Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2004 / Paper 1 / 8

On the straight path $\mathbf{r}(t)=(t,t,t)$, $0\le t\le 1$, both fields restrict to $(3t^5,2t^5,t^5)$. Dotting with ${\mathbf{r}}^{\prime }=(1,1,1)$ gives

For $\mathbf{r}(t)=(t,t^2,t^2)$, the velocity is $(1,2t,2t)$. Because $y=z$ throughout this path, both fields again agree pointwise, now as $(3t^8,2t^7,t^7)$. ThusThese equal values alone do not establish path independence for either field on all of space.

Direct differentiation gives $\boxed{\Phi =x^3{y}^{2}z+C}$ with $\nabla \Phi =\mathbf{G}$, so $\mathbf{G}$ is a conservative vector field and its integral depends only on the endpoints. In contrast,is nonzero, for example at $(1,1,1)$. Thus $\mathbf{F}$ is not conservative. Both earlier paths lie in the surface $y=z$, where the two fields happen to coincide. This illustrates why coinciding path integrals need not imply a conservative field.