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

  • question_answer The area between the parabola \[{{y}^{2}}=4ax\]and \[{{x}^{2}}=8ay\] is                                                               [RPET 1997]

    A)            \[\frac{8}{3}{{a}^{2}}\]    

    B)            \[\frac{4}{3}{{a}^{2}}\]

    C)            \[\frac{32}{3}{{a}^{2}}\]  

    D)            \[\frac{16}{3}{{a}^{2}}\]

    Correct Answer: C

    Solution :

     Required area \[A=\int_{0}^{(a{{2}^{8/3}})}{\sqrt{4ax}}dx-\int_{0}^{a{{2}^{8/3}}}{\frac{{{x}^{2}}}{8a}}\,dx=\frac{32{{a}^{2}}}{3}\]


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