The area (in sq. units) of the region {(x,y): | vedclass

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(D) The region is defined by $y^2 \geq 2x$ (outside the parabola) and $x^2+y^2 \leq 4x$ (inside the circle) in the fourth quadrant $(x \geq 0, y \leq

Vedclass · Accepted Answer

$\pi - \frac{8}{3}$

Vedclass · Answer

(D) The region is defined by $y^2 \geq 2x$ (outside the parabola) and $x^2+y^2 \leq 4x$ (inside the circle) in the fourth quadrant $(x \geq 0, y \leq 0)$.The circle equation is $(x-2)^2 + y^2 = 4$,with center $(2,0)$ and radius $2$.The intersection points of $y^2 = 2x$ and $x^2+y^2 = 4x$ are found by substituting $y^2 = 2x$ into the circle equation: $x^2 + 2x = 4x \implies x^2 - 2x = 0 \implies x(x-2) = 0$.So,$x=0$ and $x=2$. For $x=2$,$y^2 = 4 \implies y = \pm 2$. Since $y \leq 0$,the intersection point is $(2, -2)$.The area is the integral of the difference between the circle and the parabola in the fourth quadrant from $x=0$ to $x=2$.Area $= \int_{0}^{2} (\sqrt{4x-x^2} - (-\sqrt{2x})) dx = \int_{0}^{2} \sqrt{4-(x-2)^2} dx + \sqrt{2} \int_{0}^{2} x^{1/2} dx$.Using the formula $\int \sqrt{a^2-u^2} du = \frac{u}{2}\sqrt{a^2-u^2} + \frac{a^2}{2}\sin^{-1}(\frac{u}{a})$,the first part is $\pi - 2$.The second part is $\sqrt{2} [\frac{2}{3} x^{3/2}]_{0}^{2} = \sqrt{2} \cdot \frac{2}{3} \cdot 2\sqrt{2} = \frac{8}{3}$.Total Area $= (\pi - 2) - (	ext{area under parabola}) = \pi - 2 - (\pi - 2 - 8/3)$ is not correct; the region is the area of the circular sector minus the area under the parabola.The correct area is $\int_{0}^{2} (\sqrt{4x-x^2} - \sqrt{2x}) dx = (\pi - 2) - \frac{8}{3} = \pi - \frac{14}{3}$? No,let's re-evaluate: The area is $\int_{0}^{2} \sqrt{4x-x^2} dx - \int_{0}^{2} \sqrt{2x} dx = \pi - \frac{8}{3}$.

The area (in sq. units) of the region $\{(x,y):y^2 \geq 2x, x^2+y^2 \leq 4x, x \geq 0, y \leq 0 \}$ is

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