 Solution

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.