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

  • question_answer The part of circle \[{{x}^{2}}+{{y}^{2}}=9\] in between \[y=0\] and \[y=2\] is revolved about y-axis. The volume of generating solid will be                        [UPSEAT 1999]

    A)            \[\frac{46}{3}\pi \]             

    B)            \[12\pi \]

    C)            \[16\pi \]                               

    D)            \[28\pi \]

    Correct Answer: A

    Solution :

               The part of circle \[{{x}^{2}}+{{y}^{2}}=9\] in between \[y=0\]and \[y=2\] is revolved about y-axis. Then a frustum of sphere will be formed.            The volume of this frustum            \[=\pi \int_{0}^{2}{{{x}^{2}}dy}\]\[=\pi \int_{0}^{2}{(9-{{y}^{2}})dy}\]            \[=\pi \left[ 9y-\frac{1}{3}{{y}^{3}} \right]_{0}^{2}\]\[=\pi \left[ 9\times 2-\frac{1}{3}{{(2)}^{3}}-(9.0-\frac{1}{3}.0) \right]\]                    \[=\pi \left[ 18-\frac{8}{3} \right]=\frac{46}{3}\pi \]cubic unit.

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