Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2019 / Paper 1 / 12

The height is constant along each circular ring, whose area element after angular integration is $2\pi rdr$. The volume integral in cylindrical coordinates is thereforeHere $u=r/R$. Using integration by parts, ${\int }_0^{1}u{e}^{-u}du=[-(u+1)e^{-u}{]}_0^1=1-2/e$. HenceThe dimensionless coefficient is positive, and the cube root of a volume has the required physical dimension of length.

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 areaThe volume integral, followed by integration by substitution $u=z/H$, isThis 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 $2L$.

In the $x,z$ plane, the embankment lies above the parabola $z=H{x}^2/L^2$ and below $z=H$. The bore has circular cross-sectionIt 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 $H{x}^2/L^2\le z$, 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 thereforeThe disk has area $A=\pi D^2/4$ and centroid height $D/2$, so ${\iint }_{\mathcal{D}}zdxdz=AD/2$. Applying the volume of a solid with affine thickness formula givesThe width varies along the circular section; multiplying its area by the top width alone would miss this variation.