5
$\begingroup$

Mr. Hilbert was sitting alone at the dinner table of the Grand Hotel restaurant, waiting for his two friends, Mr. Euler and Mr. Langrange, to show up. It was 5 minutes past six already, but knowing the two of them, it wasn't too much of a surprise to Mr. Hilbert that their arrivals times were a bit variational.

Being slightly bored, Mr. Hilbert started to play with his napkin ring. At first, it looked just like a hollow sphere with a circular hole, but upon closer inspection, the hole seemed more like this:

Bird's eye view of the hole

where the hole is the shaded area, the intersection of four quarter-circles of radius $r$. And this hole was drilled straight through a sphere of diameter $r$:

How the hole is drilled through the napkin-ring sphere

"Hmm, I wonder how much volume is cut out of the original sphere in order to get this napkin ring?" wondered Mr. Hilbert.


Can you figure it out? Express your answer in terms of $r$.

$\endgroup$

closed as off-topic by xnor, Deusovi, Julian Rosen, f'', 2012rcampion Sep 23 '15 at 4:41

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question is off-topic as it appears to be a mathematics problem, as opposed to a mathematical puzzle. For more info, see "Are math-textbook-style problems on topic?" on meta." – xnor, Deusovi, Julian Rosen, f'', 2012rcampion
If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ $k.r^3$ for some $k$ between 0 and 1 $\endgroup$ – Dr Xorile Sep 22 '15 at 2:20
  • $\begingroup$ I did the first part of that question once before - if it helps anybody, the area of the cross-section is $\frac{\pi}{3} - \sqrt{3} + 1$ for $r = 1$. $\endgroup$ – Joe Z. Sep 22 '15 at 3:00
  • $\begingroup$ @Dr_Xorile That answer gets $\lceil 2k-1 \rceil$ upvotes from me $\endgroup$ – Hackiisan Sep 22 '15 at 3:01
  • $\begingroup$ I have a numerically well-behaved integral for $k\approx 0.280226$, but trying it symbolically, Mathematica spits it back out unevaluated. $\endgroup$ – 2012rcampion Sep 22 '15 at 4:33
  • 1
    $\begingroup$ Is this just a laborious integral or is there some nice trick? $\endgroup$ – xnor Sep 22 '15 at 23:06
2
$\begingroup$

Taking the solution by @Dr Xorile a step further, I get:

$$ V = kr^3 $$

where

$$ \begin{align} k = \int_0^{\frac{\sqrt{3}-1}{2}}&\left[\left(1-4 x^2\right) \tan ^{-1}\left(\frac{\sqrt{3-4 x (x+1)}-1}{\sqrt{4 x+2 \sqrt{3-4 x (x+1)}-3}}\right)\cdots\right. \\ &-\sqrt{4 x+2 \sqrt{3-4 x (x+1)}-3}\cdots \\ &\left.+\sqrt{(3-4 x (x+1)) \left(4 x+2 \sqrt{3-4 x (x+1)}-3\right)}\right] \, dx \\ &\approx 0.280\,225\,788\,122\,747\,024\,207\,247\,869\,453\,505\ldots \end{align} $$

$\endgroup$
  • $\begingroup$ Looks good! Can't say I've double checked it, but a tick for effort... $\endgroup$ – Dr Xorile Sep 22 '15 at 5:30
  • $\begingroup$ @Hackiisan Do you have a closed-form solution? $\endgroup$ – 2012rcampion Sep 22 '15 at 20:01
  • $\begingroup$ I do not, but I did it in a different way and the integral I have is much shorter. Our answers agree tho! $\endgroup$ – Hackiisan Sep 22 '15 at 20:28
  • $\begingroup$ Actually I'll just give you the green tick =) I'll post my answer later. $\endgroup$ – Hackiisan Sep 22 '15 at 20:35
0
$\begingroup$

Okay, I might be a bit rusty. But looking from the top, and considering just one of the quadrants (actually just the top half of one), you can integrate and get the following:

$\int^{\sqrt{3}-1}_0\int^{\sqrt{4-(x+1)^2}-1}_0\int^{\sqrt{1-x^2-y^2}}_0dzdydx$

I've set $r=2$, but I've also only considered an eighth of the volume, so the above gives the answer in units of $r$.

The first couple of integrations doesn't lead to

$\int^{\sqrt{3}-1}_0\left[y-yx^2-y^3/3\right]^{\sqrt{4-(x+1)^2}-1}_0dx$

Because I missed the sqrt in the innermost integral. Thanks for noticing

$\endgroup$
  • $\begingroup$ Shouldn't your upper limit on $z$ have a square root? $\endgroup$ – 2012rcampion Sep 22 '15 at 4:20
  • $\begingroup$ I think you're right. Grrr. That ruins things a little. $\endgroup$ – Dr Xorile Sep 22 '15 at 5:29
0
$\begingroup$

In my answer, the integral was performed in cylindrical coordinates. The tricky part is getting a simple $\rho(\theta)$, which can be calculated as shown in the following diagram:

enter image description here

This gives us: $$r^2=\rho^2 +d^2-2\rho d \cos{\left(\theta +\frac{\pi}{2}\right)}$$ $$\Rightarrow\frac{2\rho(\theta)}{r}=\sqrt{2}\left(\sqrt{1+\sin^2\theta}-\sin\theta\right)$$

Therefore:

$$\begin{align*} V_{\text{hole}}&=16\int^{\pi/2}_{\pi/4}{\int^0_{\rho(\theta)}{\int^\sqrt{(r/2)^2-\rho^2}_0}{\rho dz d\rho d\theta}} \\ \frac{V_{\text{hole}}}{r^3} &=\frac{\pi}{6}-\frac{2}{3}\int^{\pi/2}_{\pi/4}{\left[1-\left(\frac{2\rho(\theta)}{r}\right)^2\right]^{3/2} d\theta} \\ &= \frac{\pi}{6}-\frac{4\sqrt{2}}{3}\int^{\pi/2}_{\pi/4}{\left[\cos2\theta+2\sin\theta\sqrt{1+\sin^2\theta} - \frac{3}{2}\right]^{3/2} d\theta} \\ &\approx 0.280225788122747024207247869453505\cdots \end{align*}$$

$\endgroup$
  • 1
    $\begingroup$ I got to that integral and then gave up trying to evaluate it. How is this a puzzle and not just an integration exercise? $\endgroup$ – f'' Sep 23 '15 at 1:17
  • $\begingroup$ @f'' I'm sorry to hear that. As the title suggests, it grew out of a desire to see the napkin-ring puzzle generalized to other hole shapes. For good measure, I also included the link to the "a hollow sphere" puzzle in the question. $\endgroup$ – Hackiisan Sep 23 '15 at 6:47
  • $\begingroup$ I actually didn't expect puzzlers to brute-force through the Cartesian answer. The "aha" moments for me in this puzzle was 1) realizing that Cartesian sucks here, and cylindrical coordinates work better (even though the problem seemed intuitively suited for Cartesian), and 2) discovering an elegant way to represent $\rho$, which took a while for this humble author. $\endgroup$ – Hackiisan Sep 23 '15 at 6:53

Not the answer you're looking for? Browse other questions tagged or ask your own question.