# 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.

• What makes it more difficult with three holes? – M.B. Nov 5 '13 at 12:58

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 $36 \pi \approx 113.1$.

• you should know that the intersection area is not like a cube. i think you have missed to compute the additional volume to find the total volume of holes. help me if i am wrong. – Bhaskara-III Sep 4 '16 at 14:49
• @Bhaskara-III When did I say it was like a cube? My diagram clearly illustrates what the volume is. – Ron Gordon Sep 4 '16 at 15:02
• you have considered 6 cylinders each of length $2\sqrt 2-1$ outside the intersection region which have flat circular ends of radius $1$. actually, you can't separate them with flat faces, in reality each cylinder has one end as flat circle of radius $1$ while other as non planar complicated shape i.e. not as a flat circle. the volume of each cylinder is not equal to $\color{red}{\pi \cdot1^2\cdot (2\sqrt 2-1)}$ that's why your answer has error in the value. – Bhaskara-III Sep 8 '16 at 7:02
• The figure was generated in Mathematica and is based on no assumption. I dispute your assertion. If you are going to stick with it, then why don't you give us an idea of what this "complicated surface" looks like. – Ron Gordon Sep 8 '16 at 13:19
• you may have a close look on my picture added in the answer which shows that the common region of three cylinders does not have any flat face. – Bhaskara-III Sep 11 '16 at 10:35

It is just a hint to visualize the region of intersection inside the sphere of radius 3.

here is picture showing that the region of intersection of three cylinders ,each having radius 1, doesn't have any flat face. volume of intersection region has been calculated in the answer as $8(2-\sqrt 2)$. Now, how to calculate the volume of remaining region of holes? in the answer, it has been assumed that the faces of intersection region are flat. i don't know why. please someone clarify my doubt. 