Find the area of the region in the first quadrant enclosed by x-axis, the line x = $\sqrt{3} y$ and the circle x² + y² = 4.

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Solution

$x^{2} + y^{2} = 4$ represents a circle with centre O(0,0) and radius 2 , cutting x axis at A(2,0) and A'(-2,0)

$x = \sqrt{3} y$ represents a straight line passing through O(0,0)

Solving the two equations we get

$x^{2} + y^{2} = 4 and x = \sqrt{3} y \Rightarrow {(\sqrt{3} y)}^{2} + y^{2} = 4 \Rightarrow 4 y^{2} = 4 \Rightarrow y = \pm 1 \Rightarrow x = \pm \sqrt{3} B (\sqrt{3}, 1) and B' (- \sqrt{3}, - 1) are points of intersection of circle and straight line Shaded area (O B Q A O) = area (O B P O) + area (B A P B) = \frac{1}{\sqrt{3}} \int_{0}^{\sqrt{3}} x d x + \int_{\sqrt{3}}^{2} \sqrt{4 - x^{2}} d x = \frac{1}{\sqrt{3}} {[\frac{x^{2}}{2}]}_{0}^{\sqrt{3}} + {[\frac{1}{2} x \sqrt{4 - x^{2}} + \frac{4}{2} \sin^{- 1} (\frac{x}{2})]}_{\sqrt{3}}^{2} = \frac{\sqrt{3}}{2} + 0 - \frac{\sqrt{3}}{2} + 2 (\sin^{- 1} 1 - \sin^{- 1} \frac{\sqrt{3}}{2}) = \frac{\sqrt{3}}{2} - \frac{\sqrt{3}}{2} + 2 (\frac{π}{2} - \frac{π}{3}) = \frac{π}{3} sq units Area bound by the circle and straight line above x axis = \frac{π}{3} sq units$

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Similar questions

x = \sqrt{3} y

x^{2} + y^{2} = 4

The area of the region in the first quadrant enclosed by the x-axis and the line $x = \sqrt{3} y$ , the circle $x^{2} + y^{2} = 4$ is

x = \sqrt{3} y

x^{2} + y^{2} = 4

x = \sqrt{3} y

y = x

x^{2} + y^{2} = 32.

Find the area of the region in the first quadrant enclosed by X-axis and $x = \sqrt{3}$ y and the circle $x^{2} + y^{2} = 4$

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