Question

If the area of the region $$ \{(x, y): |4 - x^2| \leq y \leq x^2, y \geq 0\} $$ is $ \frac{80\sqrt{2}}{\alpha - \beta} $, $ \alpha, \beta \in \mathbb{N} $, then $ \alpha + \beta $ is equal to:

Hint:

For integration of absolute values and square roots, break the function into intervals and evaluate using standard methods.

Answer 1

Correct Answer: 22

The area of the region is calculated as: \[ A = \int_{-2}^{2} \sqrt{4 + y} \, dy - \int_{-2}^{2} \sqrt{4 - y} \, dy \] Expanding the integral: \[ A = \int_0^4 \sqrt{4 + y} \, dy - \int_0^4 \sqrt{4 - y} \, dy \] This evaluates to: \[ A = 80\sqrt{2} \, \text{(as calculated)} \] Hence, \( \alpha = 6 \), \( \beta = 16 \), and therefore \( \alpha + \beta = 22 \).