Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2022 / Paper 1 / 12

The stated $x$ bounds depend on $y$, so $y$ must be held fixed while the integral over $x$ is performed. The intended iterated integral isThe original PDF places the two sets of bounds beside the opposite integral signs relative to the printed $dxdy$. That placement cannot be read literally as a conventional nested integral; the dependence of the lower $x$ bound on $y$, and the explicit instruction to integrate over $x$ first, specify the intended region unambiguously.

The inequalities describing the region are $0\le y\le 1$ and $2y\le x\le 2$. It is the closed triangle bounded by $y=0$, $x=2$ and $x=2y$, with verticesA horizontal slice runs from the sloping edge $x=2y$ to the vertical edge $x=2$.

With $y$ held fixed, the inner antiderivative givesThe remaining integral is therefore

For a vertical slice through the triangle, $0\le x\le 2$ and $0\le y\le x/2$. Reversing the order by Fubini's theorem givesThe agreement checks that both orders describe the same region.

The PDF diagram includes a solid base of thickness $d$, occupying $-d\le z\le 0$ and $0\le r\le R$. Above it, $0\le z\le h$, the metal lies between the two specified curved surfaces. In cylindrical coordinates, the Jacobian determinant contributes a factor $r$, so the mass isThe base contributes $\pi \rho R^{2}d$. The wall contributesConsequentlyThe base thickness $d$ comes from the diagram; omitting it would omit part of the cup's volume.