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

  • question_answer Area bounded by the curve \[{{x}^{2}}=4y\] and the straight line \[x=4y-2\] is            [SCRA 1986; IIT 1981; Pb. CET 2003]

    A)            \[\frac{8}{9}\] sq. unit      

    B)            \[\frac{9}{8}\] sq. unit

    C)            \[\frac{4}{3}\] sq. unit      

    D)            None of these

    Correct Answer: B

    Solution :

     Solving the equations \[{{x}^{2}}=4y\]and \[x=4y-2\] simultaneously. The points of intersection of the parabola and the line are \[A(2,\,1)\]and \[B\left( -1,\frac{1}{4} \right)\].            \ The required area = shaded area                            \[=\left[ \int_{-1}^{2}{y\,d{{x}_{(\text{for}\,x=4y-2)}}} \right]-\left[ \int_{-1}^{2}{yd{{x}_{(\text{for}\,{{x}^{2}}=4y)}}} \right]\]                            \[=\int_{-1}^{2}{\frac{1}{4}(x+2)dx-\int_{-1}^{2}{\frac{1}{4}{{x}^{2}}dx}}\]                                        \[=\frac{1}{4}\left[ \frac{{{x}^{2}}}{2}+2x \right]_{-1}^{2}-\frac{1}{4}\left[ \frac{{{x}^{3}}}{3} \right]_{-1}^{2}=\frac{9}{8}\]sq. unit.

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