= Solution
The horizontal cross-section is a <triangle>, so its <area> is half its width times its depth:
$$
A(z)=LD\left(1-\frac zH\right)\sqrt{1-\frac z{2H}}.
$$
Thus the <volume integral>, with $t=z/H$, is
$$
V=LDH\int_0^1(1-t)\sqrt{1-t/2}\,dt.
$$
For the <substitution rule>, put $u=1-t/2$. The endpoint values are $1$ and $1/2$, and $1-t=2u-1$, giving
$$
\begin{aligned}
\frac V{LDH}&=2\int_{1/2}^1(2u-1)u^{1/2}\,du\\
&=2\left[\frac45u^{5/2}-\frac23u^{3/2}\right]_{1/2}^1\\
&=\frac{4+2\sqrt2}{15}.
\end{aligned}
$$
Therefore $\boxed{V=\dfrac{(4+2\sqrt2)LDH}{15}}$.
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