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(B) The given parabolas are $y^2=8x$ and $x^2=8y$.To find the points of intersection,substitute $y = \frac{x^2}{8}$ into $y^2=8x$:$\left(\frac{x^2}{8}

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$\frac{64}{3}$ sq. units

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(B) The given parabolas are $y^2=8x$ and $x^2=8y$.To find the points of intersection,substitute $y = \frac{x^2}{8}$ into $y^2=8x$:$\left(\frac{x^2}{8}ight)^2 = 8x \Rightarrow \frac{x^4}{64} = 8x \Rightarrow x^4 = 512x \Rightarrow x(x^3 - 512) = 0$.Thus,$x=0$ or $x=8$. The intersection points are $O(0,0)$ and $P(8,8)$.The required area $A$ is given by the integral of the upper curve minus the lower curve from $x=0$ to $x=8$:$A = \int_0^8 (\sqrt{8x} - \frac{x^2}{8}) dx = \int_0^8 (2\sqrt{2}\sqrt{x} - \frac{x^2}{8}) dx$.$A = [2\sqrt{2} \cdot \frac{x^{3/2}}{3/2}]_0^8 - [\frac{x^3}{24}]_0^8$.$A = [\frac{4\sqrt{2}}{3} \cdot x^{3/2}]_0^8 - \frac{512}{24}$.$A = \frac{4\sqrt{2}}{3} \cdot (8\sqrt{8}) - \frac{64}{3} = \frac{4\sqrt{2}}{3} \cdot (16\sqrt{2}) - \frac{64}{3} = \frac{128}{3} - \frac{64}{3} = \frac{64}{3} 	ext{ sq. units}$.

The area of the region included between the parabolas $y^2=8x$ and $x^2=8y$ is

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