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

  • question_answer If area bounded by the curves \[{{y}^{2}}=4ax\] and \[y=mx\] is \[{{a}^{2}}/3,\],  then the value of \[m\] is

    A)            2    

    B)            \[-2\]

    C)            \[\frac{1}{2}\]                      

    D)            None of these

    Correct Answer: A

    Solution :

                       The two curves \[{{y}^{2}}=4ax\] and \[y=mx\] intersect at \[\left( \frac{4a}{{{m}^{2}}},\frac{4a}{m} \right)\] and the area enclosed by the two curves is given by \[\int_{0}^{4a/{{m}^{2}}}{(\sqrt{4ax}-mx)}\,dx\].                    \[\therefore \,\,\,\int_{0}^{4a/{{m}^{2}}}{(\sqrt{4ax}-mx)}\,dx=\frac{{{a}^{2}}}{3}\]                    Þ  \[\frac{8}{3}\frac{{{a}^{2}}}{{{m}^{3}}}=\frac{{{a}^{2}}}{3}\Rightarrow {{m}^{3}}=8\Rightarrow m=2\].


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