Definite Integration Question 501
Question: The area bounded by the circle $ x^{2}+y^{2}=4, $ line $ x=\sqrt{3}y $ and $ x- $ axis lying in the first quadrant, is
[RPET 1997; Kurukshetra CEE 1998]
Options:
A) $ \frac{\pi }{2} $
B) $ \frac{\pi }{4} $
C) $ \frac{\pi }{3} $
D) $ \pi $
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Answer:
Correct Answer: C
Solution:
Required area $ =\int_0^{\sqrt{3}}{\frac{x}{\sqrt{3}}dx+\int _{\sqrt{3}}^{2}{\sqrt{4-x^{2}}dx}} $
$ =\frac{1}{\sqrt{3}}[ \frac{x^{2}}{2} ]_0^{\sqrt{3}}+[ \frac{x}{2}\sqrt{4-x^{2}}+\frac{4}{2}{{\sin }^{-1}}\frac{x}{2} ] _{\sqrt{3}}^{2} $
$ =\frac{\sqrt{3}}{2}+[ \pi -\frac{\sqrt{3}}{2}-\frac{2\pi }{3} ]=\frac{\pi }{3} $ .
Trick : Area of sector made by an arc = $ \frac{{{\theta }^{c}}R^{2}}{2} $
$ =\frac{\pi }{6}.\frac{4}{2}=\frac{\pi }{3} $ .
BETA
