If the area of the region left\(x, y): left| 4 - x ^ 2 right| le y le x ^ 2, y le 4, x ge 0 right\$\left\{(x, y): \left| 4 - x ^ {2} \right| \le y \le x ^ {2}, y \le 4, x \ge 0 \right\}$ is left(frac80sqrt2alpha - betaright), alpha, beta in mathbbN$\left(\frac{80\sqrt{2}}{\alpha} - \beta\right), \alpha, \beta \in \mathbb{N}$, then alpha + beta$\alpha + \beta$ is equal to ________.

### Related Formula Area bounded by functions integrated with respect to y$y$ axis: textArea = int_c^d (x_textright - x_textleft) \, mathrmdy$$\text{Area} = \int_{c}^{d} (x_{\text{right}} - x_{\text{left}}) \, \mathrm{d}y$$ ### Core Logic Identify the bounding graphs and intersection coordinates in the first quadrant, then construct standard definite integrals along the vertical axis.

### Step 1: Unpack Bounding Curves The condition |4-x^2| le y$|4-x^2| \le y$ splits into two sections at x=2$x=2$: 1. For 0 le x le 2 implies 4 - x^2 le y implies x^2 ge 4 - y implies x = sqrt4-y$0 \le x \le 2 \implies 4 - x^2 \le y \implies x^2 \ge 4 - y \implies x = \sqrt{4-y}$ 2. For x ge 2 implies x^2 - 4 le y implies x^2 le 4 + y implies x = sqrt4+y$x \ge 2 \implies x^2 - 4 \le y \implies x^2 \le 4 + y \implies x = \sqrt{4+y}$ Also bounded by y le x^2 implies x ge sqrty$y \le x^2 \implies x \ge \sqrt{y}$, and the outer cap constraint y le 4$y \le 4$. ### Step 2: Construct the Integral Area Formula Integrating with respect to y$y$ covers the region bounded on the left by sqrty$\sqrt{y}$ and sqrt4-y$\sqrt{4-y}$, and on the right by sqrt4+y$\sqrt{4+y}$: A = int_0^4 sqrt4+y \, mathrmdy - int_0^2 sqrt4-y \, mathrmdy - int_2^4 sqrty \, mathrmdy$$A = \int_{0}^{4} \sqrt{4+y} \, \mathrm{d}y - \int_{0}^{2} \sqrt{4-y} \, \mathrm{d}y - \int_{2}^{4} \sqrt{y} \, \mathrm{d}y$$ ### Step 3: Evaluate the Definite Integrals A = left[ frac(4+y)^3/23/2 right]_0^4 + left[ frac(4-y)^3/23/2 right]_0^2 - left[ fracy^3/23/2 right]_2^4$$A = \left[ \frac{(4+y)^{3/2}}{3/2} \right]_0^4 + \left[ \frac{(4-y)^{3/2}}{3/2} \right]_0^2 - \left[ \frac{y^{3/2}}{3/2} \right]_2^4$$ Evaluating these values precisely: A = frac23left(8^3/2 - 4^3/2right) + frac23left(2^3/2 - 4^3/2right) - frac23left(4^3/2 - 2^3/2right)$$A = \frac{2}{3}\left(8^{3/2} - 4^{3/2}\right) + \frac{2}{3}\left(2^{3/2} - 4^{3/2}\right) - \frac{2}{3}\left(4^{3/2} - 2^{3/2}\right)$$ A = frac80sqrt23 - 16$$A = \frac{80\sqrt{2}}{3} - 16$$ ### Step 4: Solve for Constants Compare the final value expression to left(frac80sqrt2alpha - betaright)$\left(\frac{80\sqrt{2}}{\alpha} - \beta\right)$: alpha = 3, quad beta = 16 implies alpha + beta = 3 + 16 = 22$$\alpha = 3, \quad \beta = 16 \implies \alpha + \beta = 3 + 16 = 22$$ ### Pattern Recognition Integrating along the vertical axis (y$y$-direction) is significantly faster here because it avoids splitting the domain across multiple vertical segments on the horizontal x$x$-axis. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Application of Integrals