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

  • question_answer Area inside the parabola \[{{y}^{2}}=4ax,\]between the lines \[x=a\]and\[x=4a\]is equal to                    [Pb. CET 2002; Karnataka CET 2005]

    A)            \[4{{a}^{2}}\]                       

    B)            \[8{{a}^{2}}\]

    C)            \[28\frac{{{a}^{2}}}{3}\]  

    D)            \[35\frac{{{a}^{2}}}{3}\]

    Correct Answer: C

    Solution :

                       We have \[{{y}^{2}}=4ax\] Þ \[y=2\sqrt{ax}\]                    We know the equations of lines \[x=a\] and \[x=4a\]            \ The area inside the parabola between the lines                  \[A=\int_{a}^{4a}{y\,dx}=\int_{a}^{4a}{2\sqrt{ax}}\,dx=2\sqrt{a}\int_{a}^{4a}{\,{{x}^{\frac{1}{2}}}dx=2\sqrt{a}\left[ \frac{{{x}^{\frac{3}{2}}}}{\frac{3}{2}} \right]}_{a}^{4a}\]             \[=\frac{4}{3}{{a}^{\frac{1}{2}}}\left[ {{(4a)}^{\frac{3}{2}}}-{{(a)}^{\frac{3}{2}}} \right]=\frac{4}{3}{{a}^{\frac{1}{2}}}{{a}^{\frac{3}{2}}}[8-1]\] \[=\frac{28}{3}{{a}^{2}}\].

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