Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2024 / Paper 2 / 13 / b

Parameterize the straight line by $\mathbf{r}(t)=(t,t,t{)}^T$, $0\le t\le 1$. Then ${\mathbf{r}}^{\prime }=(1,1,1{)}^T$ and the line integral becomes

On the circular arc, $z=0$ and $dz=0$, so the line integral reduces toThis exact differential evaluation applies to the specified arc; for example its parametric equation is $(x,y,z)=(1-\cos \theta ,\sin \theta ,0)$, $0\le \theta \le \pi /2$. On the vertical segment $x=y=1$, only $dz$ contributes, giving ${\int }_0^1(\lambda +1)dz=\lambda +1$. The whole path therefore givesTo answer the final conservativeness requirement, compute the curl:It vanishes identically exactly whenOn the simply connected whole space this condition is sufficient for a conservative vector field. Explicitly a vector-field potential isBoth path calculations then give $13/6$, equal to $\varphi (1,1,1)-\varphi (0,0,0)$ by the fundamental theorem for line integrals. Equality of just these two path values would only force $\mu =2\lambda +1$; it is the curl calculation that establishes full path independence.