What is the volume of the region in three-dimensional space defined by the inequalities $|x|+|y|+|z|\le1$ and $|x|+|y|+\left|z-\frac12\right|\le1$?

1. At height $z$ the two conditions say $|x|+|y|\le1-|z|$ and $|x|+|y|\le1-\left|z-\frac12\right|$. The region $|x|+|y|\le m$ is a square (diamond) of area $2m^2$.
2. For $-\frac12\le z\le\frac14$ the smaller bound is $m=\frac12+z$; for $\frac14\le z\le1$ it is $m=1-z$; outside $[-\frac12,1]$ the region is empty.
3. Volume $=\displaystyle\int_{-1/2}^{1/4}2\left(\tfrac12+z\right)^2dz+\int_{1/4}^{1}2(1-z)^2dz=2\cdot\frac{(3/4)^3}3+2\cdot\frac{(3/4)^3}3=\boxed{\dfrac9{16}}$.