= Solution
Use <spherical coordinates> $(r,\theta,\phi)$ 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 $r\leq R$ with $r\cos\theta\geq-d$.
For $0\leq r\leq d$, every polar <angle> $0\leq\theta\leq\pi$ is allowed. For $d<r\leq R$, the cut requires $0\leq\theta\leq\arccos(-d/r)$. Splitting the <volume integral> at $r=d$ therefore gives
$$
\begin{aligned}
M_e={}&\rho\int_0^{2\pi}\left[\int_0^d\int_0^\pi r^2\sin\theta\,d\theta\,dr+\int_d^R\int_0^{\arccos(-d/r)}r^2\sin\theta\,d\theta\,dr\right]d\phi\\
={}&2\pi\rho\left[2\int_0^d r^2\,dr+\int_d^R r^2\left(1+\frac dr\right)\,dr\right]\\
={}&2\pi\rho\left[\frac{2d^3}{3}+\frac{R^3-d^3}{3}+\frac d2(R^2-d^2)\right]\\
={}&\boxed{\pi\rho\left(\frac{2R^3}{3}+R^2d-\frac{d^3}{3}\right).}
\end{aligned}
$$
Substitution 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>.
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