Determine the area between the curves \( y = x^2 \) and \( y = 4 - x^2 \) for the interval where they intersect.

To find the area between the curves \( y = x^2 \) and \( y = 4 - x^2 \), we first need to determine their points of intersection. We set the functions equal to each other: \[ x^2 = 4 - x^2 \] Adding \( x^2 \) to both sides gives: \[ 2x^2 = 4 \] Dividing both sides by 2, we get: \[ x^2 = 2 \] Taking the square root of both sides, we find the intersection points: \[ x = \sqrt{2} \quad \text{and} \quad x = -\sqrt{2} \] Next, we'll calculate the area between the curves from \( x = -\sqrt{2} \) to \( x = \sqrt{2} \). The area \( A \) can be found using the integral: \[ A = \int_{-\sqrt{2}}^{\sqrt{2}} ((4 - x^2) - x^2) \, dx \] Simplifying the integrand: \[ A = \int_{-\sqrt{2}}^{\sqrt{2}} (4 - 2x^2) \, dx \] This integral can be separated into two parts: \[ A = \int_{-\sqrt{2}}^{\sqrt{2}} 4 \, dx - \int_{-\sqrt{2}}^{\sqrt{2}} 2x^2 \, dx \] Calculating the first integral: \[ \int_{-\sqrt{2}}^{\sqrt{2}} 4 \, dx = 4 \left[ x \right]_{-\sqrt{2}}^{\sqrt{2}} = 4 \left( \sqrt{2} - (-\sqrt{2}) \right) = 4 \cdot 2\sqrt{2} = 8\sqrt{2} \] Calculating the second integral: \[ \int_{-\sqrt{2}}^{\sqrt{2}} 2x^2 \, dx = 2 \left[ \frac{x^3}{3} \right]_{-\sqrt{2}}^{\sqrt{2}} = 2 \left( \frac{(\sqrt{2})^3}{3} - \frac{(-\sqrt{2})^3}{3} \right) = 2 \left( \frac{2\sqrt{2}}{3} - \frac{-2\sqrt{2}}{3} \right) = 2 \cdot \frac{4\sqrt{2}}{3} = \frac{8\sqrt{2}}{3} \] Now we combine the results to find the total area: \[ A = 8\sqrt{2} - \frac{8\sqrt{2}}{3} \] We can combine these by finding a common denominator: \[ A = \frac{24\sqrt{2}}{3} - \frac{8\sqrt{2}}{3} = \frac{16\sqrt{2}}{3} \] Thus, the area between the curves \( y = x^2 \) and \( y = 4 - x^2 \) is \[ \boxed{\frac{16\sqrt{2}}{3}} \]