question_answer

The area of the smaller segment cut off from the circle \[{{x}^{2}}+{{y}^{2}}=9\] by \[x=1\] is                                     [RPET 2002]

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

Correct Answer: B

Solution :

           Area of smaller part \[I=2\int_{1}^{3}{\sqrt{9-{{x}^{2}}}\,dx}\]                    \[=2.\frac{1}{2}\left[ x\sqrt{9-{{x}^{2}}}+9{{\sin }^{-1}}\frac{x}{3} \right]_{1}^{3}=\left[ 9\frac{\pi }{2}-\sqrt{8}-9{{\sin }^{-1}}\left( \frac{1}{3} \right) \right]\]            \[=\left[ 9\left( \frac{\pi }{2}-{{\sin }^{-1}}\left( \frac{1}{3} \right) \right)-\sqrt{8} \right]\]\[=\left[ 9{{\cos }^{-1}}\left( \frac{1}{3} \right)-\sqrt{8} \right]\]            = \[[9{{\sec }^{-1}}(3)-\sqrt{8}]\].