Chapter 9.1, Problem 51E

To determine

To find: To find the total area of all the semicircles.

Answer to Problem 51E

The total area of all the semicircles is $\frac{\pi }{2}$ .

Explanation of Solution

Given:

The area of semicircle is $\frac{1}{2^n}$ .

Calculation:

Number of semicircles in the first row is $2$ .

Total area occupied by the semicircle in the first row us $(2)\frac{1}{2}\pi {\left(\frac{1}{2}\right)}^2$ .

Total area occupied by the semicircle in the second row us ${(2)}^2\frac{1}{2}\pi {\left(\frac{1}{2^2}\right)}^2$

Total area occupied by the semicircle in the $n^{th}$ us ${(2)}^n\frac{1}{2}\pi {\left(\frac{1}{2}\right)}^2$

Sum of area of the entire semicircle is

  $\begin{array}{l}S={\displaystyle \sum _{n=1}^{\infty }{2}^n.\frac{1}{2}\pi {\left(\frac{1}{2^n}\right)}^2}\\ S={\displaystyle \sum _{n=1}^{\infty }\frac{1}{2}\pi {\left(\frac{1}{2}\right)}^n}\end{array}$

If the sequence of partial sum has a limit as $n\to \infty$ , the series converges otherwise the sequence is divergence. For series to be convergence the term should approach to zero. That is common ratio should be less than $1$ . That is $\left|r\right|<1$ , where $r$ is the common ratio.

Initial term is $\frac{\pi }{4}$ and common ratio is $\frac{1}{2}$ .

  $A=\frac{\frac{\pi }{4}}{1-\frac{1}{2}}=\frac{\pi }{2}$