Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2019 / Paper 2 / 13

A line integral uses $\mathbf{F}(\mathbf{r}(t))\cdot {\mathbf{r}}^{\prime }(t)$, including the tangent vector. With $\mathbf{r}(t)=(t,t,t)$, $0\le t\le 1$, the three components sum toConsequently

It is helpful first to separate an exact gradient:The fundamental theorem for line integrals makes the gradient contribution $2\alpha +{\alpha }^2$ for either path. On the curved path, $dy=2tdt$, $dz=3t^{2}dt$, $y=t^2$ and $z=t^3$, so the remaining contribution isThus

A continuously differentiable conservative vector field has zero curl. HereThis vanishes everywhere if and only if $\boxed{\beta =0}$. Since the field is smooth on the simply connected domain ${\mathbb{R}}^3$, zero curl is also sufficient; the explicit scalar potential in the next part confirms it.

With $\beta =0$, integrating ${\Phi }_x=2\alpha x+{\alpha }^{2}y+\alpha z$ givesThe other two partial derivatives require $K_y=K_z=0$. The origin condition fixes the remaining constant to zero. Hence the scalar potential isIts value at $(1,1,1)$ gives ${\alpha }^2+2\alpha =1$, so