question_answer

The area of the smaller segment cut off from the circle \[{{x}^{2}}+{{y}^{2}}=9\,\,by\,\,x=1\] is

A) \[\frac{1}{2}(9\,se{{c}^{-1}}3-\sqrt{8})\] sq. unit

B) \[(9\,se{{c}^{-1}}3-\sqrt{8})\] sq. unit

C) \[(\sqrt{8}-9\,\,se{{c}^{-1}}3)\]sq. unit

D) None of the above

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                         \[A=2\int_{1}^{3}{\sqrt{9-{{x}^{2}}}\,\,dx}=2\cdot \frac{1}{2}\left[ x\sqrt{9-{{x}^{2}}}+9\,{{\sin }^{-1}}\frac{x}{3} \right]_{1}^{3}\] \[=\left[ 3\cdot \sqrt{9-9}+9\,\,{{\sin }^{-1}}\left( \frac{3}{3} \right)-1.\sqrt{9-1}-9{{\sin }^{-1}}\left( \frac{1}{3} \right) \right]\]                   \[=[9\,\,{{\sec }^{-1}}(3)-\sqrt{8}]\] sq. unit.