= Solution
The two planes bound $z-1\le y\le1-z$, and the elliptical cross-section has $-1\le z\le1$. Under $x=2r\cos\theta$, $z=r\sin\theta$, the absolute <Jacobian determinant> for $(x,z)$ is $2r$. Therefore
$$
V=\int_0^{2\pi}\int_0^1\int_{r\sin\theta-1}^{1-r\sin\theta}2r\,dy\,dr\,d\theta
=\int_0^{2\pi}\int_0^1 4r(1-r\sin\theta)\,dr\,d\theta
=\boxed{4\pi}.
$$
As a check, the cross-sectional <ellipse> has area $2\pi$, while the part of the thickness proportional to $z$ integrates to zero by reflection. The mean thickness is therefore two.
Back to article page
