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Q33E

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Essential Calculus: Early Transcendentals
Found in: Page 370
Essential Calculus: Early Transcendentals

Essential Calculus: Early Transcendentals

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

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Short Answer

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

Step by Step Solution

TABLE OF CONTENTS :
TABLE OF CONTENTS

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

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

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