The area of the region A = {(x, y): 4x^2 + y^ | vedclass

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(B) The given region is bounded by the ellipse $4x^2 + y^2 = 8$ and the parabola $y^2 = 4x$.First, find the intersection points by substituting $y^2 =

Vedclass · Accepted Answer

$\pi+\frac{2}{3}$

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(B) The given region is bounded by the ellipse $4x^2 + y^2 = 8$ and the parabola $y^2 = 4x$.First, find the intersection points by substituting $y^2 = 4x$ into $4x^2 + y^2 = 8$:$4x^2 + 4x - 8 = 0 \implies x^2 + x - 2 = 0 \implies (x+2)(x-1) = 0$.Since $x \ge 0$ for the parabola, we have $x = 1$. At $x=1$, $y^2 = 4$, so $y = \pm 2$.The area $A$ is given by $2 \int_0^1 \sqrt{4x} dx + 2 \int_1^{\sqrt{2}} \sqrt{8-4x^2} dx$.$= 4 \int_0^1 \sqrt{x} dx + 4 \int_1^{\sqrt{2}} \sqrt{2-x^2} dx$.$= 4 \left[ \frac{2}{3} x^{3/2} \right]_0^1 + 4 \left[ \frac{x}{2} \sqrt{2-x^2} + \frac{2}{2} \sin^{-1} \left( \frac{x}{\sqrt{2}} \right) \right]_1^{\sqrt{2}}$.$= \frac{8}{3} + 4 \left[ (0 + \sin^{-1}(1)) - (\frac{1}{2} \sqrt{1} + \sin^{-1}(\frac{1}{\sqrt{2}})) \right]$.$= \frac{8}{3} + 4 \left[ \frac{\pi}{2} - \frac{1}{2} - \frac{\pi}{4} \right] = \frac{8}{3} + 2\pi - 2 - \pi = \pi + \frac{2}{3}$ sq. units.

The area of the region $A = \{(x, y): 4x^2 + y^2 \le 8 \text{ and } y^2 \le 4x\}$ is:

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