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

  • question_answer If the area bounded by \[y=a{{x}^{2}}\]and \[x=a{{y}^{2}}\], \[a>0\], is 1, then \[a=\]                    [IIT Screening 2004]

    A)            1    

    B)            \[\frac{1}{\sqrt{3}}\]

    C)            \[\frac{1}{3}\]                      

    D)            None of these

    Correct Answer: B

    Solution :

               The x-coordinate of A is \[\frac{1}{a}\]                    According to the given condition, \[1=\int_{0}^{1/a}{\left( \sqrt{\frac{x}{a}}-a{{x}^{2}} \right)}\text{ }dx\]                    Þ\[1=\frac{1}{\sqrt{a}}.\frac{2}{3}[{{x}^{3/2}}]_{0}^{1/a}-\frac{a}{3}[{{x}^{3}}]_{0}^{1/a}\]Þ \[{{a}^{2}}=\frac{1}{3}\Rightarrow a=\frac{1}{\sqrt{3}}\].


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