The area (in square units) bounded by the cur | vedclass

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

Question

(C) Given curves are $x = -2y^2$ and $x = 1 - 3y^2$.To find the intersection points,set the equations equal to each other:$-2y^2 = 1 - 3y^2$$y^2 = 1$$

Vedclass · Accepted Answer

$4/3$

Vedclass · Answer

(C) Given curves are $x = -2y^2$ and $x = 1 - 3y^2$.To find the intersection points,set the equations equal to each other:$-2y^2 = 1 - 3y^2$$y^2 = 1$$y = \pm 1$When $y = 1$,$x = -2(1)^2 = -2$. When $y = -1$,$x = -2(-1)^2 = -2$.The intersection points are $(-2, 1)$ and $(-2, -1)$.The area $A$ is given by the integral with respect to $y$ from $-1$ to $1$ of the right curve minus the left curve:$A = \int_{-1}^{1} ((1 - 3y^2) - (-2y^2)) \, dy$$A = \int_{-1}^{1} (1 - y^2) \, dy$Since the function is even,$A = 2 \int_{0}^{1} (1 - y^2) \, dy$$A = 2 [y - \frac{y^3}{3}]_{0}^{1}$$A = 2 (1 - \frac{1}{3}) = 2 (\frac{2}{3}) = \frac{4}{3}$ square units.

The area (in square units) bounded by the curves $x = -2y^2$ and $x = 1 - 3y^2$ is

Similar Questions

The area (in square units) of the region enclosed by the two circles $x^2+y^2=1$ and $(x-1)^2+y^2=1$ is

If the area of the region bounded by the curves $y=4-\frac{x^2}{4}$ and $y=\frac{x-4}{2}$ is equal to $\alpha$,then $6 \alpha$ equals

The area (in square units) of the region bounded by the curves $x=y^2$ and $x=3-2y^2$ is

The area (in sq. units) of the region $A = \{(x,y) \in R \times R | 0 \le x \le 3, 0 \le y \le 4, y \le x^2 + 3x\}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module