The area of the smaller region enclosed by th | vedclass

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(C) The given curves are $y^{2}=8x+4$ (a parabola) and $x^{2}+y^{2}+4\sqrt{3}x-4=0$ (a circle).First,we find the points of intersection by substitutin

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$\frac{1}{3}(4-12\sqrt{3}+8\pi)$

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(C) The given curves are $y^{2}=8x+4$ (a parabola) and $x^{2}+y^{2}+4\sqrt{3}x-4=0$ (a circle).First,we find the points of intersection by substituting $y^{2}=8x+4$ into the circle equation:$x^{2}+(8x+4)+4\sqrt{3}x-4=0$$x^{2}+8x+4\sqrt{3}x=0$$x(x+8+4\sqrt{3})=0$So,$x=0$ or $x=-(8+4\sqrt{3})$.For $x=0$,$y^{2}=4 \implies y=\pm 2$. The intersection points are $(0, 2)$ and $(0, -2)$.The circle equation can be rewritten as $(x+2\sqrt{3})^{2}+y^{2}=16$,which is a circle with center $(-2\sqrt{3}, 0)$ and radius $4$.Since the region is symmetric about the $x$-axis,the area is $2 	imes \int_{0}^{2} (x_{circle} - x_{parabola}) dy$.From the circle: $x = -2\sqrt{3} + \sqrt{16-y^{2}}$.From the parabola: $x = \frac{y^{2}-4}{8}$.Area $= 2 \int_{0}^{2} ((-2\sqrt{3} + \sqrt{16-y^{2}}) - (\frac{y^{2}-4}{8})) dy$$= 2 [ -2\sqrt{3}y + \frac{y}{2}\sqrt{16-y^{2}} + \frac{16}{2}\sin^{-1}(\frac{y}{4}) - \frac{y^{3}}{24} + \frac{y}{2} ]_{0}^{2}$$= 2 [ (-4\sqrt{3} + \sqrt{12} + 8\sin^{-1}(\frac{1}{2}) - \frac{8}{24} + 1) - 0 ]$$= 2 [ -4\sqrt{3} + 2\sqrt{3} + 8(\frac{\pi}{6}) - \frac{1}{3} + 1 ]$$= 2 [ -2\sqrt{3} + \frac{4\pi}{3} + \frac{2}{3} ] = \frac{1}{3} (4 - 12\sqrt{3} + 8\pi)$.

The area of the smaller region enclosed by the curves $y^{2}=8x+4$ and $x^{2}+y^{2}+4\sqrt{3}x-4=0$ is equal to.

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