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

  • question_answer The area of the curve \[x{{y}^{2}}={{a}^{2}}(a-x)\] bounded by y-axis is [RPET 1996]

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

    B)            \[2\pi {{a}^{2}}\]

    C)            \[3\pi {{a}^{2}}\]                

    D)            \[4\pi {{a}^{2}}\]

    Correct Answer: A

    Solution :

               Since the curve is symmetrical about x-axis, therefore Required area\[A=2\int_{0}^{a}{a\sqrt{\frac{a-x}{x}}\,dx}\]                    Put \[x=a{{\sin }^{2}}\theta \]                    \[\Rightarrow dx=2a\sin \theta .\cos \theta \,d\theta \]                    \[A=2\int_{0}^{\pi /2}{a\sqrt{\frac{a{{\cos }^{2}}\theta }{a{{\sin }^{2}}\theta }}}a\sin 2\theta \,d\theta \]                       \[=2{{a}^{2}}\int_{0}^{\pi /.2}{\frac{\cos \theta }{\sin \theta }2\sin \theta \cos \theta \,d\theta }\]                    \[A=4{{a}^{2}}\int_{0}^{\pi /2}{{{\cos }^{2}}\theta \,d\theta }\]Þ \[A=4{{a}^{2}}.\frac{1}{2}.\frac{\pi }{2}=\pi {{a}^{2}}\].


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