A cube $ABCDEFGH$ with edge length $6$ has bottom face $ABCD$ and top face $EFGH$, with $E$ above $A$, $F$ above $B$, $G$ above $C$, and $H$ above $D$. A plane through $A$, $C$, and the midpoint of $\overline{FG}$ cuts the cube into two pieces. What is the volume of the piece that contains vertex $B$?

1. Use coordinates $A=(0,0,0)$, $B=(6,0,0)$, $C=(6,6,0)$, $D=(0,6,0)$ with $E,F,G,H$ directly above at height $6$. The midpoint of $\overline{FG}$ is $N=(6,3,6)$.
2. The plane contains $A$ and $C$, so it has an equation $x-y+kz=0$; since it contains $N$, $6-3+6k=0$, $k=-\frac12$. The plane is $2x-2y-z=0$, and $B$ is on the side $2x-2y-z>0$.
3. The cross-section of the piece at height $z=t$ ($0\le t\le6$) is the part of the square $[0,6]^2$ with $x-y\ge\frac t2$, a right triangle with legs $6-\frac t2$ and area $\frac12\left(6-\frac t2\right)^2$.
4. Volume $=\displaystyle\int_0^6\frac12\left(6-\frac t2\right)^2dt=\left[-\frac{(6-t/2)^3}{3}\right]_0^6=\frac{6^3-3^3}3=\boxed{63}$.