How to find the volume of a rectangular prism with all chamfered edges? I know it would be the volume of the full prism minus the triangular prism shapes cut off of the edges but can't work out how to calculate the corners. The corners look like this: https://www.google.com.au/search?q=chamfered+edges+cube&client=ms-unknown&prmd=sivn&source=lnms&tbm=isch&sa=X&ved=0ahUKEwiowveT_rPaAhXGXrwKHeo1Dl8Q_AUIEigC&biw=412&bih=652&dpr=2.63#imgrc=69Xby28QdeQr6M:
 A: Let the box have sufficiently large edge lengths $a$, $b$, $c$; the original volume then is $abc$. Along the edges we cut off triangular prisms with cross section a right triangle of area ${r^2\over2}$. Not looking at the small $r^3$-cubes at the vertices this amounts to the following edge cutoff:
$$4\cdot{r^2\over2}\cdot\bigl((a-2r)+(b-2r)+(c-2r)\bigr)\ .\tag{1}$$
Now look at the cube $C=[0,r]^3$, and denote the polyhedral set cut off from $C$ by $B$. A point $(x,y,z)\in C$ belongs to $B$ iff at least one of the following is true:
$$x+y\leq r,\quad x+z\leq r,\quad y+z\leq r\ .$$
The points $(x,y,z)\in C$ satisfying $x+y\leq r$ make up half of $C$, hence
$${\rm vol}(B)={r^3\over2}+{\rm vol}(B')\ ,$$
whereby $B'$ is given by
$$B'=\bigl\{(x,y,z)\in C\,\bigm|\, x+y\geq r \ \wedge \  \ z\leq r-\min\{x,y\}\bigr\}\ .$$
Due to symmetry we only have to consider the case $y\leq x$ and obtain
$${\rm vol}(B')=2\int_{r/2}^r\int_{r-x}^x\int_0^{r-y} dz\>dy\>dx={r^3\over4}\ .$$
The total cutoff at the vertices therefore is
$$8\left({1\over2}+{1\over4}\right)\>r^3=6r^3\ .$$
Add this to $(1)$ and obtain a total cutoff given by $2(a+b+c)r^2-6r^3$, so that the remaining volume is
$$V=abc-2(a+b+c)r^2+6r^3\ .$$
