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  • Can someone explain The area (in sq. units) of the smaller portion enclosed between the curves, x2+y2=4 and y2=3x, is :

The area (in sq. units) of the smaller portion enclosed between the curves, x2+y2=4 and y2=3x, is :

 

  • Option 1)

    \frac{1}{2\sqrt{3}} +\frac{\pi }{3}

  • Option 2)

    \frac{1}{\sqrt{3}} +\frac{2\pi }{3}

  • Option 3)

    \frac{1}{2\sqrt{3}}+\frac{2\pi }{\sqrt{3}}

  • Option 4)

    \frac{1}{\sqrt{3}} + \frac{4\pi }{3}

 

Answers (1)

As we learnt in

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

 

Point of intersection

x2+3x-4=0

\Rightarrow    x2+4x-x-4=0

x=1

and  y=\pm\sqrt{3}

Area =2\left ( \int_{0}^{1}\left ( \sqrt{3}\sqrt{x} \right )dx+\int_{1}^{2}\left ( \sqrt{4-x^{2}} \right )dx\right )

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

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

=2\left ( \frac{2}{\sqrt{3}}-\frac{\sqrt{3}}{2}+\pi -\frac{\pi }{3} \right )

= 2\left ( \frac{1}{2\sqrt{3}}+\frac{2\pi }{3} \right )

=\frac{1}{\sqrt{3}}+\frac{4\pi }{3}

 

 


Option 1)

\frac{1}{2\sqrt{3}} +\frac{\pi }{3}

Incorrect option    

Option 2)

\frac{1}{\sqrt{3}} +\frac{2\pi }{3}

Incorrect option    

Option 3)

\frac{1}{2\sqrt{3}}+\frac{2\pi }{\sqrt{3}}

Incorrect option    

Option 4)

\frac{1}{\sqrt{3}} + \frac{4\pi }{3}

Correct option    

Posted by

Vakul

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