By Wieslaw Tadeusz Zelazko

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X ∂y ∂z Thus by insertion, ∂W2 ∂W1 − = Vz (x, y, z) + ∂x ∂y z γ ∂Vz z (x, y, ζ) dζ = Vz (x, y, γ) + [Vz (x, y, ζ)]ζ=γ = Vz (x, y, z). ∂z Summarizing, × W = V, and we have proved that W is a vector potential for V. Remark. The formula of this example of a vector potential in R3 is far easier to apply than the usual procedure of solution given in most textbooks. 6 Given the vector field V(x, y, z) = 2x + x2 y, y − xy 2 , 7z + 5z 3 , 1. Compute the divergence (x, y, z) ∈ R3 . · V and the rotation × V.

Since V is divergence fret, the ingoing flux through O must be equal to the outgoing flux through B(0, a), where n = (0, 0, −1), hence the flux is 2 O V · n dS = B(0,a) = − V · n dS = B(0,a) x2 y2 − 1, 1 − 2 , 1 a a B(0),a) · (0, 0, −1) dS dS = − areal B(0, a) = −πa2 . b) Alternatively it follows from 3) and Stokes’s theorem that O V · n dS = 1 3 ( O × U) · n dS = 1 3 ∂O U · t ds = B(0,a) V · n dS = · · · = −πa2 , where the dots indicate that we proceed as above. 1 c) Alternatively we compute the line integral U · t ds.

This implies the existence of the vector potentials and that one of these can be found by the formula 1 W0 (x) = where T(x) = U(x) × x. T(τ x) dτ, 0 First calculate T(x) = U(x) × x = ex ey ez −yz 0 xy x y z = −xy 2 , x2 y + yz 2 , −y 2 z . com 61 Calculus 2c-10 Vector potentials All coordinates are precisely of degree 3, thus by an integration with respect to τ , 1 W0 (x) = 1 T(τ x) dτ = T(x) 0 τ 3 dτ = 0 1 1 1 − xy 2 , (x2 + z 2 )y, − y 2 z . 4 4 4 We see that W0 (x) is a vector potential for U(x).

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