Volumes of Revolutions : Lord of the Rings

Question

Question: The "Lord of the Rings" has a collection of solid gold rings for different-sizes fingers. The cross section of each ring is a segment of a circle radius $R$ as shown in the diagram below. All rings in the collection have the same width $w$.

The "Lord of the Rings" says to Gandaulf, "although the rings have different diameters, they all contain the same amount of gold."

Is this true?. Justify your answer.

Picture:

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I am stuck setting up the volume of revolution:

I get the circle as $x^2+y^2=R^2$ but I'm trouble locating the limits of the integration and any other things I have to do...

Answer 1

There's no need to integrate; the volume of the ring is the difference between two well-known volumes. The spherical segment with height $w$ and both radii $r$ has volume $\frac{\pi w}6\left(3r^2+3r^2+w^2\right)$, and the cylinder cut out of it has volume $\pi r^2w$. The difference is

$$ \frac{\pi w}6\left(3r^2+3r^2+w^2\right)-\pi r^2w=\frac{\pi w^3}6\;. $$

So the Lord of the Rings was right.

Answer 2

If we use the Disc method and symmetry, we get

$\displaystyle V=2\int_0^{w/2}\pi\left((R(y))^2-(r(y))^2\right)dy=2\pi\int_0^{w/2}\left((R^2-y^2)-\big(R^2-\left(\frac{w}{2}\right)^2\big)\right)dy$

$\displaystyle\hspace{.2 in}=2\pi\int_0^{w/2}\left(\frac{w^2}{4}-y^2\right)dy=2\pi\left[\frac{w^2}{4}\big(y\big)-\frac{y^3}{3}\right]_0^{w/2}=\frac{\pi w^3}{6}$

(using that $R(y)=x$ and $r(y)=r$)

Answer 3

Consider the shaded area, which is a circular segment.

Let $2\theta$=angle subtended by the arc of this segment.

Area of segment, $$A=\frac 12 R^2(2\theta-\sin 2\theta)$$

Distance of centroid of circular segment from centre of ring, $$s=\frac 43R \left(\frac {\sin^3\theta}{2\theta-\sin 2\theta}\right)$$ By Pappus' centroid theorem, volume of ring is given by cross-sectional area multiplied by length travelled by centroid, i.e. $$\begin{align} V&=A\cdot 2\pi s\\ &=\frac 43 \pi(R\sin\theta)^3 \\ &=\frac 43\pi \left(\frac w2\right)^3\\ &=\frac \pi 6 w^3\end{align}$$ which is constant, as $w$ is constant.