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

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(A) The equation of the circle is $x^2 + y^2 = 4$,which represents a circle with center $(0, 0)$ and radius $r = 2$. The equation of the line is $x -

Vedclass · Accepted Answer

$\frac{\pi}{3}$

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(A) The equation of the circle is $x^2 + y^2 = 4$,which represents a circle with center $(0, 0)$ and radius $r = 2$. The equation of the line is $x - \sqrt{3}y = 0$,or $x = \sqrt{3}y$. To find the intersection point in the first quadrant,substitute $x = \sqrt{3}y$ into the circle equation: $(\sqrt{3}y)^2 + y^2 = 4 \implies 3y^2 + y^2 = 4 \implies 4y^2 = 4 \implies y^2 = 1$. Since we are in the first quadrant,$y = 1$. Then $x = \sqrt{3}(1) = \sqrt{3}$. The intersection point is $(\sqrt{3}, 1)$. The area of the region in the first quadrant bounded by the $X$-axis,the line,and the circle is given by: Area $= \int_0^{\sqrt{3}} \frac{x}{\sqrt{3}} dx + \int_{\sqrt{3}}^2 \sqrt{4 - x^2} dx$. Evaluating the first integral: $\frac{1}{\sqrt{3}} \left[ \frac{x^2}{2} \right]_0^{\sqrt{3}} = \frac{1}{\sqrt{3}} \cdot \frac{3}{2} = \frac{\sqrt{3}}{2}$. Evaluating the second integral: $\int_{\sqrt{3}}^2 \sqrt{2^2 - x^2} dx = \left[ \frac{x}{2} \sqrt{4 - x^2} + \frac{4}{2} \sin^{-1} \left( \frac{x}{2} \right) \right]_{\sqrt{3}}^2$. $= \left( \frac{2}{2} \sqrt{4 - 4} + 2 \sin^{-1}(1) \right) - \left( \frac{\sqrt{3}}{2} \sqrt{4 - 3} + 2 \sin^{-1} \left( \frac{\sqrt{3}}{2} \right) \right)$. $= (0 + 2 \cdot \frac{\pi}{2}) - (\frac{\sqrt{3}}{2} + 2 \cdot \frac{\pi}{3}) = \pi - \frac{\sqrt{3}}{2} - \frac{2\pi}{3} = \frac{\pi}{3} - \frac{\sqrt{3}}{2}$. Adding the two parts: $\frac{\sqrt{3}}{2} + (\frac{\pi}{3} - \frac{\sqrt{3}}{2}) = \frac{\pi}{3}$ 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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