Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2006 / Paper 2 / 5

In the plane $x=0$, the right half of the cross-section has coordinates $y=a(1-\cos \theta )\sin \theta$, $z=a(1-\cos \theta )\cos \theta$, $0\le \theta \le \pi$. Reflect across the $z$ axis for the other half. The cardioid has a cusp at $(y,z)=(0,0)$ for $\theta =0$, crosses $z=0$ at $(\pm a,0)$ for $\theta =\pi /2$, and reaches $(0,-2a)$ for $\theta =\pi$. The small upper lobes reach $z=a/4$; the widest section is below the equator.

For horizontal slices of a surface of revolution in spherical coordinates, use disk area times vertical thickness. Here the cylindrical radius is $\rho =a(1-\cos \theta )\sin \theta$ and the height is $z=a(\cos \theta -{\cos }^2\theta )$. For $\theta >\pi /2$, $dz/d\theta =-a(1-2\cos \theta )\sin \theta <0$. Hence a positive angular increment givesThese are horizontal disks in the lower part of the solid. A spherical angular-wedge element $r^2\sin \theta drd\theta d\varphi$ describes a different infinitesimal slice.

Integrate the horizontal-disk expression over $\pi /2<\theta <\pi$. With $u=\cos \theta$ and $\sin \theta d\theta =-du$,Evaluating the integral of this polynomial gives