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Q33E

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Found in: Page 370

### Essential Calculus: Early Transcendentals

Book edition 2nd
Author(s) James Stewart
Pages 830 pages
ISBN 9781133112280

# Find the area of the crescent-shaped region (called a lune) bounded by arcs of circles with radii and (see the figure)

The area of the crescent-shaped region is $$r\sqrt {{R^2} - {r^2}} + \frac{\pi }{2}{r^2} - {R^2}\arcsin \left( {\frac{r}{R}} \right)$$.

See the step by step solution

## General form of a circle

The general form of a circle with center at origin is given by

$${x^2} + {y^2} = {R^2}$$

Apply the condition as follows if the function $$f$$ is continuous on $$( - a,a)$$.

a) If $$f$$ is even $$(f( - x) = f(x))$$, then $$\int_{ - a}^a f (x)dx = 2\int_0^a f (x)dx$$.

b) If $$f$$ is odd $$(f( - x) = f(x))$$, then $$\int_{ - a}^a f (x)dx = 0$$.

## The equation of the second circle

Let us suppose that the equation of the bigger circle is given as (being center at the origin)

$${x^2} + {y^2} = {R^2}$$

From here, we have

\begin{aligned}{l}{y^2} = {R^2} - {x^2}\\y = \pm \sqrt {{R^2} - {x^2}} \end{aligned}

Let us suppose that the distance between the centers of the two circles is $$a$$ as shown in the given figure. Therefore, the equation of the second circle is given as

$${x^2} + {(y - b)^2} = {r^2}$$

Rewrite the equation

\begin{aligned}{l}{(y - b)^2} = {r^2} - {x^2}\\y - b = \pm \sqrt {{r^2} - {x^2}} \\y = b \pm \sqrt {{r^2} - {x^2}} \end{aligned}

## Find the crescent area

Both the curves forming the crescent are above the $$x$$ axis (having positive $$y$$ values). Therefore, the functions to calculate the crescent area are taken as $$f(x) = b + \sqrt {{r^2} - {x^2}}$$, and $$g(x) = \sqrt {{R^2} - {x^2}}$$. The crescent area is given by

\begin{aligned}{l}A = \int\limits_{ - r}^r {\left( {f\left( x \right) - g\left( x \right)} \right)} dx\\A = \int\limits_{ - r}^r {\left( {\left( {b - \sqrt {{r^2} - {x^2}} } \right) - \left( {\sqrt {{R^2} - {x^2}} } \right)} \right)} dx\\A = 2\int\limits_0^r {\left( {\left( {b - \sqrt {{r^2} - {x^2}} } \right) - \left( {\sqrt {{R^2} - {x^2}} } \right)} \right)} dx\\A = 2\int\limits_0^r {bdx + 2\int\limits_0^r {\sqrt {{r^2} - {x^2}} dx} - 2\int\limits_0^r {\sqrt {{R^2} - {x^2}} dx} } \end{aligned}

## Solve the first integral

Now, estimate the first integral

\begin{aligned}{l}2\int_0^r b dx = 2(bx)_0^r\\2\int_0^r b dx = 2(br - 0)\\2\int_0^r b dx = 2br\\2\int_0^r b dx = 2r\sqrt {{R^2} - {r^2}} \end{aligned}

## Solve the second integral

To estimate the second integral

Take \begin{aligned}{l}x = r\sin t\\dx = r\cos tdt\end{aligned},

\begin{aligned}{l}2\int_0^r {\sqrt {{r^2} - {x^2}} } dx = 2\int_0^{\frac{\pi }{2}} {\sqrt {{r^2} - {r^2}{{\sin }^2}t} } \cdot r\cos tdt\\2\int_0^r {\sqrt {{r^2} - {x^2}} } dx = 2\int_0^{\frac{\pi }{2}} r \cos t \cdot r\cos tdt\\2\int_0^r {\sqrt {{r^2} - {x^2}} } dx = 2\int_0^{\frac{\pi }{2}} {{r^2}} {\cos ^2}tdt\\2\int_0^r {\sqrt {{r^2} - {x^2}} } dx = 2\int_0^{\frac{\pi }{2}} {{r^2}} \left( {\frac{{1 + \cos 2t}}{2}} \right)dt\\2\int_0^r {\sqrt {{r^2} - {x^2}} } dx = {r^2}\int_0^{\frac{\pi }{2}} {\left( {1 + \cos 2t} \right)} dt\end{aligned}

Simplify further the integral

\begin{aligned}{l}2\int_0^r {\sqrt {{r^2} - {x^2}} } dx = {r^2}\left( {t + \frac{{\sin 2t}}{2}} \right)_0^{\frac{\pi }{2}}\\2\int_0^r {\sqrt {{r^2} - {x^2}} } dx = {r^2}\left( {\left( {\pi /2 + \frac{{\sin \pi }}{2}} \right) - \left( {0 + \frac{{\sin 0}}{2}} \right)} \right)\\2\int_0^r {\sqrt {{r^2} - {x^2}} } dx = {r^2}\left( {\frac{\pi }{2} + 0 - 0} \right)\\2\int_0^r {\sqrt {{r^2} - {x^2}} } dx = \frac{{\pi {r^2}}}{2}\end{aligned}

## Solve the third integral

Similarly, to estimate the third integral,

Let take \begin{aligned}{l}x = R\sin t\\dx = R\cos tdt\end{aligned}

\begin{aligned}{l} - 2\int_0^r {\sqrt {{R^2} - {x^2}} } dx = - 2\int_0^{\arcsin (r/R)} {\sqrt {{R^2} - {R^2}{{\sin }^2}t} } \cdot R\cos tdt\\ - 2\int_0^r {\sqrt {{R^2} - {x^2}} } dx = - 2\int_0^{\arcsin (r/R)} R \cos t \cdot R\cos tdt\\ - 2\int_0^r {\sqrt {{R^2} - {x^2}} } dx = - 2\int_0^{\arcsin (r/R)} {{R^2}} {\cos ^2}tdt\\ - 2\int_0^r {\sqrt {{R^2} - {x^2}} } dx = - 2\int_0^{\arcsin (r/R)} {{R^2}} \left( {\frac{{1 + \cos 2t}}{2}} \right)dt\end{aligned}

Solve further the integral

\begin{aligned}{l} - 2\int_0^r {\sqrt {{R^2} - {x^2}} } dx = - {R^2}\int_0^{\arcsin (\frac{r}{R})} {\left( {1 + \cos 2t} \right)} dt\\ - 2\int_0^r {\sqrt {{R^2} - {x^2}} } dx = - {R^2}\left( {t + \frac{{\sin 2t}}{2}} \right)_0^{\arcsin (\frac{r}{R})}\\ - 2\int_0^r {\sqrt {{R^2} - {x^2}} } dx = - {R^2}(t + \sin t\cos t)_0^{\arcsin (\frac{r}{R})}\end{aligned}

Further the integral can be solved as

\begin{aligned}{l} - 2\int_0^r {\sqrt {{R^2} - {x^2}} } dx = - {R^2}\left( {\left( {\arcsin \left( {\frac{r}{R}} \right) + \sin \left( {\arcsin \left( {\frac{r}{R}} \right)} \right) \cdot \cos \left( {\arcsin \left( {\frac{r}{R}} \right)} \right)} \right) - (0 + 0)} \right)\\ - 2\int_0^r {\sqrt {{R^2} - {x^2}} } dx = - {R^2}\left( {\left( {\arcsin \left( {\frac{r}{R}} \right) + \frac{r}{R} \cdot \frac{{\sqrt {{R^2} - {r^2}} }}{R}} \right)} \right)\\ - 2\int_0^r {\sqrt {{R^2} - {x^2}} } dx = - {R^2}\arcsin \left( {\frac{r}{R}} \right) - r\sqrt {{R^2} - {r^2}} \end{aligned}

Hence, the crescent area is estimated as

\begin{aligned}{l}A = 2r\sqrt {{R^2} - {r^2}} + \frac{{\pi {r^2}}}{2} - {R^2}\arcsin \left( {\frac{r}{R}} \right) - r\sqrt {{R^2} - {r^2}} \\A = r\sqrt {{R^2} - {r^2}} + \frac{{\pi {r^2}}}{2} - {R^2}\arcsin \left( {\frac{r}{R}} \right)\end{aligned}