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

  • question_answer The area bounded by the curves \[{{y}^{2}}=8x\] and \[y=x\] is

    A)            \[\frac{128}{3}\] sq. unit 

    B)            \[\frac{32}{3}\] sq. unit

    C)            \[\frac{64}{3}\] sq. unit   

    D)            32 sq. unit

    Correct Answer: B

    Solution :

               \[{{y}^{2}}=8x\] and \[y=x\Rightarrow {{x}^{2}}=8x\Rightarrow x=0\],8                    \ Required area =\[\int_{0}^{8}{(2\sqrt{2}\sqrt{x}-x)dx}\]                    \[=\left[ \frac{4\sqrt{2}}{3}{{x}^{3/2}}-\frac{{{x}^{2}}}{2} \right]_{0}^{8}=\frac{128}{3}-\frac{64}{2}=\frac{32}{3}sq.\]unit.


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