G dk dk 1 xez y 2 so z k z z dz z 2 c z dz dz 2 12 xy

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Unformatted text preview: k (z ) = y 2z + y + k (z ) giving g (x, y, z ) = exy + xez + x2 y + y 2z + y + k (z ). ∂g dk dk 1 = xez + y 2 + so = z =⇒ k (z ) = z dz = z 2 + C . ∂z dz dz 2 12 xy z 2 2 Hence g (x, y, z ) = e + xe + x y + y z + y + z + C . 2 (b) ω = F1 dx + F2 dy + F3 dz = x sin y dx + y cos z dy − z cos x dz ∂F1 ∂F2 ∂F1 ∂F2 Since = x cos y while = 0, = , hence, ω is not closed and, ∂y ∂x ∂y ∂x consequently, not exact. There is no g such that dg = ω . F3 = xez + y 2 + z = (c) ω = F1 dx+F2 dy +F3 dz = (y 2 +z ln 3)dx+(2xy +sin z )dy + y cos z +(x+1) ln 3 dz ∂F1 ∂F2 ∂F1 ∂F3 ∂F2 ∂F3 Since = 2y = , = ln 3 = and = cos z = , ω is closed...
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This document was uploaded on 03/17/2014 for the course MAT B42 at University of Toronto- Toronto.

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