Definite Integration Question 284
Question: The area of the smaller segment cut off from the circle $ x^{2}+y^{2}=9 $ by $ x=1 $ is
[RPET 2002]
Options:
A) $ \frac{1}{2}(9{{\sec }^{-1}}3-\sqrt{8}) $
B) $ 9{{\sec }^{-1}}(3)-\sqrt{8} $
C) $ \sqrt{8}-9{{\sec }^{-1}}(3) $
D) None of these
Show Answer
Answer:
Correct Answer: B
Solution:
Area of smaller part $ I=2\int_1^{3}{\sqrt{9-x^{2}}dx} $
$ =2.\frac{1}{2}[ x\sqrt{9-x^{2}}+9{{\sin }^{-1}}\frac{x}{3} ]_1^{3}=[ 9\frac{\pi }{2}-\sqrt{8}-9{{\sin }^{-1}}( \frac{1}{3} ) ] $
$ =[ 9( \frac{\pi }{2}-{{\sin }^{-1}}( \frac{1}{3} ) )-\sqrt{8} ] $
$ =[ 9{{\cos }^{-1}}( \frac{1}{3} )-\sqrt{8} ] $
= $ [9{{\sec }^{-1}}(3)-\sqrt{8}] $ .
BETA
