What is the volume of the region in three-dimensional space defined by the inequalities $|x|+|y|+|z|\le1$ and $|x|+|y|+|z-1|\le1$? $\textbf{(A)}\ \frac{1}{6}\qquad$ $\textbf{(B)}\ \frac{1}{4}\qquad$ $\textbf{(C)}\ \frac{1}{3}\qquad$ $\textbf{(D)}\ \frac{1}{2}\qquad$ $\textbf{(E)}\ 1$
Solution
For $z\in[0,1]$, the cross-sectional constraint from the first region is $|x|+|y|\le1-z$ (a diamond of area $2(1-z)^2$); from the second, $|x|+|y|\le1-|z-1|=z$ (area $2z^2$).
By symmetry about $z=\frac12$, the intersection's cross-sectional area is $2z^2$ for $z\in[0,\frac12]$ and $2(1-z)^2$ for $z\in[\frac12,1]$.
Integrating: $2\int_0^{1/2}2z^2\,dz=\frac16$. The answer is (\textbf{A}).
$\textbf{(A)}\ \frac{1}{6}\qquad$ $\textbf{(B)}\ \frac{1}{4}\qquad$ $\textbf{(C)}\ \frac{1}{3}\qquad$ $\textbf{(D)}\ \frac{1}{2}\qquad$ $\textbf{(E)}\ 1$ 