= 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 gives
$$
\boxed{dV=\pi a^3(1-\cos\theta_0)^2(1-2\cos\theta_0)\sin^3\theta_0\,d\theta.}
$$
These are horizontal disks in the lower part of the solid. A spherical angular-wedge element $r^2\sin\theta\,dr\,d\theta\,d\phi$ describes a different infinitesimal slice.
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