Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2023 / Paper 1 / 12 / b

At height $z$, the horizontal triangle has half-width $L\sqrt{1-z/(2H)}$ along $x$ and depth $D(1-z/H)$ along $y$. These shrink with $z$, so its orthogonal projection onto the $(x,y)$ plane is the base triangle with vertices $(-L,0)$, $(L,0)$ and $(0,D)$:The orthogonal projection onto the $(y,z)$ plane is the triangle with vertices $(0,0)$, $(D,0)$ and $(0,H)$:The orthogonal projection onto the $(x,z)$ plane is bounded by $z=0$, $z=H$ and the two curvesEquivalently, these curves lie on the parabola $z=2H(1-x^2/L^2)$, truncated at height $H$. In particular the top width is $\sqrt{2}L$, so this projection has a horizontal upper edge from $-L/\sqrt{2}$ to $L/\sqrt{2}$; it is not triangular. The top horizontal cross-section has zero depth, consistent with this edge.

The horizontal cross-section is a triangle, so its area is half its width times its depth:Thus the volume integral, with $t=z/H$, isFor the substitution rule, put $u=1-t/2$. The endpoint values are $1$ and $1/2$, and $1-t=2u-1$, givingTherefore $\boxed{V=\frac{(4+2\sqrt{2})LDH}{15}}$.