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

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(D) The region is bounded by $y = x^2$,$y = 8 - x^2$,and $y = 7$. First,find the intersection points: $x^2 = 8 - x^2 \implies 2x^2 = 8 \implies x^2 =

Vedclass · Accepted Answer

$20$

Vedclass · Answer

(D) The region is bounded by $y = x^2$,$y = 8 - x^2$,and $y = 7$. First,find the intersection points: $x^2 = 8 - x^2 \implies 2x^2 = 8 \implies x^2 = 4 \implies x = \pm 2$. At $x = \pm 2$,$y = 4$. Also,$x^2 = 7 \implies x = \pm \sqrt{7}$ and $8 - x^2 = 7 \implies x^2 = 1 \implies x = \pm 1$. The region is symmetric about the $y$-axis. The area is $2 	imes \int_{0}^{\sqrt{7}} (	ext{upper curve} - 	ext{lower curve}) dx$. Specifically,the upper boundary is $y = 7$ for $x \in [0, 1]$ and $y = 8 - x^2$ for $x \in [1, 2]$. The lower boundary is $y = x^2$ for $x \in [0, 2]$. Area $= 2 \left[ \int_{0}^{1} (7 - x^2) dx + \int_{1}^{2} (8 - x^2 - x^2) dx ight]$ $= 2 \left[ \int_{0}^{1} (7 - x^2) dx + \int_{1}^{2} (8 - 2x^2) dx ight]$ $= 2 \left[ (7x - \frac{x^3}{3})_{0}^{1} + (8x - \frac{2x^3}{3})_{1}^{2} ight]$ $= 2 \left[ (7 - \frac{1}{3}) + ((16 - \frac{16}{3}) - (8 - \frac{2}{3})) ight]$ $= 2 \left[ \frac{20}{3} + (\frac{32}{3} - \frac{22}{3}) ight] = 2 \left[ \frac{20}{3} + \frac{10}{3} ight] = 2 \left[ \frac{30}{3} ight] = 20$.

The area of the region $\{(x, y): x^2 \leq y \leq 8-x^2, y \leq 7\}$ is

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