Past exam of the mathematical methods II course of the University of Oxford / 2014 resit / 11 / d

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- Solution d

Solution

With those constants, integrate

ϕ_{x} = x + 3 y + 3 z

with respect to

x

to obtain

ϕ = x^{2} /2 + 3 x y + 3 x z + g (y, z)

. Matching

ϕ_{y} = 3 x + y - z

gives

g = y^{2} /2 - yz + h (z)

. Finally, matching

ϕ_{z} = 3 x - y - 2 z

gives

h = - z^{2} + C

. Thus a potential of a conservative vector field is

ϕ (x, y, z) = \frac{x ^{2}}{2} + 3 x y + 3 x z + \frac{y ^{2}}{2} - yz - z^{2} + C, F = \nabla ϕ .

This explicit polynomial establishes a global potential on all of

R^{3}

; no topological obstruction is present. Its additive constant is arbitrary.

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