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A doughnut has been partially eaten by a meticulous person so that the portion remaining is given by rotating the half-circular region shown above about the y axis. What fraction of the original doughnut remains, assuming that the original doughnut was the volume of revolution of the full circle? Give your answer in exact form in terms of r and R.

My solution: dropbox.com/s/a816w2rgmkq223r/2013-02-12%2015.19.40.jpg

So clearly, volume of a full torus over that of above is $1\over r$

Bu it's wrong? Where did I go wrong?

EDIT: Corrected some mistakes, still getting the wrong answer:

https://www.dropbox.com/s/1sf72pxcm6dk8pp/2013-02-12%2018.45.46.jpg https://www.dropbox.com/s/wlko0w958plo1pa/2013-02-12%2018.46.22.jpg

EDIT 2: I made a small mistake in the above solution as well. It should be $4 \pi $ and not $2 \pi$. Correcting that, I got the answer:

$$\left(\frac12\right) - \left(8r\over3 \pi R\right)$$

But this is also wrong! What should I do now?

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    $\begingroup$ One suggestion to go hunting: because you're trying to find a volume, your units at every step should be of the form $(\text{length})^3$, but your final result is in units of $(\text{length})^2$. Can you find the point where it shifts over? $\endgroup$ Feb 12, 2013 at 23:25
  • $\begingroup$ Your first integral became zero. Why? It is from $-r$ to $0$ , not to $r$. $\endgroup$
    – Maesumi
    Feb 12, 2013 at 23:35
  • $\begingroup$ Oh!! Right, my bad. $\endgroup$
    – None
    Feb 12, 2013 at 23:44
  • $\begingroup$ The ratio of $1/r$ needs to be checked. 1) the dimension is nnot correct 2) the denominator of the last fraction is $2$ not $r$ 3) your calculation shows $V=\pi^2 r^2 R$ and volume of torus is $2\pi^2r^2R$ so if your calculation was right the ratio would be $1/2$. $\endgroup$
    – Maesumi
    Feb 12, 2013 at 23:44
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    $\begingroup$ No, the logical answer will be less than 1/2. The outer semicircle, the eaten portion, has higher volume. If you know Pappus' theorem you can bypass the integrals or have a way of double checking them. $\endgroup$
    – Maesumi
    Feb 13, 2013 at 0:12

2 Answers 2

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Guldin's rule, also called Pappus' centroid theorem, says the following: When you rotate a shape $S$ around an exterior axis the volume $V$ of the solid generated in this way is equal to the area of the shape times the circumference of the circle described by the centroid of $S$.

It follows that the volume of the full torus is given by $V_{\rm tot}=\pi r^2\cdot 2\pi R$.

In order to compute the remaining volume we need the centroid $(0,\rho)$ of the half circle $\{(x,y)\>|\> x^2+y^2\leq r^2,\ y\geq0\}$. It is determined by the moment equation $${1\over2}\>\pi r^2\ \rho=\int_{-r}^r{\sqrt{r^2-x^2}\over2}\>\sqrt{r^2-x^2}\ dx={2\over3}r^3\ ,$$ from which we get ${\displaystyle \rho={4r\over 3\pi}}$.

Since by Guldin's rule the remaining volume is given by $V_{\rm rem}={1\over2}\pi r^2\cdot 2\pi(R-\rho)$ we now obtain $${V_{\rm rem}\over V_{\rm tot}}={1\over2}-{2\over3\pi}\>{r\over R}\ .$$

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This problem popped back up in the queue, so I had a look. (I'm assuming this was never completely resolved, to judge from OP's final comment on 13 Feb, 16:10.) In the first page image posted, the construction of the integral on line 5 is largely correct:

$$\int_{-r}^0 \ t \sqrt{r^2 - t^2} \ + \ R \sqrt{r^2 - t^2} \ \ dt \ . $$

But None has neglected the fact that since the equation for the circle is $ \ (x - R)^2 + y^2 = r^2 \ $ , which does not describe a function, the total "height" of the "vertical slices" runs from $ \ -\sqrt{r^2 - (x - R)^2} \ $ to $ \ +\sqrt{r^2 - (x - R)^2} \ $. The total volume integral for this "inner semi-torus" should thus be

$$V \ = \ 2 \cdot 2 \pi\int_{-r}^0 \ t \sqrt{r^2 - t^2} \ + \ R \sqrt{r^2 - t^2} \ \ dt \ , $$

after the first substitution. [EDIT: Now it's clear that None's EDIT 2 refers to this.]

The other issue is that the calculation of the first portion of the volume integral goes slightly astray. For the substitution $ \ u = r^2 - t^2 \ , \ du = -2t \ dt \ $ , the transformed limits are found correctly as $ \ u = 0 \ $ to $ \ u = r^2 \ $ . However, the replacement for $ \ t \ dt \ $ should have been $ \ -\frac{1}{2}\ du \ $ , making the result

$$\int_{-r}^0 \ t \sqrt{r^2 - t^2} \ dt \ \rightarrow \ -\frac{1}{2}\int_0^{r^2} \sqrt{u} \ du \ = \ -\frac{1}{3}r^3 ; $$

the other portion,

$$\int_{-r}^0 \ R \sqrt{r^2 - t^2} \ \ dt \ = \ \frac{\pi}{4} R \ r^2 $$

is determined correctly. Hence, the volume of this solid of revolution is

$$V \ = \ 4 \pi \cdot (\frac{\pi}{4} R \ r^2 \ - \ \frac{1}{3}r^3 ) \ = \ \pi^2 R \ r^2 \ - \ \frac{4 \pi}{3}r^3 \ . $$

(Assumingly, since the problem theme concerns consuming donuts, the second term suggests the volume of a spherical "donut hole".) This result is confirmed by the application of Pappus' (second) centroid theorem: a semi-circle of area $ \ \frac{1}{2} \pi \ r^2 \ $ is revolved around a circle of radius $ \ R - \frac{4}{3 \pi}r \ ^{*} $ , generating a volume of

$$ \ 2 \pi \cdot (R - \frac{4}{3 \pi}r) \cdot (\frac{1}{2} \pi \ r^2) \ = \ \pi^2 R \ r^2 \ - \ \frac{4 \pi}{3}r^3 \ . $$

$^{*}$ As the centroid of a semi-circle is a familiar result, I shan't derive it here.

To answer the question in the problem statement then, the remaining fraction of the original donut is given by

$$\frac{\pi^2 R \ r^2 - \frac{4 \pi}{3}r^3}{2 \pi^2 R \ r^2 } \ = \ \frac{1}{2} - (\frac{2}{3 \pi} \cdot \frac{r}{R}) \ , $$

so more than half of the donut has been eaten, as discussed in some of the comments.

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