For

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

, the endpoints correspond to

t = 0, 1

and

\dot{r} = (1, 2 t, 3 t^{2})

. The vector field on the path is

A = (9 t^{2}, - 14 t^{5}, 20 t^{7})

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

Including the derivative of each path coordinate is essential; merely substituting the path into the field does not supply the line element.

b

Solution

Parametrize the straight path by

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

for

0 \leq t \leq 1

. Then

\dot{r} = (1, 1, 1)

and

A = (3 t^{2} + 6 t, - 14 t^{2}, 20 t^{3})

. Thus

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

This differs from the preceding path's value. Indeed,

\nabla \times A = (14 y, - 20 z^{2}, - 6)

is not zero, so the field is not a conservative vector field and its integral need not depend only on endpoints.

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