<s>[INST] Volume of a sphere with three holes drilled in it. Suppose that the sphere $x^2+y^2+z^2=9$ has three holes of radius $1$ drilled through it. One down the $z$-axis, one along the $x$-axis, and one along the $y$-axis. What is the volume of the resulting solid? I can do it for two holes but I'm stuck on three.
 [/INST] The 3 holes such drilled form an intersection of 3 cylinders in the center, plus 6 cylinder/cap pairs.  I will treat each problem in turn.
Here is a picture of the situation:

Intersection of 3 cylinders
The problem is to find the volume of three orthogonal, intersecting cylinders:
$$\begin{align}x^2+y^2&=1\\x^2+z^2&=1\\ y^2+z^2&=1\end{align}$$
The intersection region is pictured below:

We note that there are two ways to bound the volume over $x$:
$$\begin{align}|x| &\le \sqrt{1-y^2}\\ |x| &\le \sqrt{1-z^2}\end{align}$$
Since we are computing the volume of the interior of the region defined by these bounds, it stands to reason that $|x|$ must be bounded by the smaller of these two bounds:
$$|x| \le \min{\left(\sqrt{1-y^2},\sqrt{1-z^2}\right)}$$
so that the volume integral takes the form
$$\int_{-1}^1 dz \: \int_{-\sqrt{1-z^2}}^{\sqrt{1-z^2}} dy \: \int_{-m(y,z)}^{m(y,z)} dx = 2 \int_{-1}^1 dz \: \int_{-\sqrt{1-z^2}}^{\sqrt{1-z^2}} dy \: m(y,z)$$
Below is a representation of the integration region for this integral:

The reason for the lines is because $\sqrt{1-y^2} \lt \sqrt{1-z^2}$ according to whether $|y| \gt |z|$. The integral is then symmetric over the regions bounded by the sloped lines; thus, we need only consider one such region and the others will yield the same result.  Let's then consider the region surrounding the positive $y$ (horizontal) axis in the above figure.  In this case, $m(y,z) = \sqrt{1-y^2}$; when we use polar coordinates, the integral becomes
$$\begin{align}8 \int_{-\pi/4}^{\pi/4} d\phi \: \int_0^1 d\rho \, \rho \sqrt{1-\rho^2 \cos^2{\phi}} &= 4 \frac{2}{3} \int_{-\pi/4}^{\pi/4} d\phi \: \left( 1- \left|\sin^3{\phi}\right|\right) \sec^2{\phi}\\ &= \frac{16}{3} -  \frac{16}{3}  \int_0^{\pi/4} d\phi \: \sin^3{\phi} \, \sec^2{\phi}\\ &= 8 \left (2 - \sqrt{2}\right ) \end{align}$$
Cylinder and cap
The cross-sectional geometry here corresponds to a rectangle of width $2$ inscribed symmetrically about a diameter of a circle of radius $3$, with the cap corresponding to the resulting circular segment.  The distance from center of circle to short edge of rectangle is $\sqrt{3^2-1^2}=2 \sqrt{2}$, so the height of the cylinder outside the intersection above is $2 \sqrt{2}-1$.  The volume of a cylinder is thus $\pi \cdot 1^2 \cdot (2 \sqrt{2}-1) = (2 \sqrt{2}-1) \pi$.
The volume of the cap is the difference between the volume of the spherical sector subtended by the solid angle defined by the hole and the corresponding cone.  The volume of the spherical sector is given by $\frac13 (3)^3 \Omega$, where $\Omega$ is the solid angle.  We find $\Omega$ by integrating over angle in spherical coordinates:
$$\Omega = \int_0^{\theta_0} d\theta \, \sin{\theta} \, \int_0^{2 \pi} d\phi = 2 \pi (1-\cos{\theta_0})$$
where $\sin{\theta_0} = \frac13 \implies \cos{\theta_0} = 2 \sqrt{2}/3$.  Thus the volume of a sector is $6\pi (3-2 \sqrt{2})$.
The volume of a cone is $\frac13 \pi (1^2) 2 \sqrt{2} = 2 \sqrt{2} \pi/3$.  Thus, the volume of a cap is
$$6\pi (3-2 \sqrt{2}) - \frac{2 \sqrt{2} \pi}{3} = \left (18 - \frac{38 \sqrt{2}}{3}\right )\pi$$
Putting it all together
The volume of the holes is $6$ times the sum of the volumes of the cylinder and cap, plus the volume of the intersection:
$$V_{\text{holes}} = 6 \pi (2 \sqrt{2}-1) + \left (108- 76 \sqrt{2}\right )\pi + 8 (2-\sqrt{2})= (102-64\sqrt{2})\pi + 8 (2-\sqrt{2})$$
The volume left over after drilling is therefore the volume of the sphere minus the volume of the holes, or
$$V = \frac{4 \pi}{3} (3)^3 - V_{\text{holes}} = (64 \sqrt{2}-66) \pi - 8 (2-\sqrt{2}) \approx 72.31$$
compared with the original volume of the sphere $36 \pi \approx 113.1$.
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