= Solution
Integrate by horizontal slices. At a fixed height $0\le z\le H$, the parabolic boundary gives $|x|\le L\sqrt{z/H}$; the other inequality gives $|y|\le b(2-z/H)$. Thus the cross-section is a rectangle of <area>
$$
A(z)=4Lb\sqrt{\frac zH}\left(2-\frac zH\right).
$$
The <volume integral>, followed by <integration by substitution> $u=z/H$, is
$$
\begin{aligned}
V_{\rm bank}&=\int_0^H A(z)\,dz
=4LbH\int_0^1(2u^{1/2}-u^{3/2})\,du\\
&=4LbH\left(\frac43-\frac25\right)
=\boxed{\frac{56}{15}LbH.}
\end{aligned}
$$
This has units of <volume>. As a check on the geometry, the width decreases with height while the length of the horizontal slice increases from zero to $2L$.
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