# 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

G gaurav

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

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