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 the points where these two curves intersect. This occurs when: \[ x^2 = 4 - x^2 \] Adding \( x^2 \) to both sides gives us: \[ 2x^2 = 4 \] Dividing both sides by 2 results in: \[ x^2 = 2 \] Taking the square root of both sides, we find: \[ x = \sqrt{2} \quad \text{and} \quad x = -\sqrt{2} \] Next, we need to set up the integral to find the area between the curves. The area \( A \) between the curves from \( x = -\sqrt{2} \) to \( x = \sqrt{2} \) is given by: \[ A = \int_{-\sqrt{2}}^{\sqrt{2}} ((4 - x^2) - x^2) \, dx \] This simplifies to: \[ A = \int_{-\sqrt{2}}^{\sqrt{2}} (4 - 2x^2) \, dx \] We can now compute this integral. First, let's find the antiderivative: \[ A = \int (4 - 2x^2) \, dx = 4x - \frac{2}{3}x^3 \] Now, evaluating from \( -\sqrt{2} \) to \( \sqrt{2} \): \[ A = \left[ 4x - \frac{2}{3}x^3 \right]_{-\sqrt{2}}^{\sqrt{2}} \] Calculating the upper limit: \[ A(\sqrt{2}) = 4(\sqrt{2}) - \frac{2}{3}(\sqrt{2})^3 = 4\sqrt{2} - \frac{2}{3}(2\sqrt{2}) = 4\sqrt{2} - \frac{4\sqrt{2}}{3} \] Combining these: \[ A(\sqrt{2}) = \frac{12\sqrt{2}}{3} - \frac{4\sqrt{2}}{3} = \frac{8\sqrt{2}}{3} \] Now for the lower limit: \[ A(-\sqrt{2}) = 4(-\sqrt{2}) - \frac{2}{3}(-\sqrt{2})^3 = -4\sqrt{2} + \frac{4\sqrt{2}}{3} \] Combining these: \[ A(-\sqrt{2}) = -\frac{12\sqrt{2}}{3} + \frac{4\sqrt{2}}{3} = -\frac{8\sqrt{2}}{3} \] Finally, we find the area: \[ A = A(\sqrt{2}) - A(-\sqrt{2}) = \frac{8\sqrt{2}}{3} - \left(-\frac{8\sqrt{2}}{3}\right) = \frac{8\sqrt{2}}{3} + \frac{8\sqrt{2}}{3} = \frac{16\sqrt{2}}{3} \] Thus, the area between the curves is \[ \boxed{\frac{16\sqrt{2}}{3}}. \]