= Solution
In the $x,z$ plane, the embankment lies above the <parabola> $z=Hx^2/L^2$ and below $z=H$. The bore has circular cross-section
$$
\mathcal D=\left\{(x,z):x^2+(z-D/2)^2\le D^2/4\right\}.
$$
It touches the valley floor at the origin. To check that this entire <disk> lies inside the projected embankment, its equation gives $x^2\le Dz-z^2\le Dz$. Since $HD/L^2<1$, it follows that $Hx^2/L^2\le z$, with strict inequality away from the bottom contact. Also $z\le D<H$. Thus no circular segment needs to be subtracted from the tunnel section.
For each $(x,z)$ in the <disk>, the length of bore inside the material is the local width $2b(2-z/H)$. The required <volume integral> is therefore
$$
V_{\rm cut}=\iint_{\mathcal D}2b\left(2-\frac zH\right)\,dx\,dz.
$$
The <disk> has <area> $A=\pi D^2/4$ and <centroid> height $D/2$, so $\iint_{\mathcal D}z\,dx\,dz=AD/2$. Applying the <volume of a solid with affine thickness> formula gives
$$
\boxed{V_{\rm cut}=2bA\left(2-\frac{D}{2H}\right)=\pi bD^2\left(1-\frac D{4H}\right).}
$$
The width varies along the circular section; multiplying its <area> by the top width alone would miss this variation.
\Image[/past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2019/paper-1-2019-q12-tunnel-section.png]
{title=Original cross-section of the parabolic embankment and the circular tunnel}
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