Kvant Math Problem 67
The ring is the solid obtained from a sphere by drilling a cylindrical hole through its center.
Verified: no
Verdicts: SKIP + SKIP
Solve time: 1m34s
Source on kvant.digital
Problem
A jeweler has been commissioned to make a gold ring of width $h$, shaped as the solid bounded by the surface of a sphere with center $O$ and the surface of a cylinder of radius $r$, whose axis passes through the point $O$ (Fig. 1). The craftsman made such a ring but chose $r$ too small. How much gold will he need to add if $r$ must be increased by a factor of $k$ while keeping the width $h$ unchanged?
Insert illustration
Exploration
The ring is the solid obtained from a sphere by drilling a cylindrical hole through its center. This is the classical napkin-ring solid. Let the sphere have radius $R$ and the cylindrical hole have radius $r$. The width of the ring is given to be $h$.
A meridional cross-section through the axis is a circle of radius $R$ with a central strip of width $2r$ removed. The remaining section consists of two symmetric circular arcs. The distance between the two boundary planes of the ring equals
$$h=2\sqrt{R^{2}-r^{2}}.$$
Hence
$$R^{2}-r^{2}=\frac{h^{2}}{4}.$$
The crucial observation is that when $h$ is fixed, increasing $r$ forces $R$ to increase as well, according to
$$R^{2}=r^{2}+\frac{h^{2}}4.$$
The volume of a napkin ring is famous for depending only on $h$. Since the problem asks how much gold must be added when $r$ is changed while $h$ remains fixed, it is natural to suspect that the answer is zero.
Before accepting that, compute the volume directly. Using cylindrical coordinates, the ring volume is
$$V=2\pi\int_r^R \rho\sqrt{R^2-\rho^2},d\rho.$$
Evaluating,
$$V=\frac{4\pi}{3}\left(R^2-r^2\right)^{3/2}.$$
Substituting $R^2-r^2=h^2/4$ gives
$$V=\frac{\pi h^3}{6}.$$
Indeed the volume depends only on $h$. Thus changing $r$ while preserving $h$ leaves the volume unchanged, so no gold is needed.
The step most likely to hide an error is the volume computation. It must be carried out carefully and then expressed solely through $h$.
Problem Understanding
A sphere of radius $R$ is drilled by a cylindrical hole of radius $r$ whose axis passes through the center. The remaining solid is a ring-shaped body. Its width is fixed at $h$.
The original ring was made with hole radius $r$. The customer requires the hole radius to be increased to $kr$. During the modification, the width $h$ must remain unchanged. The question asks how much additional gold must be supplied.
This is a Type C problem, because a definite quantity must be determined.
The expected answer is that no gold is needed. The reason is that the volume of a napkin ring depends only on its width $h$, not on the radius of the cylindrical hole.
Proof Architecture
First, prove that for a napkin ring of width $h$, the sphere radius and hole radius satisfy $R^2-r^2=h^2/4$; this follows from the geometry of a diametral cross-section.
Second, derive a formula for the volume of the ring,
$$V=\frac{4\pi}{3}(R^2-r^2)^{3/2},$$
by integrating the areas of circular annuli perpendicular to the axis.
Third, substitute $R^2-r^2=h^2/4$ to obtain
$$V=\frac{\pi h^3}{6},$$
showing that the volume depends only on $h$.
Finally, compare the initial and modified rings. Since both have the same width $h$, they have the same volume, hence the same amount of gold.
The most delicate lemma is the derivation of the volume formula.
Solution
Let the original sphere have radius $R$ and the cylindrical hole have radius $r$.
Consider a plane through the axis of the cylinder. The cross-section of the sphere is a circle of radius $R$, and the cylindrical hole removes the central strip between the lines at distance $r$ from the center.
If $A$ and $B$ are the two points where the boundary of the hole meets the circle, then the width of the ring equals the length of the chord segment remaining along the axis:
$$h=2\sqrt{R^2-r^2}.$$
Hence
$$R^2-r^2=\frac{h^2}{4}.$$
To compute the volume, place the $z$-axis along the axis of the cylinder. For a fixed height $z$, the sphere contributes a circular cross-section of radius
$$\sqrt{R^2-z^2}.$$
Material is present only when this radius is at least $r$, namely for
$$|z|\le \sqrt{R^2-r^2}.$$
At height $z$, the cross-sectional area of the ring is
$$A(z) =\pi\bigl(R^2-z^2\bigr)-\pi r^2 =\pi\bigl(R^2-r^2-z^2\bigr).$$
Therefore
$$V =\int_{-\sqrt{R^2-r^2}}^{\sqrt{R^2-r^2}} \pi\bigl(R^2-r^2-z^2\bigr),dz.$$
Let
$$a=\sqrt{R^2-r^2}.$$
Then
$$V =\pi\int_{-a}^{a}(a^2-z^2),dz.$$
Since the integrand is even,
$$V =2\pi\int_0^a(a^2-z^2),dz.$$
Evaluating the integral,
$$V =2\pi\left[a^2z-\frac{z^3}{3}\right]_0^a =2\pi\left(a^3-\frac{a^3}{3}\right) =\frac{4\pi}{3}a^3.$$
Substituting $a=\sqrt{R^2-r^2}$ yields
$$V=\frac{4\pi}{3}(R^2-r^2)^{3/2}.$$
Using $R^2-r^2=h^2/4$,
$$V =\frac{4\pi}{3}\left(\frac{h^2}{4}\right)^{3/2} =\frac{4\pi}{3}\cdot\frac{h^3}{8} =\frac{\pi h^3}{6}.$$
Thus the volume of the ring depends only on the width $h$.
After the modification, the hole radius becomes $kr$. Let the new sphere radius be $R_1$. Because the width remains equal to $h$,
$$R_1^2-(kr)^2=\frac{h^2}{4}.$$
The new ring therefore has volume
$$V_1=\frac{\pi h^3}{6}.$$
The original ring also has volume
$$V_0=\frac{\pi h^3}{6}.$$
Hence
$$V_1=V_0.$$
The amount of gold contained in the ring is unchanged. Consequently, no additional gold is required.
$$\boxed{0}$$
Equality holds for every value of $k$, provided the width $h$ is kept fixed.
Verification of Key Steps
The relation between $h$, $R$, and $r$ comes directly from a right triangle in the diametral cross-section. Half the width is the distance from the line representing the cylinder wall to the sphere along the axis, namely $\sqrt{R^2-r^2}$. Multiplying by $2$ gives $h$. Any mistake in identifying this segment would produce an incorrect dependence on $r$.
For the volume integral, the cross-sectional area at height $z$ is not a disk but an annulus. Its outer radius is $\sqrt{R^2-z^2}$ and its inner radius is $r$. Thus
$$A(z)=\pi(R^2-z^2)-\pi r^2.$$
Omitting the inner disk would give the sphere volume instead of the ring volume.
Substituting $R^2-r^2=h^2/4$ after integration is essential. If one substitutes prematurely without expressing the limits correctly, the cancellation responsible for the independence from $r$ may be obscured.
Alternative Approaches
A shorter approach uses the classical napkin-ring theorem. Since
$$h=2\sqrt{R^2-r^2},$$
the ring has the same volume as a sphere of radius $h/2$. The sphere volume is
$$\frac{4\pi}{3}\left(\frac h2\right)^3 =\frac{\pi h^3}{6}.$$
Hence every napkin ring of width $h$ has the same volume, regardless of the hole radius. The original and modified rings therefore contain equal amounts of gold.
The integral proof is preferable because it derives the required volume formula directly from the geometry of the problem and makes the independence from $r$ completely explicit.