Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2020 / Paper 1 / 19

In the Cartesian coordinate system, the Laplacian of a scalar field is the sum of its three unmixed second partial derivatives. Each squared coordinate here contributes one, soEquivalently $\nabla \psi =(x,y,z)$ and its divergence is three.

The sloping height relation on the cone is $z=b(1-r/a)$. A parametric equation of its curved surface is thereforeIts two tangent vectors are ${\mathbf{R}}_r=(\cos \theta ,\sin \theta ,-b/a)$ and ${\mathbf{R}}_{\theta }=r(-\sin \theta ,\cos \theta ,0)$. Multiplication by $dr$ and $d\theta$ produces the two infinitesimal sides requested in the paper. Their cross product is the vector area element:Its radial component is positive, so this normal vector points outwards from the solid cone. Taking its norm gives the scalar area elementThe parametric equation degenerates at the apex, but that single point contributes no area.

Compute the flux directly using the outward vector area element, as required. Restrict the vector field to the surface parametric equation and take the dot product:The terms quadratic in $r$ cancel. Therefore the surface integral isThis is a direct evaluation of the flux of the position vector through a cone.

On the sphere of radius $a$, the outward unit normal is $\mathbf{n}=(x,y,z)/a$. Thus $\mathbf{F}\cdot \mathbf{n}=a$ everywhere. Multiplying by the surface area $4\pi a^2$ givesThe same surface integral follows in spherical coordinates from $dS=a^2\sin \vartheta d\vartheta d\phi$.

The new cone can be parametrized byThe outward vector area element is $r(\frac{b}{a}\cos \theta ,1,\frac{b}{a}\sin \theta )drd\theta$. This is ${\mathbf{R}}_{\theta }^{\prime }\times {\mathbf{R}}_r^{\prime }drd\theta$; reversing the order is needed to retain the outward orientation after the exchange of coordinates. Its dot product with $\mathbf{F}({\mathbf{R}}^{\prime })$ is again $brdrd\theta$. HenceThis equality also follows because a rigid rotation takes the first cone into the second and leaves the radial vector field $\mathbf{F}(\mathbf{x})=\mathbf{x}$ invariant.