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

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(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

$\frac{\pi}{2}$

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(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 $= \frac{1}{2} 	imes 2 	imes \frac{\pi}{2} = \frac{\pi}{2}$.

$x$	$0$	$1$	$2$	$3$	$4$	$5$
$f(x)$	$2$	$3$	$6$	$11$	$18$	$27$

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

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