The area of the region S = {(x, y) : y^{2} le | vedclass

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(B) The region $S$ is bounded by the parabola $y^{2} = 8x$ and the line $y = \sqrt{2}x$ for $x \geq 1$.First,find the intersection points of $y^{2} =

Vedclass · Accepted Answer

$\frac{11 \sqrt{2}}{6}$

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(B) The region $S$ is bounded by the parabola $y^{2} = 8x$ and the line $y = \sqrt{2}x$ for $x \geq 1$.First,find the intersection points of $y^{2} = 8x$ and $y = \sqrt{2}x$:$(\sqrt{2}x)^{2} = 8x \Rightarrow 2x^{2} = 8x \Rightarrow 2x(x - 4) = 0$.Thus,the curves intersect at $x = 0$ and $x = 4$.Since the region is defined for $x \geq 1$,the limits of integration are from $x = 1$ to $x = 4$.The area $A$ is given by:$A = \int_{1}^{4} (\sqrt{8x} - \sqrt{2}x) \, dx$$A = \int_{1}^{4} (2\sqrt{2}\sqrt{x} - \sqrt{2}x) \, dx$$A = 2\sqrt{2} \left[ \frac{x^{3/2}}{3/2} \right]_{1}^{4} - \sqrt{2} \left[ \frac{x^{2}}{2} \right]_{1}^{4}$$A = \frac{4\sqrt{2}}{3} (4^{3/2} - 1^{3/2}) - \frac{\sqrt{2}}{2} (4^{2} - 1^{2})$$A = \frac{4\sqrt{2}}{3} (8 - 1) - \frac{\sqrt{2}}{2} (16 - 1)$$A = \frac{4\sqrt{2}}{3} (7) - \frac{\sqrt{2}}{2} (15)$$A = \frac{28\sqrt{2}}{3} - \frac{15\sqrt{2}}{2} = \frac{56\sqrt{2} - 45\sqrt{2}}{6} = \frac{11\sqrt{2}}{6}$.

The area of the region $S = \{(x, y) : y^{2} \leq 8x, y \geq \sqrt{2}x, x \geq 1\}$ is

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