Find the area of the region in the first quad | vedclass

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

Vedclass · Accepted Answer

$\frac{\pi}{3}$

Vedclass · Answer

(A) The area of the region bounded by the circle $x^{2}+y^{2}=4$,the line $x=\sqrt{3} y$,and the $x-$ axis in the first quadrant is the area of the region $OAB$ as shown in the figure.The point of intersection of the line $x=\sqrt{3} y$ and the circle $x^{2}+y^{2}=4$ in the first quadrant is found by substituting $x=\sqrt{3} y$ into the circle equation:$(\sqrt{3} y)^{2} + y^{2} = 4 \implies 3y^{2} + y^{2} = 4 \implies 4y^{2} = 4 \implies y = 1$.Thus,$x = \sqrt{3}(1) = \sqrt{3}$. The intersection point is $A(\sqrt{3}, 1)$.Area $OAB = 	ext{Area}(	riangle OCA) + 	ext{Area}(	ext{region } ACB)$.Area of $	riangle OCA = \frac{1}{2} 	imes OC 	imes AC = \frac{1}{2} 	imes \sqrt{3} 	imes 1 = \frac{\sqrt{3}}{2}$.Area of region $ACB = \int_{\sqrt{3}}^{2} y \, dx = \int_{\sqrt{3}}^{2} \sqrt{4-x^{2}} \, dx$.Using the formula $\int \sqrt{a^{2}-x^{2}} \, dx = \frac{x}{2} \sqrt{a^{2}-x^{2}} + \frac{a^{2}}{2} \sin^{-1}(\frac{x}{a})$:$= \left[ \frac{x}{2} \sqrt{4-x^{2}} + \frac{4}{2} \sin^{-1}(\frac{x}{2}) ight]_{\sqrt{3}}^{2}$$= \left( \frac{2}{2} \sqrt{4-4} + 2 \sin^{-1}(1) ight) - \left( \frac{\sqrt{3}}{2} \sqrt{4-3} + 2 \sin^{-1}(\frac{\sqrt{3}}{2}) ight)$$= (0 + 2 	imes \frac{\pi}{2}) - (\frac{\sqrt{3}}{2} 	imes 1 + 2 	imes \frac{\pi}{3})$$= \pi - \frac{\sqrt{3}}{2} - \frac{2\pi}{3} = \frac{\pi}{3} - \frac{\sqrt{3}}{2}$.Total Area $OAB = \frac{\sqrt{3}}{2} + (\frac{\pi}{3} - \frac{\sqrt{3}}{2}) = \frac{\pi}{3}$ square units.

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

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