Real Functions of Several Variables Examples of Nabla by Leif Mejlbro

By Leif Mejlbro

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Read or Download Real Functions of Several Variables Examples of Nabla Calculus,Vector Potentials, Green’s Identities and Curvilinear Coordinates,Electromagnetism and Various other Types Calculus 2c-10 PDF

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Extra resources for Real Functions of Several Variables Examples of Nabla Calculus,Vector Potentials, Green’s Identities and Curvilinear Coordinates,Electromagnetism and Various other Types Calculus 2c-10

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A) Stokes’s theorem. The square K lies in the Y Z-plane, and the unit normal vector is in the chosen orientation given by n = (1, 0, 0). Then by 1) and Stokes’s theorem, K V · t ds = n · rot V dS = ˜ K π π = ˜ K (− cos y + sin z) dS π (− cos y) dy dz + 0 0 0 dy · π sin z dz 0 = 0 + π · [− cos z]π0 = 2π. b) Straight forward computation of the line integral. The curve K is composed of the curves K1 : r1 (t) = (0, t, 0), t ∈ [0, π], t = (0, 1, 0), K2 : r2 (t) = (0, π, t), t ∈ [0, π], t = (0, 0, 1), K3 : r3 (t) = (0, π − t, π), t ∈ [0, π], t = (0, −1, 0), K4 : r4 = (0, 0, π − t), t ∈ [0, π], t = (0, 0, −1).

2) Now ∂ {z M (x, y)} = M (x, y) = ex + x(ey − exy ) ∂z and ∂ ∂ {L(x, y) + M (x, y)} = yex − y 2 exy + ex − exy − xyexy = {z M (x, y)}, ∂x ∂z so the necessary conditions of a gradient field are not satisfied, and U is not a gradient field. Clearly, U is of class C ∞ in all of R3 , and R3 is star shaped. ) As (L, M ) is a gradient field, we have in particular ∂L ∂M = , ∂y ∂x thus div U = z ∂M ∂L −z +0=z ∂x ∂y ∂M ∂L − ∂x ∂y = 0, and U is divergence free and defined in a star shaped domain. Therefore U has a vector potential.

Then ⎛ ⎞ y2 a2 y 2 z2 ⎜ −1 + z 2 − z 4 ln 1 + a2 ⎟ ⎜ ⎟ ⎜ ⎟ 1 ⎜ ⎟ 1 2 2 2 2 ⎟. x a z x τ V(τ x) dτ = ⎜ ⎜ 2 ⎜ 1 − 2 + 4 ln 1 + 2 , ⎟ 0 ⎟ z z a ⎜ ⎟ ⎝ ⎠ 1 We now find W0 by 1 W0 (x) = 0 τ V(τ x) dτ × x e1 = 1 2 −1+ e2 y 2 a2 y 2 z2 − 4 ln 1+ 2 2 z z a 1− z2 x2 a2 x2 + 4 ln 1+ 2 z z a x ⎛ = ⎜ ⎜ ⎜ ⎜ 1⎜ ⎜ 2⎜ ⎜ ⎜ ⎜ ⎝ y z−y− x2 a2 x2 z2 + 3 ln 1 + 2 z z a x+z− a2 y 2 z2 y2 + 3 ln 1 + 2 z z a −x − y + e3 1 z ⎞ a2 z2 y 3 + x3 − 4 (x3 + y 3 ) ln 1 + 2 2 z z a ⎟ ⎟ ⎟ ⎟ ⎟ ⎟, ⎟ ⎟ ⎟ ⎟ ⎠ z = 0. For z = 0 the result is obtained by taking the limit.

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