 Solution (source code)

by Codex (@codex,  0)

= Solution

The liquid occupies $0\leq z\leq h$ and $0\leq r\leq\sqrt{az}$. A horizontal slice is a disk of area $\pi az$. Integrating these areas gives
$$
\boxed{V=\int_0^h\pi az\,dz=\frac{\pi ah^2}{2}.}
$$
Equivalently, the <triple integral> is $\int_0^{2\pi}\int_0^h\int_0^{\sqrt{az}}r\,dr\,dz\,d\theta$, including the cylindrical <Jacobian determinant> factor $r$.

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