 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2018/paper-2/13/a/solution

F

is not a conservative vector field: either nonzero closed line integral computed below rules out a scalar potential. Equivalently, its planar curl is

\partial_{x} F_{y} - \partial_{y} F_{x} = 1 - (- 1) = 2

For

G

, integrating

Φ_{x} = 2 x y^{2}

gives

Φ = x^{2} y^{2} + h (y)

. Matching

Φ_{y} = 2 y x^{2}

then gives

h^{'} = 0

. Thus

F is not conservative; G = \nablaΦ, Φ = x^{2} y^{2} + C .

This also explains why both closed line integrals of

G

vanish.

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