Napkin Ring Problem

## **Problem**

Figure 1 shows an animated central cross-section of a sphere of radius

r

{\displaystyle r}

 through which a centrally placed cylinder of radius

a

{\displaystyle a}

 has been cored out (drilled out and the material removed). The remaining shape is called a **napkin ring**. Determine the volume of the napkin ring.

## **Solution**

### **Method 1: Washer method**

Consider the diagram of the cross-section of the napkin ring (Figure 2). Let the radius of the sphere be

r

{\displaystyle r}

. Let the radius of the cylindrical hole be

a

{\displaystyle a}

, and half the height of the cylindrical hole be

h

{\displaystyle h}

. Note that

a
2

=

r
2

−

h
2

{\displaystyle a^2 = r^2-h^2 }

.

The volume of the napkin ring

V

{\displaystyle V}

 is equal to the volume of the sphere minus the volume of the cylinder and two spherical caps.

# V

V

s
p
h
e
r
e

−

V

c
y
l
i
n
d
e
r

−
2

V

s
p
h
e
r
i
c
a
l

c
a
p

{\displaystyle V = {V}\_{sphere} - {V}\_{cylinder} - 2 {V}\_{spherical \; cap} }

The volume of the sphere and the cylinder are well known:

V

s
p
h
e
r
e

=

4
π

3

r

3

{\displaystyle {V}\_{sphere} = \frac{4\pi}{3} {r}^{3} }

V

c
y
l
i
n
d
e
r

=
2
π
h

a

2

.

{\displaystyle {V}\_{cylinder} = 2\pi h {a}^{2}. }

Use integration to compute the volume of a spherical cap:

V

s
p
h
e
r
i
c
a
l

c
a
p

=

∫

h

r

π

(

r

2

−

y

2

)

d
y

=
π

[

r
2

y
−

y
3

3

h

r

]

=
π

(

2

r

3

3

−

r

2

h

h

3

3

)

.

{\displaystyle
\begin{align}
{V}\_{spherical \; cap} &= \int\_{h}^{r} \pi \left({r}^{2} - {y}^{2} \right) dy \\[5 pt]
& = \pi \left[r^2y - \frac{y^3}{3}\Biggr|\_{h}^{r}\right] \\[5 pt]
&= \pi \left(\frac{2{r}^{3}}{3} - {r}^{2} h + \frac{{h}^{3}}{3} \right).
\end{align} }

Hence, the volume of the napkin ring is

V

=

4
π

3

r

3

−
2
π
h

a

2

−
2
π

(

2

r

3

3

−

r

2

h

h

3

3

)

=

4
π

3

r

3

−
2
π
h

a

2

−

4
π

3

r

3

2
π

r

2

h
−

2
π

h

3

3

=
−
2
π
h

a

2

2
π

r

2

h
−

2
π

h

3

3

.

{\displaystyle
\begin{align}
V &= \frac{4\pi}{3} {r}^{3} - 2\pi h {a}^{2} - 2 \pi \left(\frac{2{r}^{3}}{3} - {r}^{2} h + \frac{{h}^{3}}{3} \right) \\[5 pt]
&= \frac{4\pi}{3} {r}^{3} - 2\pi h {a}^{2} - \frac{4\pi}{3}{r}^{3} + 2\pi {r}^{2} h - \frac{2\pi{h}^{3}}{3} \\[5 pt]
&= - 2\pi h {a}^{2} + 2\pi {r}^{2} h - \frac{2\pi{h}^{3}}{3}.
\end{align} }

Since

a
2

=

r
2

−

h
2

{\displaystyle a^2 = r^2-h^2 }

, we get

V

=
−
2
π
h

(

r

2

−

h

2

)

2
π

r

2

h
−

2
π

h

3

3

=
2
π

h

3

−

2
π

h

3

3

=

4
π

3

h

3

.

{\displaystyle \begin{align}
V &= - 2\pi h \left({r}^{2} - {h}^{2}\right) + 2\pi {r}^{2} h - \frac{2\pi{h}^{3}}{3} \\[5 pt]
&= 2\pi {h}^{3} - \frac{2\pi{h}^{3}}{3} \\[5 pt]
&= \frac{4\pi}{3} {h}^{3}.
\end{align} }

Expressing the volume of the napkin ring in terms of the height of the cylindrical hole

# H
2
h

{\displaystyle H = 2h }

 yields:

# V

π
6

H

3

.

{\displaystyle V = \frac{\pi}{6} {H}^{3}. }

### **Method 2: Shell method**

The shell method is much faster for this problem. By revolving the function

# y

r
2

−

x
2

{\displaystyle y = \sqrt{r^2-x^2}}

around the y-axis, and applying the shell method formula, we get the upper half of the napkin ring. Doubling this gives the total volume:

V

=
2

∫

a

r

2
π
x

r
2

−

x
2

d
x

=
4
π

[

−

1
3

(

r
2

−

x
2

)

3

/

2

a

r

]

=

4
π

3

(

r
2

−

a
2

)

3

/

2

.

{\displaystyle \begin{align}
V &= 2\int\_{a}^{r} 2\pi x \sqrt{r^2 - x^2} dx \\[5 pt]
&= 4\pi \left[-\frac{1}{3} (r^2 - x^2)^{3/2}\Biggr|\_{a}^{r}\right] \\[5 pt]
&=\frac{4\pi}{3} (r^2 - a^2)^{3/2}.
\end{align}}

Since

a
2

=

r
2

−

h
2

{\displaystyle a^2 = r^2-h^2 }

, we get

V

=

4
π

3

(

r
2

−
(

r
2

−

h
2

)

)

3

/

2

=

4
π

3

(

h
2

)

3

/

2

=

4
π

3

h

3

.

{\displaystyle \begin{align}
V &= \frac{4\pi}{3} (r^2 - (r^2-h^2))^{3/2} \\[5 pt]
&= \frac{4\pi}{3} (h^2)^{3/2} \\[5 pt]
&= \frac{4\pi}{3} {h}^{3}.
\end{align} }

Expressing the volume of the napkin ring in terms of the height of the cylindrical hole

# H
2
h

{\displaystyle H = 2h }

 yields:

# V

π
6

H

3

.

{\displaystyle V = \frac{\pi}{6} {H}^{3}. }

Who would have guessed that this volume is independent of the radius of the sphere! This means that if you core out any sphere of any size so that the remaining napkin rings have the same height, those napkin rings will also have the same volume! This looks unbelievable at first, but is in fact true!

## **Historical Remark**

The napkin ring problem dates back to Edo Japan. Seki Kowa (1642 - 1708 AD), the leading Japanese mathematician at the time, was the first person to have solved this problem using a form of integral calculus called *Enri*. Seki Kowa called the shape *kokan* (弧環, lit. "arc ring").