= Solution
For the one-to-one transformation $(u,v,w)\mapsto(x,y,z)$, the <Jacobian determinant> used in the stated formula is
$$
J=\det\frac{\partial(x,y,z)}{\partial(u,v,w)}.
$$
Locally, the derivative maps a small rectangular box of volume $du\,dv\,dw$ to a parallelepiped of volume $|J|\,du\,dv\,dw$. Partitioning the region into such boxes and passing to the <Riemann integral> gives the <change of variables formula>
$$
\int_Df(x,y,z)\,dx\,dy\,dz
=\int_\Delta |J|f(x(u,v,w),y(u,v,w),z(u,v,w))\,du\,dv\,dw.
$$
In the plane $y=0$, the region is inside the unit circle and on the right-hand side of the hyperbola
$$
x^2+z^2\leq1,
\qquad
\frac{x^2}{\alpha^2}-\frac{z^2}{\alpha^2\gamma^2}\geq1.
$$
It is nonempty exactly when $\alpha\leq1$; it has positive area when $\alpha<1$. In $x>0$, its boundaries are $x=\sqrt{1-z^2}$ and $x=\alpha\sqrt{1+z^2/\gamma^2}$. The minimum and maximum $x$ values are $\alpha$ and $1$, both at $z=0$. The largest absolute $z$ occurs where the boundaries meet:
$$
z_0=\gamma\sqrt{\frac{1-\alpha^2}{1+\gamma^2}},
\qquad
x_0=\sqrt{\frac{1+\alpha^2\gamma^2}{1+\gamma^2}}.
$$
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2019/ia/paper-3-region-cross-section.png]
{title=Cross-section of the integration region}
Now set $u=x$, $v=y/k$, and $w=z$. The inverse <Jacobian determinant> is $k$, and in cylindrical coordinates $r^2=u^2+v^2$ the transformed region is
$$
r^2+w^2\leq1,
\qquad
r^2\geq\alpha^2+\frac{w^2}{\gamma^2}.
$$
Thus $|w|\leq z_0$, and each horizontal cross-section is an annulus. Its volume is
$$
\begin{aligned}
\operatorname{Vol}(D)
&=k\pi\int_{-z_0}^{z_0}
\left[1-w^2-\alpha^2-\frac{w^2}{\gamma^2}\right]dw\\
&=\frac{4\pi k}{3}(1-\alpha^2)z_0.
\end{aligned}
$$
Therefore, for $\alpha<1$,
$$
\boxed{\operatorname{Vol}(D)=
\frac{4\pi k\gamma}{3\sqrt{1+\gamma^2}}(1-\alpha^2)^{3/2}}.
$$
For $\alpha=1$ the region has zero volume.
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