= Solution
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 satisfies
$$
-d\leq s\leq R,\qquad0\leq r\leq\sqrt{R^2-s^2},\qquad0\leq\phi<2\pi.
$$
In 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> $r$, is
$$
\begin{aligned}
M_e&=\rho\int_{-d}^{R}\int_0^{2\pi}\int_0^{\sqrt{R^2-s^2}}r\,dr\,d\phi\,ds\\
&=\pi\rho\int_{-d}^{R}(R^2-s^2)\,ds\\
&=\pi\rho\left(\frac{2R^3}{3}+R^2d-\frac{d^3}{3}\right).
\end{aligned}
$$
Thus each end piece has <mass>
$$
\boxed{M_e=\pi\rho a^3\left[\frac{2\beta^3}{3}+\beta^2(\alpha-\gamma)-\frac{(\alpha-\gamma)^3}{3}\right].}
$$
Equivalently this is the <volume of a spherical cap> with height $R+d$, multiplied by the uniform <mass density>.
\Image[/past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2024/paper-1-q12-integration-region.png]
{title=The retained end piece and the two radial ranges used in spherical integration}
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