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.

• A

  27
• Correct answer
  B

  33
• C

  15
• D

  18

Solution

$\begin{aligned} & \text { Required area }=\pi-4 \int_0^1\left(1-x^2\right) d x \\ & =\pi-4\left[x-\frac{x^3}{3}\right]_0^1 \\ & =\pi-4 \times \frac{2}{3}=\pi-\frac{8}{3} \\ & \therefore \alpha=\pi-\frac{8}{3} \\ & 9 \alpha=9 \pi-24 \rightarrow \beta=9, \gamma=-24 \\ & |\beta-\gamma|=|9+24|=33\end{aligned}$

Hence, the correct answer is option: 'B'.