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Area lying in the first quadrant and bounded by the circle x2 + y2 = 4, the line x=\sqrt{3}y and
x-axis is

  • Option 1)

    \pi

  • Option 2)

    \pi/2

  • Option 3)

    \pi/3

  • Option 4)

    none of these

 

Answers (1)

best_answer

As we learnt

Area along y axis -

Let y_{1}= f_{1}(x)\, and \, y_{2}= f_{2}(x) be two curve, then area bounded by the curves and the lines

y = a and y = b is

A=\int_{a}^{b}\left ( x_{2}-x_{1} \right )dy

- wherein

 

 Required \: Area=\int_{0}^{1}\left ( x_{2}-x_{1} \right )dy

=\int_{0}^{1}\left (\sqrt{4-y^{2}}-\sqrt{3} y \right )dy

=\left [ \frac{1}{2}y\sqrt{4-y^{2}}+\frac{1}{2}\left ( 4 \right ) \sin^{-1}\frac{y}{2}-\frac{\sqrt{3}y^{2}}{2}\right ]_{0}^{1}

=\frac{\sqrt{3}}{2}+2\sin^{-1}\left ( \frac{1}{2} \right )-\frac{\sqrt{3}}{2}-2\sin^{-1}0=\frac{\sqrt{3}}{2}+2\left ( \frac{\pi }{6} \right )-\frac{\sqrt{3}}{2}=\frac{\pi }{3}

 


Option 1)

\pi

Option 2)

\pi/2

Option 3)

\pi/3

Option 4)

none of these

Posted by

gaurav

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