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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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

Similar Questions

The area of the region $R = \{(x, y) : 5x^2 \leq y \leq 2x^2 + 9\}$ is ........ $\text{square units}$. (in $\sqrt{3}$)

The area of the region bounded by the curves $y=x^3$, $y=x^2$ and the lines $x=0$ and $x=2$ is

The area of the region given by $\{(x, y): xy \leq 8, 1 \leq y \leq x^2\}$ is :

The area of the region bounded by the curves $y=9x^2$ and $y=5x^2+4$ (in sq units) is

If the area bounded by the curves $y=ax^2$ and $x=ay^2$ $(a>0)$ is $3$ sq units, then the value of $a$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module