The area of the region lying between the curv | vedclass

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Question

(C) The given curves are $y = \sqrt{4-x^2}$ (which is $x^2 + y^2 = 4$ for $y \ge 0$) and $y^2 = 3x$. To find the intersection point,substitute $x = \f

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$\frac{\pi}{3}+\frac{1}{2\sqrt{3}}$

Vedclass · Answer

(C) The given curves are $y = \sqrt{4-x^2}$ (which is $x^2 + y^2 = 4$ for $y \ge 0$) and $y^2 = 3x$. To find the intersection point,substitute $x = \frac{y^2}{3}$ into $x^2 + y^2 = 4$: $(\frac{y^2}{3})^2 + y^2 = 4 \implies \frac{y^4}{9} + y^2 - 4 = 0$. Let $u = y^2$,then $u^2 + 9u - 36 = 0 \implies (u+12)(u-3) = 0$. Since $u = y^2 \ge 0$,we have $y^2 = 3$,so $y = \sqrt{3}$ (in the first quadrant). At $y = \sqrt{3}$,$x = \frac{3}{3} = 1$. The area bounded by the curves and the $Y$-axis is given by $\int_{0}^{\sqrt{3}} (x_{circle} - x_{parabola}) dy = \int_{0}^{\sqrt{3}} (\sqrt{4-y^2} - \frac{y^2}{3}) dy$. $= [\frac{y}{2}\sqrt{4-y^2} + \frac{4}{2}\sin^{-1}(\frac{y}{2}) - \frac{y^3}{9}]_{0}^{\sqrt{3}}$. $= (\frac{\sqrt{3}}{2}\sqrt{4-3} + 2\sin^{-1}(\frac{\sqrt{3}}{2}) - \frac{3\sqrt{3}}{9}) - (0)$. $= \frac{\sqrt{3}}{2} + 2(\frac{\pi}{3}) - \frac{\sqrt{3}}{3} = \frac{2\pi}{3} + \frac{3\sqrt{3}-2\sqrt{3}}{6} = \frac{2\pi}{3} + \frac{\sqrt{3}}{6} = \frac{2\pi}{3} + \frac{1}{2\sqrt{3}}$. Wait,re-evaluating the integral: $\int_{0}^{\sqrt{3}} \sqrt{4-y^2} dy = [\frac{y}{2}\sqrt{4-y^2} + 2\sin^{-1}(\frac{y}{2})]_0^{\sqrt{3}} = \frac{\sqrt{3}}{2} + 2(\frac{\pi}{3}) = \frac{\sqrt{3}}{2} + \frac{2\pi}{3}$. Subtracting $\int_0^{\sqrt{3}} \frac{y^2}{3} dy = [\frac{y^3}{9}]_0^{\sqrt{3}} = \frac{3\sqrt{3}}{9} = \frac{\sqrt{3}}{3}$. Result: $\frac{2\pi}{3} + \frac{\sqrt{3}}{2} - \frac{\sqrt{3}}{3} = \frac{2\pi}{3} + \frac{\sqrt{3}}{6} = \frac{2\pi}{3} + \frac{1}{2\sqrt{3}}$. Given the options,the correct choice is $C$.

The area of the region lying between the curves $y=\sqrt{4-x^2}$,$y^2=3x$ and the $Y$-axis is

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