Napkin ring problem
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geometry Geometry (; ) is, with arithmetic, one of the oldest branches of mathematics. It is concerned with properties of space such as the distance, shape, size, and relative position of figures. A mathematician who works in the field of geometry is ...
, the napkin-ring problem involves finding the volume of a "band" of specified height around a
sphere A sphere () is a geometrical object that is a three-dimensional analogue to a two-dimensional circle. A sphere is the set of points that are all at the same distance from a given point in three-dimensional space.. That given point is th ...
, i.e. the part that remains after a hole in the shape of a circular cylinder is drilled through the center of the sphere. It is a counterintuitive fact that this volume does not depend on the original sphere's
radius In classical geometry, a radius ( : radii) of a circle or sphere is any of the line segments from its center to its perimeter, and in more modern usage, it is also their length. The name comes from the latin ''radius'', meaning ray but also the ...
but only on the resulting band's height. The problem is so called because after removing a cylinder from the sphere, the remaining band resembles the shape of a napkin ring.


Statement

Suppose that the axis of a
right circular cylinder A cylinder (from ) has traditionally been a three-dimensional solid, one of the most basic of curvilinear geometric shapes. In elementary geometry, it is considered a prism with a circle as its base. A cylinder may also be defined as an infin ...
passes through the center of a sphere of radius R and that h represents the height (defined as the distance in a direction
parallel Parallel is a geometric term of location which may refer to: Computing * Parallel algorithm * Parallel computing * Parallel metaheuristic * Parallel (software), a UNIX utility for running programs in parallel * Parallel Sysplex, a cluster of ...
to the axis) of the part of the cylinder that is inside the sphere. The "band" is the part of the sphere that is outside the cylinder. The volume of the band depends on h but not on R: V=\frac. As the radius R of the sphere shrinks, the diameter of the cylinder must also shrink in order that h can remain the same. The band gets thicker, and this would increase its volume. But it also gets shorter in circumference, and this would decrease its volume. The two effects exactly cancel each other out. In the extreme case of the smallest possible sphere, the cylinder vanishes (its radius becomes zero) and the height h equals the diameter of the sphere. In this case the volume of the band is the volume of the whole sphere, which matches the formula given above. An early study of this problem was written by 17th-century Japanese mathematician Seki Kōwa. According to , Seki called this solid an arc-ring, or in
Japanese Japanese may refer to: * Something from or related to Japan, an island country in East Asia * Japanese language, spoken mainly in Japan * Japanese people, the ethnic group that identifies with Japan through ancestry or culture ** Japanese diaspor ...
''kokan'' or ''kokwan''.


Proof

Suppose the radius of the sphere is R and the length of the cylinder (or the tunnel) is h. By the Pythagorean theorem, the radius of the cylinder is \sqrt,\qquad\qquad(1) and the radius of the horizontal cross-section of the sphere at height y above the "equator" is \sqrt.\qquad\qquad(2) The cross-section of the band with the plane at height y is the region inside the larger circle of radius given by (2) and outside the smaller circle of radius given by (1). The cross-section's area is therefore the area of the larger circle minus the area of the smaller circle: \begin & \quad \pi(\text)^2 - \pi(\text)^2 \\ & = \pi\left(\sqrt\right)^2 - \pi\left(\sqrt\,\right)^2 = \pi\left(\left(\frac\right)^2 - y^2\right). \end The radius ''R'' does not appear in the last quantity. Therefore, the area of the horizontal cross-section at height y does not depend on R, as long as y\le\tfrac\le R. The volume of the band is : \int_^ (\texty) \, dy, and that does not depend on R. This is an application of Cavalieri's principle: volumes with equal-sized corresponding cross-sections are equal. Indeed, the area of the cross-section is the same as that of the corresponding cross-section of a sphere of radius h/2, which has volume \frac\pi\left(\frac\right)^3 = \frac.


See also

*
Visual calculus Visual calculus, invented by Mamikon Mnatsakanian (known as Mamikon), is an approach to solving a variety of integral calculus problems. Many problems that would otherwise seem quite difficult yield to the method with hardly a line of calculatio ...
, an intuitive way to solve this type of problem, originally applied to finding the area of an annulus, given only its chord length * String girdling Earth, another problem where the radius of a sphere or circle is counter-intuitively irrelevant


References

* * * * Problem 132 asks for the volume of a sphere with a cylindrical hole drilled through it, but does not note the invariance of the problem under changes of radius. *. Levi argues that the volume depends only on the height of the hole based on the fact that the ring can be swept out by a half-disk with the height as its diameter. *. Reprint of 1935 edition. A problem on page 101 describes the shape formed by a sphere with a cylinder removed as a "napkin ring" and asks for a proof that the volume is the same as that of a sphere with diameter equal to the length of the hole. *. Reprint of 1954 edition. *. Republished by Dover, 2004, . Smith and Mikami discuss the napkin ring problem in the context of two manuscripts of Seki on the mensuration of solids, ''Kyuseki'' and ''Kyuketsu Hengyo So''.


External links

* {{mathworld, title=Spherical Ring, urlname=SphericalRing Volume Japanese mathematics Recreational mathematics Articles containing proofs Mathematical problems