Calculate the area enclosed by the curves x^2 | vedclass

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

Question

(C) The given curves are $x^2 = 2 - y$ (which is $y = 2 - x^2$) and $x^2 = y$.To find the intersection points,set $2 - x^2 = x^2$.$2x^2 = 2 \Rightarro

Vedclass · Accepted Answer

$\frac{8}{3}$ square units

Vedclass · Answer

(C) The given curves are $x^2 = 2 - y$ (which is $y = 2 - x^2$) and $x^2 = y$.To find the intersection points,set $2 - x^2 = x^2$.$2x^2 = 2 \Rightarrow x^2 = 1 \Rightarrow x = \pm 1$.The area $A$ is given by the integral of the upper curve minus the lower curve from $x = -1$ to $x = 1$.$A = \int_{-1}^{1} ((2 - x^2) - x^2) dx = \int_{-1}^{1} (2 - 2x^2) dx$.Since the function is even,$A = 2 \int_{0}^{1} (2 - 2x^2) dx$.$A = 4 \int_{0}^{1} (1 - x^2) dx = 4 [x - \frac{x^3}{3}]_{0}^{1}$.$A = 4 (1 - \frac{1}{3}) = 4 (\frac{2}{3}) = \frac{8}{3}$ square units.

Calculate the area enclosed by the curves $x^2 = 2 - y$ and $x^2 = y$.

Similar Questions

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 enclosed by the curves $y=\sin x+\cos x$ and $y=|\cos x-\sin x|$ and the lines $x=0, x=\frac{\pi}{2}$ is:

The area between the curves $y = x^3$ and $y = \sqrt{x}$ is

The parabola $y^2=4x$ divides the area of the circle $x^2+y^2=5$ into two parts. The area of the smaller part is equal to:

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

Vedclass Test Series

Exam Paper Generator

Online Exam Module