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12th Class Mathematics Definite Integrals Question Bank Critical Thinking

  • question_answer
    What is the area bounded by the curves \[{{x}^{2}}+{{y}^{2}}=9\] and \[{{y}^{2}}=8x\] is [DCE 1999]

    A) 0    

    B) \[\frac{2\sqrt{2}}{3}+\frac{9\pi }{2}-9{{\sin }^{-1}}\left( \frac{1}{3} \right)\]

    C) \[16\,\pi \]                            

    D) None of these

    Correct Answer: B

    Solution :

    • Solving the equations,            \[{{x}^{2}}+{{y}^{2}}=9\]  ?..(i)           
    • \[{{y}^{2}}=8x\]     ?..(ii) Put \[{{y}^{2}}=8x\] in (i), \[{{x}^{2}}+8x-9=0\]           
    • Þ \[(x+9)(x-1)=0\] i.e., \[x=-\,9\]or 1 \[x=-9\] gives imaginary value of y for equation (ii) hence neglected.           
    • \[\therefore A\,\equiv \,(1,\,0)\] and \[B\equiv (3,\,0)\]           
    • \[\therefore \]Required area = 2 times the hatched areas           
    • \[=2\left[ \int_{0}^{1}{y\,dx\,\,}\text{for}\,\text{(ii)}\,+\int_{1}^{3}{y\,dx\,\,\text{for}\,\,\text{(i)}} \right]\,\]           
    • \[=2\left[ \int_{0}^{1}{2}\sqrt{2}\,\,{{(x)}^{1/2}}\,\,dx+\int_{1}^{3}{\sqrt{{{3}^{2}}-{{x}^{2}}}\,dx} \right]\]           
    • \[=2\left[ 2\sqrt{2}\times \left( \frac{{{x}^{3/2}}}{3/2} \right)_{0}^{1}+\left( \frac{x\sqrt{9-{{x}^{2}}}}{2}+\frac{9}{2}{{\sin }^{-1}}\left( \frac{x}{3} \right) \right)_{1}^{3} \right]\]           
    • \[=2\,\left[ \frac{4\sqrt{2}}{3}+\frac{9}{2}\times \frac{\pi }{2}-\frac{\sqrt{8}}{2}-\frac{9}{2}{{\sin }^{-1}}\left( \frac{1}{3} \right) \right]\]                   
    • \[=\frac{2\sqrt{2}}{3}+\frac{9\pi }{2}-9{{\sin }^{-1}}\left( \frac{1}{3} \right)\].


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