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 region \[\{\,(x,\,y):{{x}^{2}}+{{y}^{2}}\le 1\le x+y\}\] is                                                                                   [Kerala (Engg.) 2002]

    A)            \[\frac{{{\pi }^{2}}}{5}\]  

    B)            \[\frac{{{\pi }^{2}}}{2}\]

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

    D)            \[\frac{\pi }{4}-\frac{1}{2}\]

    Correct Answer: D

    Solution :

               \[{{x}^{2}}+{{y}^{2}}=1,x+y=1\]meet when            \[{{x}^{2}}+{{(1-x)}^{2}}=1\Rightarrow {{x}^{2}}+1+{{x}^{2}}-2x=1\]                        \[\Rightarrow 2{{x}^{2}}-2x=0\,\,\Rightarrow 2x(x-1)=0\]            \[\Rightarrow x=0,\,x=1\] Þ \[y=1,\,y=0\], i.e., \[A\,(1,0);\,\,B\,(0,1)\]            Required area \[=\int_{0}^{1}{[\sqrt{1-{{x}^{2}}}-(1-x)]}\]dx                               \[=\left[ \frac{x\sqrt{1-{{x}^{2}}}}{2}+\frac{1}{2}{{\sin }^{-1}}x-x+\frac{{{x}^{2}}}{2} \right]_{0}^{1}\]                              \[=\frac{1}{2}.\frac{\pi }{2}-1+\frac{1}{2}=\frac{\pi }{4}-\frac{1}{2}\].


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