Past exam of the mathematics course of the University of Oxford / 2015 / Prelims / Mathematics V resit / Question 3

The annulus is $1\le r\le 2$ in polar coordinates. Their Jacobian determinant supplies $dxdy=rdrd\theta$, so

The original double integral covers the triangle $0\le y\le x\le 1$. By Fubini's theorem, after reversing the order, $y$ runs from $0$ to $1$ and $x$ from $y$ to $1$. Since a continuous function is integrable on this bounded region,For the square pyramid, a horizontal section at height $z$ is a square of side $2(1-z)$: each vertex moves linearly from the base towards $(1,1,1)$. Hence its cross-sectional area is $4(1-z{)}^2$, and its volume is $4{\int }_0^1(1-z{)}^{2}dz$. To use the identity just proved, choose $f(y)=1-y$:

Start with the solid between the elliptic paraboloid $z=x^2+y^2$ and the upper plane $z=10-2y$. Its projection is the disksince their vertical separation is $11-x^2-(y+1{)}^2$. The second lower bound, $z\ge 2-2x$, removes the part between the elliptic paraboloid and this lower plane whereverCall this second disk $E$. Its centre is distance $\sqrt{2}$ from the centre of $D$, andbecause $5+2\sqrt{6}<11$. Thus $E$ lies wholly inside $D$. In particular the lower plane never exceeds the upper plane on $E$, so the subtraction describes exactly the requested solid.

The nested-disk subtraction for paraboloid volumes therefore gives a difference of two double integrals:Translate polar coordinates to the centre of each disk. A general integral of this form isUsing $A=11$ and $A=3$ gives