# Exercise 3.8Z: Circle (Ring) Area

We consider circles of different sizes:

• The radius  $r$  and the area  $A$  can be thought of as interdependent random variables.
• It is assumed that the radius is restricted to the area  $6 \le r \le 8$  .

In the sketch above,  the area in which such circles  $($all with center at coordinate origin$)$  can lie is marked in yellow.  Furthermore,  it can be assumed that the radius in this interval is uniformly distributed:

$$f_r(r)=\left\{ \begin{array}{*{4}{c}} 0.5 & \rm for\hspace{0.2cm}{\rm 6\le \it r \le \rm 8}, \\\rm 0 & \rm else. \end{array} \right.$$

From subtask  (5)  narrow circular rings with center radius  $r$  and width  $b$  are considered  $($lower sketch$)$:

• The area of such a circular ring is denoted by  $R$.
• The possible center radii  $r$  are again uniformly distributed between  $6$  and  $8$.
• The circular ring width is  $b = 0.1$.

Hints:

### Questions

1

Give the transformation characteristic curve  $A = g(r)$  analytically.  What is the minimum value of the random variable $A$?

 $A_\text{min} \ = \$

2

What is the maximum value of the random variable $A$?

 $A_\text{max} \ = \$

3

What value  $m_{A} = {\rm E}\big[A\big]$  results for the  "mean"  circular area?

 $m_{ A} \ = \$

4

Calculate the probability density function of the random variable $A$.  What is the probability that the area  $A> 150$ ?

 ${\rm Pr}(A > 150) \ = \$ $\ \%$

5

What is the PDF of the random variable  $R$  $($area of the circular rings according to the sketch below$)$?  What is its minimum value?  Let  $b = 0.1$.

 $R_\text{min} \ = \$

6

It continues to apply  $b = 0.1$.  What is the maximum value of the random variable  $R$?

 $R_\text{max} \ = \$

7

What is the expected value of the random variable  $R$  for  $b = 0.1$?

 ${\rm E}\big[R\big] \ = \$

### Solution

#### Solution

(1)  The equation of the circular area is at the same time the transformation characteristic:   $A = \pi \cdot r^2$.

• From this,  with  $r = 6$  for the minimum value:
$$A_\text{min} \hspace{0.15cm}\underline {= 113.09}.$$

(2)  Correspondingly,  with  $r = 8$  for the maximum value:

$$A_\text{max} \hspace{0.15cm}\underline {= 201.06}.$$

(3)  The simplest way to solve this problem is as follows:

$$m_{\rm A}={\rm E}\big[A\big]={\rm E}\big[g(r)\big]=\int_{ -\infty}^{+\infty}g(r)\cdot f_r(r) {\rm d}r.$$
• With  $g(r) = \pi \cdot r^2$  and  $f_r(r) = 1/2$  in the range of  $6$ ... $8$  obtains:
$$m_{\rm A}=\int_{\rm 6}^{\rm 8}1/2 \cdot\pi\cdot r^{\rm 2}\, {\rm d} \it r=\frac{\pi}{\rm 6}\cdot \rm ( 8^3-6^3) \hspace{0.15cm}\underline{=\rm 154.98}.$$

(4)  The PDF of the transformed random variable  $A$  is:

$$f_A(A)=\frac{f_r(r)}{|g\hspace{0.05cm}'(r)|}\Bigg|_{r=h(y) = \sqrt{A/ \pi }}.$$
• In the range between  $A_\text{min} {= 113.09}$  and  $A_\text{max} {= 201.06}$  then holds:
$$f_A(A)=\frac{\rm 1/2}{\rm 2\cdot \pi\cdot\it r}\Bigg|_{\it r=\sqrt{\it A/\rm \pi}}=\frac{\rm 1}{\rm 4\cdot\sqrt{\it A\cdot\rm \pi}}.$$
• The probability we are looking for is obtained by integration:
$${\rm Pr}(A> 150)=\int_{\rm 150}^{\it A_{\rm max}}\frac{\rm 1}{\rm 4\cdot\sqrt{\it A\cdot\rm \pi}} \; \rm d \it A= \frac{\rm 2\cdot\sqrt{\it A}}{\rm 4\cdot\sqrt{\pi}}\Big|_{\rm 150}^{\it A_{\rm max}}.$$
• The upper limit of integration yields the value  $4$  and the lower limit  $3.455$.  This yields the probability we are looking for:
$${\rm Pr}(A> 150) \hspace{0.15cm}\underline {=54.5\%}.$$

(5)  For the circular ring area  $R$  holds for a given radius  $r$:

$$R=\left (r+{b}/{\rm 2} \right)^{\rm 2}\cdot \rm\pi-\left ({\it r}-{\it b}/{\rm 2} \right)^{\rm 2}\cdot \rm\pi= \rm2\cdot\pi\cdot\it r \cdot b.$$
• There is thus a linear relationship between  $R$  and  $r$  .
• That is:  $R$  is also uniformly distributed independently of the width  $b$  as long as  $b \ll r$.
• For the minimum value holds:
$$R_{\rm min}=\rm 2\pi\cdot 6\cdot 0.1\hspace{0.15cm}\underline{\approx3.77}.$$

(6)  Accordingly,  the maximum value is:

$$R_{\rm max}=\rm 2\pi\cdot 8\cdot 0.1\hspace{0.15cm}\underline{\approx 5.03}.$$

(7)  Due to the linear relationship between  $R$  and  $r$  the mean radius  $r = 7$  also leads to the mean circular ring area:

$${\rm E}\big[R\big]=\rm 2\pi\cdot 7\cdot 0.1\hspace{0.15cm}\underline{\approx 4.4}.$$