The area (in sq. units) of the smaller portio | vedclass

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(D) Given curves are $x^2 + y^2 = 4$ (a circle with center $(0,0)$ and radius $2$) and $y^2 = 3x$ (a parabola).To find the intersection points,substit

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$\frac{1}{\sqrt{3}} + \frac{4\pi}{3}$

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(D) Given curves are $x^2 + y^2 = 4$ (a circle with center $(0,0)$ and radius $2$) and $y^2 = 3x$ (a parabola).To find the intersection points,substitute $y^2 = 3x$ into $x^2 + y^2 = 4$:$x^2 + 3x - 4 = 0$$(x+4)(x-1) = 0$Since $x \ge 0$ for the parabola,we have $x = 1$.At $x = 1$,$y^2 = 3(1) = 3$,so $y = \pm\sqrt{3}$.The area of the smaller portion is symmetric about the $x$-axis.Area $= 2 	imes \left[ \int_{0}^{1} \sqrt{3x} \, dx + \int_{1}^{2} \sqrt{4-x^2} \, dx ight]$$= 2 	imes \left[ \sqrt{3} \left( \frac{x^{3/2}}{3/2} ight)_0^1 + \left( \frac{x}{2}\sqrt{4-x^2} + \frac{4}{2}\sin^{-1}\left(\frac{x}{2}ight) ight)_1^2 ight]$$= 2 	imes \left[ \sqrt{3} \left( \frac{2}{3} ight) + \left( (0 + 2\sin^{-1}(1)) - (\frac{1}{2}\sqrt{3} + 2\sin^{-1}(1/2)) ight) ight]$$= 2 	imes \left[ \frac{2\sqrt{3}}{3} + \left( 2 \cdot \frac{\pi}{2} - \frac{\sqrt{3}}{2} - 2 \cdot \frac{\pi}{6} ight) ight]$$= 2 	imes \left[ \frac{2}{\sqrt{3}} - \frac{\sqrt{3}}{2} + \pi - \frac{\pi}{3} ight] = 2 	imes \left[ \frac{4-3}{2\sqrt{3}} + \frac{2\pi}{3} ight]$$= 2 	imes \left[ \frac{1}{2\sqrt{3}} + \frac{2\pi}{3} ight] = \frac{1}{\sqrt{3}} + \frac{4\pi}{3}$

The area (in sq. units) of the smaller portion enclosed between the curves,$x^2 + y^2 = 4$ and $y^2 = 3x$,is

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