Past exam of the mathematics course of the University of Oxford / 2023 / Prelims / Mathematics V / 2 / b

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

0 \leq t \leq 1

F (r (t)) = (9 t^{2}, - 14 t^{5}, 20 t^{7}), r^{'} (t) = (1, 2 t, 3 t^{2}) .

Consequently the line integral is

\int_{C} F \cdot d r = \int_{0}^{1} (9 t^{2} - 28 t^{6} + 60 t^{9}) d t = 3 - 4 + 6 = 5.

ii

Compute the line integral on each straight segment. On the first,

y = z = 0

and only

d x

contributes; on the second,

x = 1, z = 0

and only

d y

contributes; on the third,

x = y = 1

and only

d z

contributes. Therefore

\int_{C} F \cdot d r = \int_{0}^{1} 3 x^{2} d x + \int_{0}^{1} 0 d y + \int_{0}^{1} 20 z^{2} d z = 1 + 0 + \frac{20}{3} = \frac{23}{3} .

Although

F_{x} = 9

on the second segment, that term does no work there because

d x = 0

Parametrize the straight segment by

r (t) = (t, t, t)

0 \leq t \leq 1

. Then the line integral is

\int_{C} F \cdot d r = \int_{0}^{1} (3 t^{2} + 6 t - 14 t^{2} + 20 t^{3}) d t = 3 - \frac{11}{3} + 5 = \frac{13}{3} .

The three different answers are possible because this vector field is not a conservative vector field; for example

\partial_{y} F_{x} = 6

whereas

\partial_{x} F_{y} = 0

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