Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2003 / Paper 1 / 7 / iii / Solution

The field is smooth on the simply connected domain ${\mathbb{R}}^3$. Its curl is $\nabla \times \mathbf{F}=(2\lambda -2,2\mu ,0)$, so a necessary and sufficient condition for a conservative vector field isAt those values it is explicitly the gradient of $\Phi =xy-x^2/2-yz+z^2/2$. Equality of the two previous path integrals would impose only $\lambda -\mu =1$, which is weaker: coinciding path integrals need not imply a conservative field. For example $\lambda =2,\mu =1$ makes those two integrals equal but leaves nonzero curl.