Let the area enclosed between the curves $|y|=1-x^2$ and $x^2+y^2=1$ be $\alpha$. If $9 \alpha=\beta \pi+\gamma ; \beta, \gamma$ are integers, then the value of $|\beta-\gamma|$ equals.

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Aicharya · Accepted Answer

$\begin{aligned} & 	ext { Required area }=\pi-4 \int_0^1\left(1-x^2ight) d x \ & =\pi-4\left[x-\frac{x^3}{3}ight]_0^1 \ & =\pi-4 	imes \frac{2}{3}=\pi-\frac{8}{3} \ & 	herefore \alpha=\pi-\frac{8}{3} \ & 9 \alpha=9 \pi-24 ightarrow \beta=9, \gamma=-24 \ & |\beta-\gamma|=|9+24|=33\end{aligned}$

Let the area enclosed between the curves $|y|=1-x^2$ and $x^2+y^2=1$ be $\alpha$. If $9 \alpha=\beta…

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