Solution
= Solution
Parametrize the straight path by $\mathbf r(t)=(t,t,t)$ for $0\leq t\leq1$. Then $\dot{\mathbf r}=(1,1,1)$ and $\mathbf A=(3t^2+6t,-14t^2,20t^3)$. Thus
$$
\boxed{\int_C\mathbf A\cdot d\mathbf r=\int_0^1(6t-11t^2+20t^3)dt=3-\frac{11}{3}+5=\frac{13}{3}.}
$$
This 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.