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

  • question_answer The area enclosed between the parabolas \[{{y}^{2}}=4x\] and \[{{x}^{2}}=4y\] is              [Karnataka CET 1999, 2003]

    A)            \[\frac{14}{3}\] sq. unit   

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

    C)            \[\frac{3}{16}\] sq. unit   

    D)            \[\frac{16}{3}\] sq. unit

    Correct Answer: D

    Solution :

               Equations of curves \[{{y}^{2}}=4x\] and \[{{x}^{2}}=4y.\]The given equations may be written as \[y=2\sqrt{x}\] and \[y=\frac{{{x}^{2}}}{4}.\]            We know that area enclosed by the parabolas \[=\int_{\,0}^{\,4}{2\sqrt{x}\,dx-}\int_{\,0}^{\,4}{\frac{{{x}^{2}}}{4}dx=\frac{32}{3}-\frac{16}{3}=\frac{16}{3}}\]sq. unit.

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