G f1 y 2 z ln 3 so g x y z y 2 z ln 3 dx

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Unformatted text preview: . ∂y ∂x ∂z ∂x ∂z ∂y Since the domain of ω is R3 , which is simply connected, ω closed =⇒ ω exact. There is a function g : R3 → R such that dg = ω . We now find g . ∂g F1 = y 2 + z ln 3 = , so g (x, y, z ) = (y 2 + z ln 3) dx + h(y, z ) = xy 2 + xz ln 3 + ∂x h(y, z ) ∂h ∂h ∂g = 2xy + so = sin z =⇒ h(y, z ) = sinz dy + F2 = 2xy + sin z = ∂y ∂y ∂y k (z ) = y sin z + k (z ) giving g (x, y, z ) = xy 2 + xz ln 3 + y sin z + k (z ). MATB42H page 2 Solutions # 5 ∂g dk dk = x ln 3 + y cos z + so = ln 3 =⇒ k (z ) = ∂z dz dz ln 3 dz = z ln 3 + C . Hence g (x, y, z ) = xy 2 + xz ln 3 + y sin z + z ln 3 + C , F3 = y cos z + (x + 1) ln 3 = where C is some constant. 2. We note...
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This document was uploaded on 03/17/2014 for the course MAT B42 at University of Toronto.

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