Chapter 6.1, Problem 31E

To determine

Find the arc of the crescent -shaped region bounded by the given arcs of circles.

Answer to Problem 31E

The volume of the circle is $3r\sqrt{R^2-r^2}+\frac{\pi r^2}{2}-R^2{\sin }^{-1}\frac{r}{R}$

Explanation of Solution

Given information: Given circles $r\text{ and }R$

Calculation:

The lune occupies only the top half of the smaller circle, so its area A is obtained by simply integrate

The distance between the curves with respect to x, from $\text{x}=-r\text{ to x}=r.$ the area is symmetric across the $y-$ axis.

Set at the origin the center of the larger circle, whose equation is $x^2+y^2=R^2\text{ or }y=\sqrt{R^2-x^2}.$

The $y-$ coordinate of the center of the smaller circle is $\sqrt{R^2-r^2}$ by the Pythagorean Theorem.

The smaller circle is thus described by the equation $x^2+{\left(y-\sqrt{R^2-r^2}\right)}^2=r^2\text{, or }$ equivalently

  $\begin{array}{l}y=\sqrt{r^2-x^2}+\sqrt{R^2-r^2}\text{.}\\ A=2{\displaystyle \underset{0}{\overset{r}{\int }}\left[\sqrt{r^2-x^2}+\sqrt{R^2-r^2}-\sqrt{R^2-x^2}\right]}dx\\ =2r\sqrt{R^2-r^2}+2{\displaystyle \underset{0}{\overset{r}{\int }}\left[\sqrt{r^2-x^2}-\sqrt{R^2-x^2}\right]}dx\end{array}$

Let‘s do a generic integral: $\begin{array}{l}{\displaystyle \underset{0}{\overset{r}{\int }}\sqrt{u^2-x^2}}dx\Rightarrow \text{ let }x=u\sin \theta \text{ then }dx=u\cos \theta d\theta :\text{when }x=0,\theta =0\end{array}$

  $\text{when x}=r,\theta ={\sin }^{-1}\frac{r}{u};\text{let }b={\sin }^{-1}\frac{r}{u};$ And draw a right triangle with hypotenuse of length u and a leg of length x opposite angle $\theta$ , so the other leg will have a length of

  $\begin{array}{l}\sqrt{u^2-x^2},\cos \theta =\frac{\sqrt{u^2-x^2}}{u}\\ \Rightarrow {\displaystyle \underset{0}{\overset{r}{\int }}\sqrt{u^2-x^2}dx}\\ ={\displaystyle \underset{0}{\overset{b}{\int }}u\cos \theta u\cos \theta }d\theta \\ ={\displaystyle \underset{0}{\overset{b}{\int }}{u}^2{\cos }^2\theta }d\theta \\ =\frac{u^2}{2}{\displaystyle \underset{0}{\overset{b}{\int }}\left(1-\cos 2\theta \right)}d\theta \\ =\frac{u^2}{2}{\left(\theta -\frac{\sin 2\theta }{2}\right)}_0^b\\ =\frac{u^2}{2}\left(b-\frac{\sin 2b}{2}\right)\\ =\frac{u^2}{2}\left(b-\sin b\cos b\right)\\ =\frac{u^2}{2}\left({\sin }^{-1}\frac{r}{u}-\frac{r}{u}\frac{\sqrt{u^2-x^2}}{u}\right)\\ =\frac{u^2}{2}{\sin }^{-1}\frac{r}{u}-\frac{r}{2}\sqrt{u^2-x^2}\end{array}$

  $\begin{array}{l}A=2r\left[\sqrt{R^2-r^2}+2{\displaystyle \underset{0}{\overset{r}{\int }}\left[\sqrt{r^2-x^2}-\sqrt{R^2-x^2}\right]}\right]dx\\ =2r\sqrt{R^2-r^2}+2\left(\frac{r^2}{2}{\sin }^{-1}\frac{r}{r}\right)-2\left(\frac{R^2}{2}{\sin }^{-1}\frac{r}{R}-\frac{r}{2}\sqrt{R^2-r^2}\right)\\ =2r\sqrt{R^2-r^2}+r^2{\sin }^{-1}1-R^2{\sin }^{-1}\frac{r}{R}+r\sqrt{R^2-r^2}\\ =3r\sqrt{R^2-r^2}+\frac{\pi r^2}{2}-R^2{\sin }^{-1}\frac{r}{R}\end{array}$

Thus, the volume of the solid is $3r\sqrt{R^2-r^2}+\frac{\pi r^2}{2}-R^2{\sin }^{-1}\frac{r}{R}$