The area (in sq. units) of the region lying i | vedclass

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Question

(A) The given equations are $y = \frac{1}{\sqrt{3}}x$ and $x^2 + y^2 = 4$.Solving these,we find the intersection point $A$ in the first quadrant: $x^2

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$\frac{\pi}{3}$

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(A) The given equations are $y = \frac{1}{\sqrt{3}}x$ and $x^2 + y^2 = 4$.Solving these,we find the intersection point $A$ in the first quadrant: $x^2 + (\frac{x}{\sqrt{3}})^2 = 4 \implies x^2 + \frac{x^2}{3} = 4 \implies \frac{4x^2}{3} = 4 \implies x^2 = 3 \implies x = \sqrt{3}$.Then $y = \frac{\sqrt{3}}{\sqrt{3}} = 1$. So,$A = (\sqrt{3}, 1)$.The required area is the sum of the area of triangle $OAB$ (where $B$ is $(\sqrt{3}, 0)$) and the area under the circle from $x = \sqrt{3}$ to $x = 2$.Area of $\Delta OAB = \frac{1}{2} 	imes 	ext{base} 	imes 	ext{height} = \frac{1}{2} 	imes \sqrt{3} 	imes 1 = \frac{\sqrt{3}}{2}$.Area under circle $= \int_{\sqrt{3}}^{2} \sqrt{4 - x^2} dx = [\frac{x}{2}\sqrt{4 - x^2} + \frac{4}{2} \sin^{-1}(\frac{x}{2})]_{\sqrt{3}}^{2}$.$= [0 + 2 \sin^{-1}(1)] - [\frac{\sqrt{3}}{2} \sqrt{4 - 3} + 2 \sin^{-1}(\frac{\sqrt{3}}{2})]$.$= [2 	imes \frac{\pi}{2}] - [\frac{\sqrt{3}}{2} + 2 	imes \frac{\pi}{3}] = \pi - \frac{\sqrt{3}}{2} - \frac{2\pi}{3} = \frac{\pi}{3} - \frac{\sqrt{3}}{2}$.Total Area $= \frac{\sqrt{3}}{2} + (\frac{\pi}{3} - \frac{\sqrt{3}}{2}) = \frac{\pi}{3} 	ext{ sq. units}$.

The area (in sq. units) of the region lying in the first quadrant and enclosed by the $X$-axis,the straight line $x - \sqrt{3}y = 0$ and the circle $x^2 + y^2 = 4$ is

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