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The area lying in the first quadrant and bounded by the circle $x^{2}+y^{2}=4$ and the lines $x= 0$ and $x=1$ is given by

  1. $\frac{\pi}{3}+\frac{\sqrt{3}}{2}$
  2. $\frac{\pi}{6}+\frac{\sqrt{3}}{4}$
  3. $\frac{\pi}{3}-\frac{\sqrt{3}}{2}$
  4. $\frac{\pi}{6}+\frac{\sqrt{3}}{2}$
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3 Answers

Best answer
6 votes
6 votes

We can write the equation as $y=\sqrt{4-x^{2}}$

$\int_{0}^{1}\sqrt{4-x^{2}}dx$

Substituting $x=2\sin\theta$

$dx=2\cos\theta d\theta$

$\implies \theta=\sin^{-1}\left ( \frac{x}{2} \right )$

$\int_{0}^{\frac{\pi }{6}}\sqrt{4-4\sin^{2}\theta}*2\cos\theta d\theta$

$\int_{0}^{\frac{\pi }{6}}2\cos\theta * 2\cos\theta d\theta$

$4\int_{0}^{\frac{\pi }{6}}\cos^{2}\theta d\theta$

$4\int_{0}^{\frac{\pi }{6}}\frac{1+\cos2\theta}{2}d\theta$

$2\int_{0}^{\frac{\pi }{6}}1+\cos2\theta d\theta$

$2\int_{0}^{\frac{\pi }{6}}d\theta +2\int_{0}^{\frac{\pi }{6}}\cos2\theta d\theta$

$\frac{\pi}{3}+2\int_{0}^{\frac{\pi }{6}}\cos2\theta d\theta$

Substituting $2\theta  =y,$

$d\theta=\frac{y}{2}$

$\frac{\pi}{3}+\int_{0}^{\frac{\pi }{3}}\cos ydy$

$\frac{\pi}{3}+\frac{\sqrt{3}}{2}$

This is our final answer because we changed function $x^{2}+y^{2}=4 \rightarrow y=\sqrt{4-x^{2}}$ we eliminated the

fourth quadrant and by integrating it over 0 to 1 we further bound it to first quadrant itself

Correct Answer: $A$

edited by
7 votes
7 votes

Area$=\int \limits_{0}^{1}\sqrt{4-x^{2}}\,dx$

We know,

$\int\sqrt{a^{2}-x^{2}}\,dx=$ $\dfrac{x}{2}\sqrt{a^2-x^2}\,+\dfrac{a^2}{2}\sin \dfrac{x}{a}\,+\,c$

So Area $=\left[\dfrac{x}{2}\sqrt{4-x^2}\,+\,2\sin \dfrac{x}{2}\right]_{0}^{1}$

$\hspace{1.4 cm}=\dfrac{\sqrt{3}}{2}+\dfrac{2\pi}{6}+0$

$\hspace{1.4 cm}=\dfrac{\pi}{3}+ \dfrac{\sqrt{3}}{2}$

Hope this help.

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3 votes

                                                   

$Required\,Area\,=\,Area\left \{  Sector(OBC)\,+\,Triangle(OBE)\right \}$

$Area\,of\,sector\,=\,\frac{1}{2}\times r^{2}\times\theta\,(\,\theta\, is\, in\, Radian)$

In, $\bigtriangleup OAB,\,sin(y) = \frac{AB}{OB} = \frac{1}{2}, (y) = sin^{-1}(\frac{1}{2}), (y)=30^{\circ}$

$(y) = 30\times (\frac{\pi}{180}) = \frac{\pi}{6}$ (Radian)

$Area\,of\,sector\,(OABC)=\,\frac{1}{2}\times r^{2}\times(y)=\,\frac{1}{2}\times 2^{2}\times(\frac{\pi}{6})=\,\frac{\pi}{3}$

$BE =\sqrt{OB^{2} - OE^{2}} = \sqrt{2^{2}-1^{2}} = \sqrt{3}$

$Area\,of\,\bigtriangleup OBE = \frac{1}{2}\times BE\,\times OE = \frac{\sqrt{3}}{2}$

$Required\,Area = \frac{\pi}{3} + \frac{\sqrt{3}}{2}$

Answer:

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