Definite Integration Question 132
Question: The area of the smaller segment cut off from the circle $ x^{2}+y^{2}=9byx=1 $ is
Options:
A) $ \frac{1}{2}(9se{c^{-1}}3-\sqrt{8}) $ sq. unit
B) $ (9se{c^{-1}}3-\sqrt{8}) $ sq. unit
C) $ (\sqrt{8}-9se{c^{-1}}3) $ sq. unit
D) None of the above
Show Answer
Answer:
Correct Answer: B
Solution:
[b] Given, equation of the circle is $ x^{2}+y^{2}=9 $ .
$ \therefore $
Area of the smaller segment cut off from the circle $ x^{2}+y^{2} $ = 9 by x = 1, is given by
$ A=2\int_1^{3}{\sqrt{9-x^{2}}dx}=2\cdot \frac{1}{2}[ x\sqrt{9-x^{2}}+9{{\sin }^{-1}}\frac{x}{3} ]_1^{3} $
$ =[ 3\cdot \sqrt{9-9}+9{{\sin }^{-1}}( \frac{3}{3} )-1.\sqrt{9-1}-9{{\sin }^{-1}}( \frac{1}{3} ) ] $
$ =[9{{\sec }^{-1}}(3)-\sqrt{8}] $ sq. unit.
BETA
