Chapter 12.3, Problem 83E

Geometry A circular disk of radius R is cut out of paper, as shown in figure (a). Two disks of radius $\frac{1}{2}R$ are cut out of paper and placed on top of the first disk, as in figure (b), and then four disks of radius $\frac{1}{4}R$ are placed on these two disks, as in figure (c). Assuming that this process can be repeated indefinitely, find the total area of all the disks.

To determine

To find: The total area of all disks which radiuses are in geometric series.

Answer to Problem 83E

The total area of all disks is $2\pi R^2$.

Explanation of Solution

Formula used:

The formula of sum of infinite geometric series when common factor is smaller than 1 is given by, $S=\frac{a}{1-r}$,where S is the sum of the geometric series, a is the first term of the series and r is the common factor of the series.

Given:

A circular disc is cut out of paper with radius R.

Two disks of radius $\frac{R}{2}$ are cut out of paper and placed on top of the disk of radius R.

Four disks of radius $\frac{R}{4}$ are placed on these two.

This process is repeated indefinitely.

Calculation:

The area of a circular disk is,

$\text{Area}=\pi \cdot {\left(\text{radius}\right)}^2$

The area of circular disk of radius R is, $A_1=\pi \cdot R^2$

The total area of three circular disks is sum of area of disk of radius R and area of two circular disks of radius $\frac{R}{2}$.

$\begin{array}{c}{A}_2=\pi \cdot R^2+2\cdot \pi \cdot {\left(\frac{R}{2}\right)}^2\\ =\pi \cdot R^2+2\cdot \pi \cdot \frac{R^2}{4}\\ =\pi \cdot R^2+\pi \cdot \frac{R^2}{2}\end{array}$

The total area of seven circular disks is sum of area of disk of radius R and area of two circular disks of radius $\frac{R}{2}$ and area of four circular disks of radius $\frac{R}{4}$,

$\begin{array}{c}{A}_3=\pi \cdot R^2+\pi \cdot \frac{R^2}{2}+4\cdot \pi \cdot {\left(\frac{R}{4}\right)}^2\\ =\pi \cdot R^2+\pi \cdot \frac{R^2}{2}+4\cdot \pi \cdot \frac{R^2}{16}\\ =\pi \cdot R^2+\pi \cdot \frac{R^2}{2}+\pi \cdot \frac{R^2}{4}\end{array}$

So, as this process is repeated indefinitely. So the total area of all disks is,

$\begin{array}{c}A=\pi \cdot R^2+\pi \cdot \frac{R^2}{2}+\pi \cdot \frac{R^2}{4}+\cdots \\ =\pi \cdot R^2\left[1+\frac{1}{2}+\frac{1}{4}+\cdots \right]\end{array}$

The series from above equation $\left[1+\frac{1}{2}+\frac{1}{4}+\cdots \right]$ is a geometric series because it increases by a common factor of $\frac{1}{2}$.

The first term of series is 1 and the common factor is $\frac{1}{2}$, so substitute 1 for a and $\frac{1}{2}$ for r in the formula $S=\frac{a}{1-r}$,

$\begin{array}{c}S=\frac{1}{1-\frac{1}{2}}\\ =\frac{1}{\frac{1}{2}}\\ =2\end{array}$

So, the sum of the infinite series $\left[1+\frac{1}{2}+\frac{1}{4}+\cdots \right]$ is 2.

So, the total area of all the disks is $A=\pi R^2\left(2\right)=2\pi R^2$.

Thus, the total area of all the disks is $2\pi R^2$.