Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2014 / Paper 2 / 16

The three Cartesian contributions to the divergence are $2(y+z)$, $-2y$, and $-2z$. Hence

Integrate the normal component on each face directly, with outward normals. Each unlisted pair of coordinates ranges independently from zero to one:For example, on $x=1$ the integral is ${\int }_0^1{\int }_0^{1}2(y+z)dydz=2$. Adding all six contributions givesThis calculation uses the six explicit surface integrals, independently of the divergence theorem.

The first field has vanishing curl and is the gradient of a monomial:For the second field, direct component differentiation givesIt is not identically zero. Thus ${\mathbf{F}}_1$ is conservative and ${\mathbf{F}}_2$ is not on ${\mathbb{R}}^3$. The conditional path-integral requests therefore apply to ${\mathbf{F}}_1$.

Parametrize the straight path by $\mathbf{r}(t)=(t,t,2t)$ for $0\le t\le 1$. The conservative field along it is $(6t^5,4t^5,t^5)$ and the tangent is $(1,1,2)$. ThusThe associated potential has ${\varphi }_1(B)-{\varphi }_1(A)=2$, in agreement with the fundamental theorem for line integrals.

For $\mathbf{r}(t)=(t,t,2t^3)$, the conservative field is $(6t^7,4t^7,t^5)$ and the tangent is $(1,1,6t^2)$. ThereforeThe two different paths give the same value because their line integrals depend only on the potential difference ${\varphi }_1(B)-{\varphi }_1(A)=2$. There is no scalar potential for ${\mathbf{F}}_2$ to which the conditional request could be applied.