F = ρ- 5 (- 3 x 2- 3 y 2- 3 z 2 + 3 ρ 2 ) = ρ- 5 (- 3 ρ 2 + 3 ρ 2 ) = 0 . c) The outward unit normal n is in the same direction as F , so F · n = | F || n | = | F | = 1 /ρ 2 . Thus ZZ S F · S = ZZ S F · n dS = ZZ 1 ρ 2 dS = 1 ρ 2 Area( S ) = 1 ρ 2 4 πρ 2 = 4 π. 2 d) Suppose that the region T enclosed by S does not include the origin. Then div F = 0 everywhere in T , so the divergence theorem gives ZZ S F · S = ZZZ T div F dV = 0 ....
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h8s - 18.02 HOMEWORK#8 DUE DECEMBER 2 2010 BJORN POONEN 1...

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