Past exam of the mathematical methods II course of the University of Oxford / 2013 / 3 / b / Solution

Parametrize the straight path by $\mathbf{r}(t)=(t,t,t)$ for $0\le t\le 1$. Then $\dot{\mathbf{r}}=(1,1,1)$ and $\mathbf{A}=(3t^2+6t,-14t^2,20t^3)$. ThusThis differs from the preceding path's value. Indeed, $\nabla \times \mathbf{A}=(14y,-20z^2,-6)$ is not zero, so the field is not a conservative vector field and its integral need not depend only on endpoints.