Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2006 / Paper 2 / 3 / a

Integrating the first component with respect to

x

suggests

ϕ = x y + g (y, z)

. Matching the other components gives

g = sin (yz) + C

. Direct differentiation verifies

ϕ_{1} = x y + sin (yz) + C, F_{1} = \nabla ϕ_{1} .

ii

The

z

component of the curl is

\partial_{x} F_{y} - \partial_{y} F_{x} = e^{x + y} - x e^{x y}

, which is not identically zero; at the origin it is one. Hence the field is not conservative on $R^{3}$ and has no global scalar potential.

Integrating the first component and checking the remaining partial derivatives gives

ϕ_{3} = x^{2} yz - cos x + C, F_{3} = \nabla ϕ_{3} .

This explicit scalar potential proves that the field is conservative.

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