How can we calculate the volume of a BCC Wigner-Seitz Cell? (Based on a imaginary cube) 
Hello. As you can see in the picture, there’s this shape and shape’s surface consists of 6 square and 8 hexagon parts and I would like to know its volume but I don’t know where to start. The only information given is that the imaginary cube which surrounds the shape has a length of “$a$” for its edges, thus the imaginary cube’s volume is $a^3$. So, how can we find the volume of that shape? (For the edge of square in the shape, I think it has a length of $\sqrt{a}/2$ but I’m not sure...)
 A: My understanding is the hexagon face is regular (all sides are equal). 
Therefore, this thing is obtained by taking an octahedron $P$, cut off one small pyramid at each of its six vertices. Assume the side length of $P$ is $x$, in order to make regular hexagon, the cutting points on each side is $\frac x3$ from vertices. 
For example, let $o$ be a vertex of $P$, and $oa, ob, oc, od$ are the four sides from $o$, then take $a'$ on $oa$ with $oa'=\frac 13 oa$,  take $b'$ on $ob$ with $ob'=\frac 13 ob$, take $c'$ on $oc$ with $oc'=\frac 13 oc$,  take $d'$ on $od$ with $od'=\frac 13 od$, the cut away the small pyramid $o$-$a'b'c'd'$, and the new exposed face $a'b'c'd'$ is a square face in your object. 
Now an easy computation shows the distance between two opposite vertices in $P$ is $\sqrt 2 x$. Now cut away the small pyramids, we see the distance between two opposite square faces is $\frac 23\sqrt x$: in fact the octahedron $P$ is obtained by gluing two big pyramids with side length $x$ together, each of them is similar to the small pyramids you cut away, the similar ratio is $3:1$. Anyway, this distance $\frac 23\sqrt x$ is just the side length (i.e. distance between two opposite faces) of your imaginary cube, therefore 
$$
a=\frac 23 \sqrt 2 x.
$$
Now the volume of the octahedron $P$ is $2\cdot \frac 13 x^2\cdot\frac {\sqrt 2 x}{2}=\frac{\sqrt 2 x^3}{3}$ (these computations are quite easy, just figure out the altitude, etc.). The volume of each of the small pyramids you cut away is $\frac 13\cdot(\frac x3)^2\cdot (\frac x3\cdot\frac{\sqrt 2}{2})=\frac{\sqrt 2 x^3}{162}$
; there are six of them, therefore your object has volume 
$\frac{\sqrt 2 x^3}{3}-6\cdot \frac{\sqrt 2 x^3}{162}=\frac{8\sqrt 2 x^3}{27}$. Plug in $x=\frac{3a}{2\sqrt 2}$, the answer is 
$$
\frac{a^3}{2}.
$$
By the way the side length of the squares in your object is $\frac x3=\frac{a}{2\sqrt 2}$. 
