Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2021 / Paper 1 / 12

In spherical coordinates, with polar angle $0\le \theta \le \pi$ and azimuth $0\le \phi <2\pi$, the volume element isThe mass density is radial, so integrating over both angles givesWriting $q=r_0/R>0$, this becomesAlthough the mass density diverges like $A/(r_{0}r)$ near the origin, multiplication by the spherical volume element gives an integrable expression. Thus the central singularity does not produce infinite mass.

For $q=r_0/R\to \infty$, expand the exponential function in powers of $1/q$:The constant and linear terms cancel in $1-q+q{e}^{-1/q}$, givingFor $q\to 0^{+}$, $e^{-1/q}$ is exponentially small:The small-$q$ correction $q{e}^{-1/q}$ is smaller than every positive power of $q$ beyond the displayed linear term. In dimensional terms, the large-$q$ leading mass is $2\pi A{R}^2/r_0$, while the small-$q$ limit is $4\pi AR$.

The positive-domain graph starts with limiting value $f(0^{+})=1$ and limiting slope $-1$, then decreases towards the horizontal asymptote $f=0$, with $f(q)\sim 1/(2q)$ for large $q$. In fact direct differentiation givesThe first inequality follows from $e^s>1+s$ for $s>0$. Thus the graph is strictly decreasing and convex upward, with $0<f(q)<1$ and no stationary points. The integral representation $f(q)={\int }_0^1(1-e^{-s/q})ds$ also shows its positivity.

These properties belong to the reusable exponential spherical mass factor.

Use cylindrical coordinates $x=r\cos \theta$, $y=r\sin \theta$. The Jacobian determinant givesThe inequalities imply $-d<z<0$ and $0\le r<a\sqrt{1+z/d}$. Thus the volume isEquivalently, integrating vertically first gives $V={\int }_0^{2\pi }{\int }_0^a{\int }_{d(r^2/a^2-1)}^{0}rdzdrd\theta$. The region is bounded by an elliptic paraboloid and its top disc; its volume is half that of the cylinder with base radius $a$ and height $d$. Including or excluding the boundary surfaces does not change the volume.