Definite Integration Question 469
Question: The area of smaller part between the circle $ x^{2}+y^{2}=4 $ and the line $ x=1 $ is
[RPET 1999]
Options:
A) $ \frac{4\pi }{3}-\sqrt{3} $
B) $ \frac{8\pi }{3}-\sqrt{3} $
C) $ \frac{4\pi }{3}+\sqrt{3} $
D) $ \frac{5\pi }{3}+\sqrt{3} $
Show Answer
Answer:
Correct Answer: B
Solution:
Area of smaller part $ =2\int_1^{2}{\sqrt{4-x^{2}}}dx $
$ =2[ \frac{x}{2}\sqrt{4-x^{2}}+2{{\sin }^{-1}}\frac{x}{2} ]_1^{2} $
$ =2[ 2.\frac{\pi }{2}-[ \frac{\sqrt{3}}{2}-2.\frac{\pi }{6} ] ] $
$ =2[ \pi -[ \frac{\sqrt{3}}{2}-\frac{\pi }{3} ] ] $
$ =\frac{8\pi }{3}-\sqrt{3} $ .
BETA
