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 smaller part between the circle \[{{x}^{2}}+{{y}^{2}}=4\]and the line \[x=1\] is           [RPET 1999]

    A)            \[\frac{4\pi }{3}-\sqrt{3}\]     

    B)            \[\frac{8\pi }{3}-\sqrt{3}\]     

    C)            \[\frac{4\pi }{3}+\sqrt{3}\]    

    D)            \[\frac{5\pi }{3}+\sqrt{3}\]

    Correct Answer: B

    Solution :

                       Area of smaller part \[=2\,\int_{1}^{2}{\sqrt{4-{{x}^{2}}}}\,dx\]            \[=2\,\left[ \frac{x}{2}\sqrt{4-{{x}^{2}}}+2{{\sin }^{-1}}\frac{x}{2} \right]_{1}^{2}\]\[=2\,\left[ 2.\frac{\pi }{2}-\left[ \frac{\sqrt{3}}{2}-2.\frac{\pi }{6} \right] \right]\]                    \[=2\,\left[ \pi -\left[ \frac{\sqrt{3}}{2}-\frac{\pi }{3} \right] \right]\]\[=\frac{8\pi }{3}-\sqrt{3}\].

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