Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2024 / Paper 1 / 12

Take the dumbbell axis to be the $z$ axis, with its midpoint at $z=0$. The right-hand sphere has centre $z=\alpha a$ and radius $R=\beta a$, while its junction with the cylinder is at $z=\gamma a$. Put $d=(\alpha -\gamma )a$, so $0<d<R$. The Pythagorean theorem in a meridional cross-section gives the handle radius $H$ from $H^2+d^2=R^2$. Thus

Use cylindrical coordinates $(r,\varphi ,z)$ about the dumbbell axis. Their Jacobian determinant contributes the volume factor $r$, so the handle's mass is the volume integralInserting the radius from part (a) yields

Centre cylindrical coordinates on the right-hand sphere, with the axial coordinate $s=z-\alpha a$ positive outwards. Using $R=\beta a$ and $d=(\alpha -\gamma )a$, the retained solid satisfiesIn particular, it contains more than a hemisphere; it is the small cap on the inward side that has been removed. The volume integral for its mass, including the Jacobian determinant $r$, isThus each end piece has massEquivalently this is the volume of a spherical cap with height $R+d$, multiplied by the uniform mass density.

Use spherical coordinates $(r,\theta ,\varphi )$ centred on the right-hand sphere, with polar axis pointing outwards along $s$. The Jacobian determinant is $r^2\sin \theta$, and the retained region is $r\le R$ with $r\cos \theta \ge -d$.

For $0\le r\le d$, every polar angle $0\le \theta \le \pi$ is allowed. For $d<r\le R$, the cut requires $0\le \theta \le \arccos (-d/r)$. Splitting the volume integral at $r=d$ therefore givesSubstitution of $R=\beta a$ and $d=(\alpha -\gamma )a$ reproduces the mass in part (c). The radial split avoids integrating over a polar angle range independent of $r$, which would incorrectly include part of the removed spherical cap.

The handle and the two end pieces have disjoint interiors; their common boundary circles contribute no volume. Hence the total mass is $M=M_h+2M_e$. With $\delta =\alpha -\gamma$, the results above giveEquivalently, since $\gamma +\delta =\alpha$,