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We get by straightforward computations, div V rot V = 0 + x ey sin z − x ey sin z = 0, ex ey ez ⎛ ∂ ∂x ∂ ∂y ∂ ∂z ⎜ ⎜ =⎜ ⎜ ⎝ ey sin z xey sin z xey cos z = xey cos z − xey cos z ey cos z − ey cos z ⎞ ⎟ ⎟ ⎟ ⎟ ⎠ ey sin z − ey sin z div W = 0, = 1 + 2ey cos z − 2ey cos z − 1 + 3z 2 = 3z 2 , ex ey ez rot W = ∂ ∂x ∂ ∂y ∂ ∂z x + 2xey cos z −2ey cos z −z + z 3 = (0 − 2ey sin z, −2xey sin z − 0, 0 − 2xey cos z) = −2 (ey sin z, xey sin z, xey cos z) = −2V. 1 − W = V, and it follows immediately that 2 1 1 − W is a vector potential for V, and that α = − .

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.

Surface integral, spherical coordinates. In spherical coordinates a parametric description of the surface is given by ⎧ x = a sin θ cos ϕ, ⎪ ⎪ ⎪ ⎪ ⎨ π y = a sin θ sin ϕ, θ ∈ 0, , ϕ ∈ [0, 2π]. ⎪ 2 ⎪ ⎪ ⎪ ⎩ z = a cos θ, Thus the normal vector becomes e1 N(θ, ϕ) a cos θ cos ϕ = e2 e3 a cos θ sin ϕ −a sin θ −a sin θ sin ϕ a sin θ cos ϕ 0 = a2 (sin2 θ cos ϕ, sin2 θ sin ϕ, sin θ cos θ) = a2 sin θ (sin θ cos ϕ, sin θ sin ϕ, cos θ), and we note that the z-component is positive, showing that we have obtained the right orientation.

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Calculus 2c-10, Examples of Nabla Calculus, Vector Potentials, Green's Identities and Curvilinear Coordinates, Electromagnetism and Various other Types by Mejlbro L.


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