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Consider a sphere of radius $a$ with 2 cylindrical holes of radius $b<a$ drilled such that both pass through the center of the sphere and are orthogonal to one another. What is the volume of the remaining solid?

Can someone help me at least setting up the integral? I know that there is a similar problem but it was a sphere with one hole.

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WOLOG, choose the coordinate system such that the sphere of radius $a$ is centered at origin $O$ and the axes of the two holes are aligned along the $x$ and $y$ axis. Let

$$c = \sqrt{a^2 - b^2},\quad d = \begin{cases}\sqrt{b^2 - c^2},& b > \frac{a}{\sqrt{2}}\\0,& \text{otherwise}\end{cases} \quad\text{ and }\quad e = \min(b,c) $$ When one intersect the sphere with two holes with a plane of $z = const$, one find:

  • $|z| > a \leadsto$ the intersection is empty.
  • $a \ge |z| \ge b \leadsto$ the intersection is a circle of radius $\rho = \sqrt{a^2-z^2}$.
  • $b \ge |z| \ge d \leadsto$ the part of circle within a distance $\lambda = \sqrt{b^2 - z^2}$ from the $x$ or $y$-axis has been removed. The intersection split into $4$ pieces and the shaded area $ABC$ in following figure is one of these pieces:
    $\hspace0.8in$ a piece of the the cross section
    It is clear $$\begin{align} \text{Area}(ABC) &= \text{Area}(OBC) - (\text{Area}(OBA) + \text{Area}(OAC))\\ &= \frac12\rho^2(2\phi) - 2(\frac12\lambda(c-\lambda)) = \rho^2\phi - \lambda(c - \lambda)\end{align}$$ where $\phi$ is the half angle of the circular arc $BC$ with respect to $O$.

  • Finally, when $b > \frac{a}{\sqrt{2}}$, $d > 0$ and for those $|z| < d$, the intersection is empty.

If we extend the definition of $\phi$ to $\frac{\pi}{4}$ and $\lambda$ to $0$ for $|z| \in [b,a]$, the volume of the sphere with two holes can be expressed as $$\text{Vol} = 8 \int_d^a \left( \rho^2 \phi - \lambda ( c - \lambda )\right) dz \tag{*1}$$

It is clear $\frac{d\phi}{dz} = 0$ for $z > b$. Let $\theta = \frac{\pi}{4} - \phi$ be the angle between $OB$ and the $x$-axis. For $z \in [b,d]$, we have: $$\begin{align} \lambda = c \tan\theta & \implies d\lambda = c (\tan^2\theta + 1)d\theta\\ & \implies d\phi = -d\theta = -\frac{c d\lambda}{\lambda^2 + c^2} = \frac{cz dz}{\lambda(\lambda^2+c^2)} \end{align} $$ Notice $\rho^2 dz = d\Delta$ where $\displaystyle\;\Delta(z) = (a^2 - \frac{z^2}{3})z\;$, we can integrate $(*1)$ by part and get

$$\begin{align} \text{Vol} &= 8 \left\{\left[\phi(z)\Delta(z)\right]_d^a - \int_d^a \left(\Delta(z)\frac{d\phi}{dz} + \lambda(c - \lambda)\right) dz \right\}\\ &= \frac{4\pi a^3}{3} + \frac{8}{3}\underbrace{\left(3b^2(b-d) - (b^3-d^3)\right)}_{\text{comes from }\int_d^b \lambda^2 dz} - 8b^2c\mathcal{O} \end{align}\tag{*2} $$ where $$\mathcal{O} = \frac{1}{b^2}\int_d^b \left[\left(a^2 - \frac{z^2}{3}\right)\frac{z^2}{(a^2 - z^2)\sqrt{b^2-z^2}} + \sqrt{b^2 - z^2}\right] dz$$

Let

  • $z = b\cos\psi$ and $\psi_0 = \cos^{-1}\frac{d}{b}$.
  • $t = \tan\psi$ and $t_0 = \tan\psi_0 = \frac{\sqrt{b^2-d^2}}{d} = \frac{e}{d}$,

We find $$\begin{align} \mathcal{O} &= \int_0^{\psi_0}\left[ \left(a^2 - \frac{b^2}{3}\cos^2\psi\right) \frac{\cos^2\psi}{(a^2-b^2\cos^2\psi)} + \sin^2\psi \right] d\psi\\ &= \int_0^{\psi_0} \left( \frac{2a^2}{3}\frac{\cos^2\psi}{a^2-b^2\cos^2\psi} + \frac13\cos^2\psi + \sin^2\psi \right) d\psi\\ &= \frac23\psi_0 -\frac13\sin\psi_0\cos\psi_0 + \frac{2a^2}{3} \int_0^{t_0} \frac{dt}{(c^2+a^2t^2)(1+t^2)}\\ &= \frac23\psi_0 -\frac13\sin\psi_0\cos\psi_0 + \frac{2a^2}{3b^2} \left(\frac{a}{c}\tan^{-1}\left(\frac{at_0}{c}\right) - \psi_0\right)\\ &= \frac{2a^3}{3b^2c}\tan^{-1}\left(\frac{ae}{dc}\right) - \frac{2c^2}{3b^2}\cos^{-1}\left(\frac{d}{b}\right) - \frac{de}{3b^2} \end{align} $$ Substitute this into $(*2)$, we find Vol is equal to $$ \frac{4\pi a^3}{3} + \frac{8}{3}\left(2b^3 - 3b^2d + d^3 + cde\right) - \frac{16a^3}{3}\tan^{-1}\left(\frac{ae}{dc}\right) + \frac{16c^3}{3}\cos^{-1}\left(\frac{d}{b}\right) $$

Notice

  • when $b > \frac{a}{\sqrt{2}}$, $e = c$,
  • when $b > \frac{a}{\sqrt{2}}$, $d = 0$

We can use this to simplify above expression and get

$$ \bbox[8pt,border:1px solid blue]{ \text{Vol} = \frac{16}{3} \times \begin{cases} \displaystyle\;a^3\left(\tan^{-1}\left(\frac{d}{a}\right) - \frac{\pi}{4}\right) + b^2(b-d) + c^3\cos^{-1}\left(\frac{d}{b}\right), & b > \frac{a}{\sqrt{2}}\\ \\ \displaystyle\;-\frac{\pi}{4}a^3 + b^3 + \frac{\pi}{2} c^3,& b < \frac{a}{\sqrt{2}} \end{cases} } $$

The formula for $b < \frac{a}{\sqrt{2}}$ has a simple geometric interpretation. We can rewrite it as $$\text{Vol}_{small} = \underbrace{\frac{4\pi}{3}a^3}_{I} - \underbrace{2 \times \frac{4\pi}{3}(a^3 - c^3)}_{II} + \underbrace{\frac{16}{3} b^3}_{III}$$

One can show that if we drill a single cylindrical hole of radius $b$ from a sphere of radius $a$. The volume of remaining sphere is $\frac{4\pi}{3}c^3$. This means the volume of the cylinder removed is $\frac{4\pi}{3} (a^3 - c^3)$. Now, let us consider the intersection of two such cylinders, one aligned along the $x$-axis, the other along the $y$-axis. If we intersect this intersection by a plane with $|z| < b$, we will obtain a square of side $2\sqrt{b^2 - z}$. This means the volume of the intersection of two such cylinders is given by $$\int_{-b}^b 4(b^2 - z^2) dz = \frac{16}{3}b^3.$$

Compare this with $(*4)$, we find $\text{Vol}_{small}$ can be calculated as:

  • start with $I$, the volume of the sphere.
  • subtract $II$, the volume of the 2 cylinder removed.
  • add back $III$, the volume of the intersection of the two cylinders which has been over-subtracted in previous step.

This is the inclusion-exclusion mentioned in Christian-Blatter's answer.

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This answer combines the ideas of the two previous answers:

It implements Christian Blatter's suggestion, and makes extensive use of Achille Hui's techniques.


Let $V$ be the volume of the sphere with 2 holes, let $V_{H}$ be the volume of each of the cylindrical holes, and let $V_{I}$ be the volume of the intersection of the sphere and the two cylindrical holes.

Let $c=\sqrt{a^2-b^2}$, and let $d=\sqrt{2b^2-a^2}$ when $a\le\sqrt{2}b$, so $c^2=b^2-d^2$ in this case.

Then $V_{H}=\displaystyle2\int_0^{b}2\pi x\sqrt{a^2-x^2}dx=2\pi\int_{c^2}^{a^2}\sqrt{u}du=\frac{4}{3}\pi(a^3-c^3)$ using the shell method,

so $\hspace{.9 in}\displaystyle \color{red}{V=\frac{4}{3}\pi a^3-2V_{H}+V_{I}=\frac{8}{3}\pi c^3-\frac{4}{3}\pi a^3+V_{I}}.$


$\textbf{I)}$ If $a\ge\sqrt{2}b$, the cross-sections of the intersection perpendicular to the z-axis are the squares defined by

$\;\;\;\;|x|\le\sqrt{b^2-z^2},\;\; |y|\le\sqrt{b^2-z^2}\;\;$; so in this case

$\displaystyle V_{I}=\int_{-b}^{b}\left(2\sqrt{b^2-z^2}\right)^2dz=8\int_0^{b}(b^2-z^2)dz=\frac{16}{3}b^3$ and

$\hspace{1 in}\displaystyle \color{blue}{V=\frac{8}{3}\pi c^3-\frac{4}{3}\pi a^3+\frac{16}{3}b^3=\frac{4}{3}\left(2\pi c^3-\pi a^3+4b^3\right)}.$


$\textbf{II)}$ If $b<a\le\sqrt{2}b$, the cross-sections of the intersection perpendicular to the z-axis are

the squares defined by $\;\;\;|x|\le\sqrt{b^2-z^2},\;\; |y|\le\sqrt{b^2-z^2}\;\;\;\;$ if $|z|\ge d$,

and the portions of these squares inside the circle $x^2+y^2=a^2-z^2\;\;\;\;$if $|z|<d$.

Dividing these squares into 8 equal pieces by drawing the diagonals, we get

$V_{I}=\displaystyle 16\int_0^{d}\left(\frac{1}{2}(a^2-z^2)\left[\frac{\pi}{4}-\cot^{-1}\frac{\sqrt{b^2-z^2}}{c}\right]+\frac{1}{2}c\sqrt{b^2-z^2}\right)dz+8\int_{d}^{b}(b^2-z^2)dz$

$\displaystyle\;\;\;\;=8\int_0^{d}\frac{1}{2}\left(a^2-z^2\right)\left[\frac{\pi}{4}-\cot^{-1}\frac{\sqrt{b^2-z^2}}{c}\right]dz+8c\int_0^{d}\sqrt{b^2-z^2}dz+8\int_d^{b}(b^2-z^2)dz$

(where the first integral corresponds to a circular sector and the second corresponds to a triangle).

$\textbf{A)}$ $\;\;\displaystyle8\int_d^{b}(b^2-z^2)dz=8\left[b^{2}z-\frac{z^3}{3}\right]_d^{b}=8\left(\frac{2}{3}b^3-b^{2}d+\frac{1}{3}d^3\right)=\frac{8}{3}(2b^3-3b^{2}d+d^3)$

$\textbf{B)}\;\;\;$ Using $z=b\sin\theta, \;\;dz=b\cos\theta d\theta$ and $\gamma=\sin^{-1}\frac{d}{b}$,

$\displaystyle8c\int_0^{d}\sqrt{b^2-z^2}dz=8c\int_0^{\gamma}b^2\cos^{2}\theta d \theta=8b^2c\int_0^{\gamma}\frac{1}{2}(1+\cos2\theta)d\theta=4b^2c\left[\theta+\sin\theta\cos\theta\right]_0^{\gamma}$

$\displaystyle\hspace{1.4 in}=4b^{2}c\left(\sin^{-1}\frac{d}{b}+\frac{cd}{b^2}\right)=4b^{2}c\sin^{-1}\frac{d}{b}+4c^{2}d$.

$\textbf{C)}\;\;\;$Using integration by parts with $\displaystyle u=\frac{\pi}{4}-\cot^{-1}\frac{\sqrt{b^2-z^2}}{c}, \;\;dv=(a^2-z^2)dz$,

$\displaystyle\;\;\;8\int_0^{d}\frac{1}{2}\left(a^2-z^2\right)\left[\frac{\pi}{4}-\cot^{-1}\frac{\sqrt{b^2-z^2}}{c}\right]dz=$

$\displaystyle8\left[\left(\frac{\pi}{4}-\cot^{-1}\frac{\sqrt{b^2-z^2}}{c}\right)\left(a^2z-\frac{z^3}{3}\right)\right]_0^d+8\int_0^d\left(a^2z-\frac{z^3}{3}\right)\left(\frac{cz}{a^2-z^2}\frac{1}{\sqrt{b^2-z^2}}\right)dz$

$\displaystyle=\frac{8c}{3}\int_0^d\frac{3a^{2}z^2-z^4}{a^2-z^2}\frac{1}{\sqrt{b^2-z^2}}dz=\frac{8c}{3}\int_0^d\left(z^2-2a^2+\frac{2a^4}{a^2-z^2}\right)\frac{1}{\sqrt{b^2-z^2}}dz$.

Using $z=b\sin\theta, \;\;dz=b\cos\theta d\theta$ and $\gamma=\sin^{-1}\frac{d}{b}$, we obtain

$\displaystyle\frac{8c}{3}\int_0^{\gamma}\left(b^2\sin^2\theta-2a^2+\frac{2a^4}{a^2-b^2\sin^2\theta}\right)d\theta$

$=\displaystyle\frac{8c}{3}\left[\frac{b^2}{2}(\theta-\sin\theta\cos\theta)-2a^2\theta\right]_0^{\gamma}+\frac{16a^4c}{3}\int_0^{\gamma}\frac{\sec^2\theta}{c^2\tan^2\theta+a^2}d\theta$

$=\displaystyle\frac{4}{3}b^2c\sin^{-1}\frac{d}{b}-\frac{4}{3}c^2d-\frac{16}{3}a^2c\sin^{-1}\frac{d}{b}+\frac{16}{3}a^4c\left[\frac{1}{ac}\tan^{-1}\left(\frac{c\tan\theta}{a}\right)\right]_0^{\gamma}$

$\displaystyle=\frac{4}{3}b^2c\sin^{-1}\frac{d}{b}-\frac{4}{3}c^2d-\frac{16}{3}a^2c\sin^{-1}\frac{d}{b}+\frac{16}{3}a^3\tan^{-1}\frac{d}{a}$.

Then $\displaystyle V_I=\left[\left(\frac{4}{3}b^2c-\frac{16}{3}a^2c\right)\sin^{-1}\frac{d}{b}-\frac{4}{3}c^2d+\frac{16}{3}a^3\tan^{-1}\frac{d}{a}\right]+\left[4c^2d+4b^2c\sin^{-1}\frac{d}{b}\right]+\;\;\;\;\;\;\left[\frac{8}{3}(2b^3-3b^2d+d^3)\right]=-\frac{16}{3}c^3\sin^{-1}\frac{d}{b}+\frac{16}{3}a^3\tan^{-1}\frac{d}{a}+\frac{8}{3}(c^2d+2b^3-3b^2d+d^3),$


so $V=\displaystyle\frac{8}{3}\pi c^3-\frac{4}{3}\pi a^3-\frac{16}{3}c^3\left(\frac{\pi}{2}-\cos^{-1}\frac{d}{b}\right)+\frac{16}{3}a^3\tan^{-1}\frac{d}{a}+\frac{8}{3}\left(2b^2(b-d)\right)$

$\hspace{1 in}=\displaystyle\color{blue}{-\frac{4}{3}\pi a^3+\frac{16}{3}c^3\cos^{-1}\frac{d}{b}+\frac{16}{3}a^3\tan^{-1}\frac{d}{a}+\frac{16}{3}b^2(b-d)}.$

$\hspace{1 in}=\displaystyle\color{blue}{\frac{16}{3}\left(c^3\cos^{-1}\frac{d}{b}+a^3\tan^{-1}\frac{d}{a}-\frac{\pi}{4}a^3+b^2(b-d)\right)}.$

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  • $\begingroup$ wow, thanks for the bounty! $\endgroup$ – achille hui Mar 6 '15 at 16:15
  • $\begingroup$ You're welcome -- it was a very nice solution! $\endgroup$ – user84413 Mar 6 '15 at 17:08
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(Edit: This answer does not cover the case of large holes; see the comments by coffeemath.)

A hint:

If $b<{a\over\sqrt{2}}$ then the intersection $B$ of the two cylinders is completely in the interior of the sphere. In this case you can proceed as follows:

Do the problem with $1$ hole, then compute the volume of $B$, and use inclusion-exclusion.

For the computation of ${\rm vol}(B)$ consider $$B_1:=\{(x,y,z)\>|\>0\leq y\leq x, \ x^2+z^2\leq b^2\}\ ,$$ whose volume is ${1\over16}$ of the volume of $B$.

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  • 1
    $\begingroup$ If the sphere radius is $a=1$ and the drilled hole radius is $b$ then in case $b>1/\sqrt{2}$ the intersection of the cylinders (each viewed as extended), which you call $B$, is not entirely inside the sphere. One can take the equations of those cylinders as $x^2+y^2=a^2,\ x^2+z^2=a^2,$ and the intersection then contains the point $(0,a,a)$ lying outside the unit sphere when $2a^2>1.$ $\endgroup$ – coffeemath Jan 25 '15 at 14:02
  • $\begingroup$ @coffeemath: You are right. The answer will be deleted in due course. $\endgroup$ – Christian Blatter Jan 25 '15 at 14:54
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    $\begingroup$ IMO the answer should not be deleted, only the restriction added about the cylinder radius not being too large in comparison with the sphere radius. The problem would still have interest, since one might reasonably expect the drilled holes weren't so wide as to make the intersection of cylinders go outside the sphere. (+1 for now, in case you don't end up deleting) $\endgroup$ – coffeemath Jan 25 '15 at 15:29

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