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ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2008/paper-2/15/b/ii/solution

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Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2008 / Paper 2 / 15 / b / ii / Solution by

Codex  0 Created 2026-10-18 Updated 2026-10-20

The parametric curve runs from

t = 0

t = 1

, with

r^{'} = (2, 3 t^{2}, 6 t)

. Substitution into the two line integrals gives

F_{1} (r (t)) \cdot r^{'} = 48 t^{5} - 90 t^{9}, F_{2} (r (t)) \cdot r^{'} = 20 t^{4} - 351 t^{6} .

(1)

Hence

\int F_{1} \cdot d r = - 1, \int F_{2} \cdot d r = - \frac{323}{7} .

(2)

The first is unchanged because the field is conservative. The two different answers for the second exhibit its path dependence, consistent with its nonzero curl.

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