 Solution

ID: past-exam-of-the-mathematical-methods-ii-course-of-the-university-of-oxford/2015/12/c/i/solution

On the specified parametrized curve,

r (t) = (t^{2}, 2 t, t^{3})

and

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

. Substitution into the field gives

F (r (t)) = (4 t^{7}, 2 t^{8}, 8 t^{6})

. Hence

F \cdot r^{'} = 8 t^{8} + 4 t^{8} + 24 t^{8} = 36 t^{8} .

The line integral is therefore

\int_{C} F \cdot d r = \int_{0}^{1} 36 t^{8} d t = 4 .

Both the field and the tangent vector must be evaluated along the curve; integrating one field component alone would miss two contributions.

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