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

  • question_answer Area bounded by curves \[y={{x}^{2}}\] and \[y=2-{{x}^{2}}\] is                                                                                         [Orissa JEE 2005]

    A)            8/3

    B)            3/8

    C)            3/2

    D)            None of these

    Correct Answer: A

    Solution :

               \[y={{x}^{2}}\]                                                       .....(i)                    \[y=2-{{x}^{2}}\]                                           .....(ii)                    \[\therefore \] By equation (i) and (ii) , we get,              \[x=\pm 1\]                    \[\therefore \] \[y=\pm 1\]                        \[\therefore \] Required area \[=2\left[ \int_{0}^{1}{(2-{{x}^{2}})dx-\int_{0}^{1}{{{x}^{2}}dx}} \right]\]                    \[=2\,\left[ 2x-\frac{2{{x}^{3}}}{3} \right]_{0}^{1}=4\left[ x-\frac{{{x}^{3}}}{3} \right]_{0}^{1}=4\left( \frac{2}{3} \right)=\frac{8}{3}\].     

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