JEE Main & Advanced Mathematics Definite Integration Question Bank Area Bounded by Region, Volume and Surface Area of Solids of Revolution

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

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