Let the area enclosed between the curves |y|=1-x^2 and x^2+y^2=1 be .

Question

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:

Accepted Answer

$$\begin{aligned}
& C_1:|y|=1-x^2 \
& C_2: x^2+y^2=1
\end{aligned}$$

$\therefore$ Required Area

$=\alpha=4\left[\right.$ Area of circle in $1^{\text {st }}$ quad. $\left.-\int_0^1\left(1-\mathrm{x}^2\right) \mathrm{dx}\right]$

$$\begin{aligned}
& =4\left[\frac{\pi}{4}-\left[\mathrm{x}-\frac{\mathrm{x}^3}{3}\right]_0^1\right] \
& \alpha=\pi-\frac{8}{3} \
& \therefore 3 \alpha=3 \pi-8 \
& \therefore 9 \alpha=9 \pi-24 \
& \therefore \beta=9, \gamma=-24 \
& \therefore|\beta-\gamma|=33
\end{aligned}$$

Solutions · 1