The area (in sq. units) of the region A = {(x | 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 intersection points, set $\frac{y^2}{2} = y + 4$.$y

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 intersection points, set $\frac{y^2}{2} = y + 4$.$y^2 = 2y + 8 \Rightarrow y^2 - 2y - 8 = 0$.$(y - 4)(y + 2) = 0$, so $y = 4$ and $y = -2$.For $y = 4$, $x = 8$. For $y = -2$, $x = 2$.The area $A$ is given by the integral $\int_{-2}^{4} (x_{	ext{right}} - x_{	ext{left}}) dy$.$A = \int_{-2}^{4} (y + 4 - \frac{y^2}{2}) dy$.$A = \left[ \frac{y^2}{2} + 4y - \frac{y^3}{6} ight]_{-2}^{4}$.$A = (\frac{16}{2} + 16 - \frac{64}{6}) - (\frac{4}{2} - 8 - \frac{-8}{6})$.$A = (8 + 16 - \frac{32}{3}) - (2 - 8 + \frac{4}{3}) = (24 - \frac{32}{3}) - (-6 + \frac{4}{3})$.$A = \frac{72 - 32}{3} - \frac{-18 + 4}{3} = \frac{40}{3} - (-\frac{14}{3}) = \frac{54}{3} = 18$ $sq. units$.

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

Similar Questions

The area of the region included between the parabola $y^{2}=x$ and the line $x+y=2$ in the first quadrant is

The area (in sq. units) of the smaller region lying above the $X$-axis and bounded between the circle $x^2+y^2=2ax$ and the parabola $y^2=ax$ is

Find the area bounded by the curve $x^{2}=4y$ and the line $x=4y-2$. (in $\pi$)

The area (in sq. units) of the region described by $A = \{ (x,y) | y \ge x^2 - 5x + 4, x + y \ge 1, y \le 0 \}$ is:

If the area enclosed between the curves $y^2 = 4kx$ and $y = kx$ for $k > 0$ is $\frac{2}{3}$ sq. units, then $k =$ ?

Vedclass Test Series

Exam Paper Generator

Online Exam Module