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

### Related Formula Area enclosed between two symmetric functions across four quadrants: alpha = 4 times left( textArea of circle in 1^textst text quadrant - int_0^1 y_textparabola \, dx right)$$\alpha = 4 \times \left( \text{Area of circle in } 1^{\text{st}} \text{ quadrant} - \int_{0}^{1} y_{\text{parabola}} \, dx \right)$$ ### Core Logic The curves are symmetric across the axes. The region is enclosed between the unit circle x^2 + y^2 = 1$x^2 + y^2 = 1$ and the parabolas y = pm(1 - x^2)$y = \pm(1 - x^2)$.

By symmetry, the area alpha$\alpha$ in the first quadrant is the area under the circle minus the area under the parabola from x=0$x=0$ to x=1$x=1$. ### Step 1: Calculate the Area Components Area of a circle quadrant with r=1$r=1$ is fracpi4$\frac{\pi}{4}$. Area under the parabola: int_0^1 (1 - x^2) \, dx = left[ x - fracx^33 right]_0^1 = 1 - frac13 = frac23$$\int_{0}^{1} (1 - x^2) \, dx = \left[ x - \frac{x^3}{3} \right]_{0}^{1} = 1 - \frac{1}{3} = \frac{2}{3}$$ ### Step 2: Evaluate the Total Area alpha = 4 left[ fracpi4 - frac23 right] = pi - frac83$$\alpha = 4 \left[ \frac{\pi}{4} - \frac{2}{3} \right] = \pi - \frac{8}{3}$$ Multiplying both sides by 9: 9alpha = 9pi - 24$$9\alpha = 9\pi - 24$$ Comparing with 9alpha = betapi + gamma$9\alpha = \beta\pi + \gamma$, we get: beta = 9, quad gamma = -24$$\beta = 9, \quad \gamma = -24$$ Hence, |beta - gamma| = |9 - (-24)| = 33$$|\beta - \gamma| = |9 - (-24)| = 33$$ ### Pattern Recognition Symmetry helps divide the integration work by 4. Recognizing standard shapes (like a circle sector) allows you to bypass complex integration entirely for half of the problem. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Integral Calculus