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Q.
The area bounded by the circle $x^2+y^2=8$, the parabola $x^2=2 y$ and the line $y=x$ in $y \geq 0$ is
Application of Integrals
Solution:
$ x^2+y^2=8 $
$x ^2=2 y$
$\Rightarrow y ^2+2 y -8=0 \Rightarrow( y +4)( y -2)=0 $
$\Rightarrow y =-4 \text { (rejected), } y =2 \Rightarrow x = \pm 2 $
$A=\int\limits_{-2}^2 \sqrt{8-x^2} d x-\int\limits_{-2}^0 \frac{x^2}{2} d x-\int\limits_0^2 x d x$
$=2 \int\limits_0^2 \sqrt{8-x^2} d x-\frac{1}{6}\left(x^3\right)_{-2}^0-\left(\frac{x^2}{2}\right)_0^2=2\left[\frac{x}{2} \sqrt{8-x^2}+\frac{8}{2} \sin ^{-1}\left(\frac{x}{2 \sqrt{2}}\right)\right]_0^2-\frac{(0+8)}{6}-\frac{(4-0)}{2} $
$=2\left[2+4 \sin ^{-1}\left(\frac{1}{\sqrt{2}}\right)-0\right]-\frac{4}{3}-2=4+\frac{8 \pi}{4}-\frac{10}{3}=2 \pi+\frac{2}{3}$
