Calculate leftover volume of a drilled out cube

Solution 1:

As suggested by Ted Shifrin in comments, place the center of the cube at the origin with its faces parallel to coordinate planes.

Then the equation of drills' cylindrical surfaces are,

$x^2 + y^2 = r^2$
$y^2 + z^2 = r^2$
$z^2 + x^2 = r^2$

Now take example of the left-out piece of solid on the right bottom corner (corner $2$) -

If you see the face shaded in dark blue and dark grey (corner $2$), that is to the right of $x^2 + z^2 = r^2$. Now at the intersection of other two cylinders,

$x^2 + y^2 = y^2 + z^2 = r^2 \implies x = \pm z$. As it's corner $2$, $x = - z$

So for $x \leq - z$, $y$ is bound below by the surface shaded in dark blue and above by the cylinder $x^2 + y^2 = r^2$.

Now at the intersection of $x = -z$ and $x^2 + z^2 = r^2, z = \pm \frac {r}{\sqrt2}$. As it's corner $2$, $z = - \frac{r}{\sqrt2}$

That leads to the bounds,

$-r \leq y \leq - \sqrt{r^2 - x^2}$
$\sqrt{r^2 - z^2} \leq x \leq - z$
$- r \leq z \leq - \frac{r}{\sqrt2}$

for the order of integration $dy ~ dx ~ dz$.

For $x \geq - z, $ you will have the same volume and then there are $8$ such corners.

So once you evaluate the integral, multiply by $16$.

Given your quest for further clarity on bounds, let me provide a bit more detail. The upper bound of $z$ is $ - \frac {r}{\sqrt2}$ because we divided the volume of corner $2$ into two. We did so because we did not have a single bound of $y$ for the whole face of corner 2 which is in the plane $y = -r$. For $x \leq - z$, $y$ is bound above by the cylinder $x^2 + y^2 = r^2$ and for $x \geq -z$, $y$ is bound above by the cylinder $y^2 + z^2 = r^2$. The bounds that I wrote in my answer above is for the part shaded in dark blue. For the part above plane $x = -z$ (shaded in dark grey), the bounds will be -

$- r \leq y \leq - \sqrt{r^2-z^2}$
$- x \leq z - \sqrt{r^2 - x^2}$
$\frac{r}{\sqrt2} \leq x \leq r$

for the order of integration $dy ~ dz ~ dx$. But the volume is same by symmetry so you can just evaluate the first integral I originally wrote and multiply it by $2$ to find the volume of corner $2$. Then multiply by $8$ for $8$ corners.

To further show what's going on, let me take corner $4$.

The surface marked in yellow is part of cube drilled by $x^2 + z^2 = r^2$; the part marked in light blue is by the drill $x^2 + y^2 = r^2$ and the part marked in green is by the drill $y^2 + z^2 = r^2$. As you can see, the plane $z = x$ is the intersection of drills $x^2 + y^2 = r^2$ and $y^2 + z^2 = r^2$ and should be used to split the volume integral to get right bounds of $y$ for the face of corner $4$, which is in the plane $y = r$.

Solution 2:

From your picture, we can actually break each corner further into half, and multiply the following set up by $16$

$$V = 16\int_{\frac{r}{\sqrt{2}}}^r\int_{\sqrt{r^2-x^2}}^x\int_{-r}^{-\sqrt{r^2-y^2}}dzdydx$$

This integral gives the correct answer, but I can't calculate properly. @Intelligenti has the proper calculation from Mathematica as

$$2(4+4\sqrt{2}-3\pi)r^3$$