The area of the region bounded by y^{2}=8x an | vedclass

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(C) To find the area of the region bounded by the parabolas $y^{2}=8x$ and $y^{2}=16(3-x)$,we first find their points of intersection.Equating the two

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$16$

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(C) To find the area of the region bounded by the parabolas $y^{2}=8x$ and $y^{2}=16(3-x)$,we first find their points of intersection.Equating the two expressions for $y^{2}$:$8x = 16(3-x)$$8x = 48 - 16x$$24x = 48$$x = 2$Substituting $x=2$ into $y^{2}=8x$,we get $y^{2}=16$,so $y = \pm 4$.The region is symmetric about the $x$-axis. The area can be calculated by integrating with respect to $y$ from $-4$ to $4$:$	ext{Area} = \int_{-4}^{4} (x_{R} - x_{L}) dy$From $y^{2}=8x$,$x_{L} = \frac{y^{2}}{8}$.From $y^{2}=16(3-x)$,$x_{R} = 3 - \frac{y^{2}}{16}$.$	ext{Area} = \int_{-4}^{4} \left(3 - \frac{y^{2}}{16} - \frac{y^{2}}{8}ight) dy = 2 \int_{0}^{4} \left(3 - \frac{3y^{2}}{16}ight) dy$$= 2 \left[ 3y - \frac{3y^{3}}{16 	imes 3} ight]_{0}^{4} = 2 \left[ 3y - \frac{y^{3}}{16} ight]_{0}^{4}$$= 2 \left( 3(4) - \frac{4^{3}}{16} ight) = 2 (12 - 4) = 2(8) = 16$.

The area of the region bounded by $y^{2}=8x$ and $y^{2}=16(3-x)$ is equal to

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