Chapter 5, Problem 12P

To determine

To calculate:

The area of the region

Answer to Problem 12P

The area of the region is $\frac{4}{3}\left(4\sqrt{2}-5\right)$

Explanation of Solution

Given information:

Consider the diagram given in the question

Calculation:

The point whose value is to find, assume that point as R. This point $R$ consist all point that are closer to the center than to the side, so if this point is multiplied by 4 then the area can be found that is asked in the problem.

The distance of the point $\left(x,y\right)$ from the side is $\left(1-y\right)$ and its distance from the center is $\sqrt{x^2+y^2}$

So, the distance are equal if

  $\begin{array}{l}\sqrt{x^2+y^2}=1-y\\ x^2+y^2=1-2y+y^2\\ y=\frac{1}{2}\left(1-x^2\right)\end{array}$

The area $R$ is equal to the area of the triangle plus a crescent shaped area

  $R=A_c+A_r$

To find these areas it is necessary, it is necessary to find the y-coordinate $h$ of the horizontal line separating them.

From the diagram,

  $\begin{array}{l}1-h=\sqrt{2h}\\ h=\frac{1}{1+\sqrt{2}}=\sqrt{2}-1\end{array}$

The area of the triangle is

  $A_r=\frac{1}{2}\left(2h\right)\left(h\right)=h^2$

The area of the crescent shaped section is

  $\begin{array}{l}{A}_c={\displaystyle {\int }_{-h}^h\left[\frac{1}{2}\left(1-x^2\right)-h\right]}dx\\ =2{\displaystyle {\int }_0^h\left[\frac{1}{2}-h-\frac{1}{2}{x}^2\right]}dx\\ =2{\left[\left(\frac{1}{2}-h\right)x-\frac{1}{6}{x}^3\right]}_0^h\\ =h-2h^2-\frac{1}{3}{h}^3\end{array}$

Now, the area of whole region $\left(4R\right)$ can be obtained as,

  $\begin{array}{l}4R=4\left[\left(h-2h^2-\frac{1}{3}{h}^3\right)+h^2\right]\\ 4R=4h\left(1-h-\frac{1}{3}{h}^2\right)\\ 4R=4\left(\sqrt{2}-1\right)\left[1-\left(\sqrt{2}-1\right)-\frac{1}{3}{\left(\sqrt{2}-1\right)}^2\right]\\ 4R=4\left(\sqrt{2}-1\right)\left(1-\frac{1}{3}\sqrt{2}\right)\\ 4R=\frac{4}{3}\left(4\sqrt{2}-5\right)\end{array}$

Therefore, the area of the region is $\frac{4}{3}\left(4\sqrt{2}-5\right)$