= Solution
Use <cylindrical coordinates> $x=r\cos\theta$, $y=r\sin\theta$. The <Jacobian determinant> gives
$$
\boxed{dV=r\,dr\,d\theta\,dz.}
$$
The inequalities imply $-d<z<0$ and $0\le r<a\sqrt{1+z/d}$. Thus the <volume> is
$$
\begin{aligned}
V&=\int_{-d}^0\int_0^{2\pi}\int_0^{a\sqrt{1+z/d}}r\,dr\,d\theta\,dz\\
&=\pi a^2\int_{-d}^0\left(1+\frac zd\right)dz\\
&=\boxed{\frac12\pi a^2d.}
\end{aligned}
$$
Equivalently, integrating vertically first gives $V=\int_0^{2\pi}\int_0^a\int_{d(r^2/a^2-1)}^0r\,dz\,dr\,d\theta$. The region is bounded by an <elliptic paraboloid> and its top disc; its <volume> is half that of the cylinder with base radius $a$ and height $d$. Including or excluding the boundary surfaces does not change the <volume>.
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