Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2013 / Paper 2 / 16

Differentiate the scalar field with respect to each Cartesian coordinate:In particular its value at the point used below is $(24,3,-6)$.

The cylinder is a level set of $h=x^2+y^2-8$, so its outward normal vector is parallel to $\nabla h=(2x,2y,0)$. At the point this is $(4,4,0)$, with norm $4\sqrt{2}$. Choosing the outward orientation givesIts negative is the equally valid inward unit normal vector; the next directional derivative then changes sign.

Use the gradient value from (i) and the outward normal vector from (ii). The rate of change per unit distance is

Put $R=\sqrt{8}$ and parametrize the circle counterclockwise, as viewed from positive $z$, by $\mathbf{r}(\varphi )=(R\cos \varphi ,R\sin \varphi ,3)$ for $0\le \varphi \le 2\pi$. The restricted field isThe PDF does not specify the differential in this line integral. An oriented vector line integral weighted by a scalar field uses $d\mathbf{r}=(-R\sin \varphi ,R\cos \varphi ,0)d\varphi$. Its two nonzero components areAll other terms vanish by sine-cosine symmetry. Thus this interpretation gives

If the intended convention is the scalar arc length integral, then $ds=Rd\varphi$ andThat scalar integral is independent of traversal direction. Giving both values retains the actual distinction left unspecified by the wording.

Assume the components of $\mathbf{F}$ have continuous second partial derivatives. Expanding the divergence of its curl givesEach positive term cancels a negative term because the corresponding mixed derivatives commute. Consequently