The area (in sq. units) of the region given b | vedclass

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(B) The region is bounded by the parabola $x = \frac{y^2}{2}$ and the line $x = y + 4$. To find the points of intersection, set $\frac{y^2}{2} = y + 4

Vedclass · Accepted Answer

$18$

Vedclass · Answer

(B) The region is bounded by the parabola $x = \frac{y^2}{2}$ and the line $x = y + 4$. To find the points of intersection, set $\frac{y^2}{2} = y + 4$. $y^2 = 2y + 8 \implies y^2 - 2y - 8 = 0$. $(y - 4)(y + 2) = 0$, so $y = 4$ and $y = -2$. The area $A$ is given by the integral $\int_{-2}^{4} (x_{	ext{right}} - x_{	ext{left}}) \, dy$. $A = \int_{-2}^{4} (y + 4 - \frac{y^2}{2}) \, dy$. $A = [\frac{y^2}{2} + 4y - \frac{y^3}{6}]_{-2}^{4}$. $A = (\frac{16}{2} + 16 - \frac{64}{6}) - (\frac{4}{2} - 8 - \frac{-8}{6})$. $A = (8 + 16 - \frac{32}{3}) - (2 - 8 + \frac{4}{3})$. $A = (24 - \frac{32}{3}) - (-6 + \frac{4}{3}) = \frac{40}{3} - (-\frac{14}{3}) = \frac{54}{3} = 18$ sq. units.

The area (in sq. units) of the region given by $R = \{(x, y) : \frac{y^2}{2} \leq x \leq y + 4\}$ is

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