The area bounded by the curves x = sqrt{2 - y | vedclass

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

Question

(A) The curve $x = \sqrt{2 - y^2}$ represents the right half of the circle $x^2 + y^2 = 2$,which has a radius $r = \sqrt{2}$.The curves $|x| = |y|$ re

Vedclass · Accepted Answer

$\frac{\pi}{2}$

Vedclass · Answer

(A) The curve $x = \sqrt{2 - y^2}$ represents the right half of the circle $x^2 + y^2 = 2$,which has a radius $r = \sqrt{2}$.The curves $|x| = |y|$ represent the lines $y = x$ and $y = -x$.These lines intersect the circle at points where $x^2 + x^2 = 2$,which gives $2x^2 = 2$,so $x^2 = 1$,meaning $x = 1$ (since $x \ge 0$).At $x = 1$,$y = 1$ and $y = -1$. The angle subtended by the arc at the center is from $	heta = -45^\circ$ to $	heta = 45^\circ$,which is $90^\circ$ or $\frac{\pi}{2}$ radians.The area of a circular sector is given by $\frac{1}{2} r^2 	heta$.Substituting $r^2 = 2$ and $	heta = \frac{\pi}{2}$,we get Area $= \frac{1}{2} 	imes 2 	imes \frac{\pi}{2} = \frac{\pi}{2}$.

The area bounded by the curves $x = \sqrt{2 - y^2}$ and $|x| = |y|$ is -

Similar Questions

The area bounded by the curve $x=4-y^{2}$ and the $Y$-axis is

The area bounded by the parabola $y^2=4ax$ and its latus-rectum $x=a$ is

The area between the curve $y^2 (a + x) = (a - x)^3$ and its vertical asymptote is

The value of $a$ $(a > 0)$ for which the area bounded by the curves $y = \frac{x}{6} + \frac{1}{x^2}$,$y = 0$,$x = a$ and $x = 2a$ has the least value,is

The area of the region bounded by the curve $y = \tan x$,the tangent drawn to the curve at $x = \frac{\pi}{4}$,and the $x$-axis is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module