Q. 70 - Calculus

Question

Prove the Napkin Ring Theorem: If a napkin ring is made by removing a cylinder of height h from a sphere, then the volume of the resulting shape does not depend on the radius of the sphere! Look at the figures in Exercise 62 to see why this result is surprising. 

Step-by-Step Solution

Answer

A napkin ring's volume is only dependent on its height; it is not affected by its radius.

Step 1: Given information.

Consider the given information.

The cross section of a napkin ring is depicted in the schematics above.

Step 2: Calculation

The goal is to demonstrate that a sphere's radius has no bearing on the volume of the sphere produced by cutting away a cylinder with height h.

Given that the cross-section is symmetric about the x-axis, the volume of the napkin ring may be calculated by rotating the circular arc $AB$ about the y-axis and multiplying by 2.

The ring is of height h. 

Therefore, the height of the circular section shown above is$\frac{h}{2}$. The circular arc AB is part of the circle of radius r centered at the origin.

Therefore, the equation of the arc AB is given by,

$\begin{array}{rcl}{x}^2+y^2& =& r^2\\ y& =& \sqrt{r^2-x^2}\end{array}$

Find the intersection point of the circle $x^2+y^2=r^2$with the line$y=\frac{h}{2}$ to find the coordinates of point A as,

$\begin{array}{rcl}{x}^2+\frac{h^2}{4}& =& r^2\\ x^2& =& r^2-\frac{h^2}{4}\\ x& =& \sqrt{r^2-\frac{h^2}{4}}\end{array}$

Therefore, x varies from $\sqrt{r^2-\frac{h^2}{4}}$ to r.

Step 3: Further calculation

The volume of the napkin ring is calculated as,

$=4\pi {\int }_{\sqrt{r^2\frac{h^2}{4}}}x\sqrt{r^2-x^2}dx$

Substitute $r^2-x^2=t$ then,$-2xdx=dt$.

The limits are,

$\begin{array}{rcl}t& =& r^2-\left(r^2-\frac{h^2}{4}\right)\\ & =& \frac{h^2}{4}\\ t& =& r^2-r^2\\ & =& 0\end{array}$

Therefore, the integral $V=4\pi {\int }_{\sqrt{r^2-\frac{h^2}{4}}}^{2}x\sqrt{r^2-x^2}dx$ becomes,

$\begin{array}{rcl}V& =& -2\pi {\int }_{n^{4/4}}^0\sqrt{i}dt\\ & =& -\frac{4\pi }{3}{\left[f^{3/2}\right]}_{{\pi }^3/4}^0\\ & =& \frac{\pi h^3}{6}\end{array}$

Thus, A napkin ring's volume is only dependent on its height; it is not affected by its radius.