The area of the region {(x, y): x^2 leq y leq | vedclass

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(B) The region is defined by $x^2 \leq y \leq |x^2-4|$ and $y \geq 1$. Due to symmetry about the $y$-axis,the total area is twice the area in the firs

Vedclass · Accepted Answer

$\frac{4}{3}(4 \sqrt{2}-1)$

Vedclass · Answer

(B) The region is defined by $x^2 \leq y \leq |x^2-4|$ and $y \geq 1$. Due to symmetry about the $y$-axis,the total area is twice the area in the first quadrant. For $x \geq 0$,the curves are $y = x^2$ and $y = |x^2-4|$. Intersection points: $x^2 = 4-x^2 \implies 2x^2 = 4 \implies x^2 = 2 \implies x = \sqrt{2}$. At $x = \sqrt{2}$,$y = 2$. The region is bounded by $y=1$ at the bottom. For $0 \leq x \leq 1$,$y$ ranges from $1$ to $x^2$ (not possible as $x^2 \leq 1$). Actually,the region is bounded by $x^2 \leq y \leq 4-x^2$ for $0 \leq x \leq \sqrt{2}$. With $y \geq 1$,we split the integral with respect to $y$: For $1 \leq y \leq 2$,$x^2 \leq y \implies x \leq \sqrt{y}$. For $2 \leq y \leq 4$,$y \leq 4-x^2 \implies x^2 \leq 4-y \implies x \leq \sqrt{4-y}$. Area $= 2 \left[ \int_{1}^{2} \sqrt{y} \, dy + \int_{2}^{4} \sqrt{4-y} \, dy \right]$. $= 2 \left[ \left( \frac{2}{3} y^{3/2} \right)_{1}^{2} + \left( -\frac{2}{3} (4-y)^{3/2} \right)_{2}^{4} \right]$. $= 2 \left[ \frac{2}{3} (2\sqrt{2} - 1) + \frac{2}{3} (2)^{3/2} \right] = 2 \left[ \frac{4\sqrt{2}}{3} - \frac{2}{3} + \frac{4\sqrt{2}}{3} \right] = 2 \left[ \frac{8\sqrt{2}-2}{3} \right] = \frac{4}{3}(4\sqrt{2}-1)$.

The area of the region $\{(x, y): x^2 \leq y \leq |x^2-4|, y \geq 1\}$ is

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