Past exam of the mathematics course of the University of Oxford / 2015 / Prelims / Mathematics V resit / Question 7 / b

Write the observation point as $(0,0,h)$ with $h>0$, reserving source coordinates for the integral. For the straight segment, $d{s}^{\prime }=d{x}^{\prime }$ and the separation is $\sqrt{x^{\prime 2}+h^2}$. Therefore its Newtonian gravitational potential isEquivalently $\varphi (h)=-2G\gamma \mathrm{arsinh}(a/h)$. At large $h$ it approaches $-G(2a\gamma )/h$, the Newtonian potential of a point mass with the segment's total mass.

Use source polar coordinates $x^{\prime }=s\cos \theta$, $y^{\prime }=s\sin \theta$ on the semicircular disk, with $0\le s\le a$ and $0\le \theta \le \pi$. Its surface area element is $sdsd\theta$, and the separation from $(0,0,h)$ is $\sqrt{s^2+h^2}$. ThusThis Newtonian gravitational potential is half that of the full disk because the separation is independent of the polar angle. At large $h$ it approaches $-G(\pi a^2\sigma /2)/h$, with the correct total mass.

Use source spherical coordinates with radius $s$, polar angle $\alpha$ measured from the positive $z$-axis, and azimuth $\theta$. The lower half of the spherical shell has $a\le s\le 2a$, $\pi /2\le \alpha \le \pi$, and $0\le \theta \le 2\pi$. Its volume element is $s^2\sin \alpha dsd\alpha d\theta$. At the observation point $(0,0,h)$, $h>0$, the separation is $\sqrt{s^2+h^2-2sh\cos \alpha }$, soThe angular integral is elementary: put $q=\cos \alpha$ to obtainWe used $\sqrt{(s+h{)}^2}=s+h$ since both $s,h$ are positive. HenceThus the axial potential of a uniform hemispherical shell, at every point of the positive $z$-axis, isThere is no need to split into $h<a$, $a<h<2a$, and $h>2a$: the observation point is outside the lower hemisphere in all three cases. As checks, $\varphi (h)\to -3\pi G\rho a^2$ as $h\downarrow 0$, and at large $h$,The last factor is exactly the total mass of the lower half of the spherical shell.