Let the area enclosed between the curves |y| = 1 – x² and x² + y² = 1 be $\mathrm{\alpha }$. If $9\mathrm{\alpha }=\mathrm{\beta \pi }+\mathrm{\gamma };\mathrm{\beta },\mathrm{\gamma }$ are integers, then the value of |$\mathrm{\beta }-\mathrm{\gamma }$| equals

C₁ : |y| = 1 – x²
C₂ : x² + y² = 1
 

$\therefore$Required Area
 $\mathrm{\alpha }$ = 4[Area of circle in 1^st quadrant– $\left.{\int }_0^1 \left(1-{\mathrm{x}}^2\right)\mathrm{dx}\right]$
$\begin{array}{l}=4\left[\frac{\mathrm{\pi }}{4}-{\left[\mathrm{x}-\frac{{\mathrm{x}}^3}{3}\right]}_0^1\right]\\ \mathrm{\alpha }=\mathrm{\pi }-\frac{8}{3}\\ \therefore 3\mathrm{\alpha }=3\mathrm{\pi }-8\\ \therefore 9\mathrm{\alpha }=9\mathrm{\pi }-24\\ \therefore \mathrm{\beta }=9,\mathrm{\gamma }=-24\\ \therefore |\mathrm{\beta }-\mathrm{\gamma }|=33\end{array}$