question_answer

Area of small part between the circle\[{{x}^{2}}+{{y}^{2}}=9\]and the line\[x=1\]is

A)  \[{{\sec }^{-1}}3-\sqrt{8}\]

B)  \[9{{\sec }^{-1}}3-\sqrt{8}\]

C)  \[9\cos e{{c}^{-1}}3-\sqrt{8}\]

D)  None of these

Correct Answer: B

Solution :

 Required area \[=2\int_{1}^{3}{\sqrt{9-{{x}^{2}}}}dx\] \[=2.\frac{1}{2}\left[ x\sqrt{9-{{x}^{2}}}+9{{\sin }^{-1}}\frac{x}{3} \right]_{1}^{3}\] \[=3\sqrt{9-9}+9{{\sin }^{1}}\left( \frac{3}{3} \right)-\sqrt{9-1}-9{{\sin }^{-1}}\left( \frac{1}{3} \right)\] \[=9.\frac{\pi }{2}-\sqrt{8}-9{{\sin }^{-1}}\left( \frac{1}{3} \right)\] \[=9\left[ \frac{\pi }{2}-{{\sin }^{-1}}\left( \frac{1}{3} \right) \right]-\sqrt{8}\] \[=9{{\cos }^{-1}}\left( \frac{1}{3} \right)-\sqrt{8}=9{{\sec }^{-1}}(3)-\sqrt{8}\]