1. **Problem statement:** Find the area of the region bounded by the curves $y = x^2 + 2$, $y = 6 - x^2$, and the line $y = 3$. 2. **Identify intersection points:** - Find where $y = x^2 + 2$ intersects $y = 3$: $$x^2 + 2 = 3 \implies x^2 = 1 \implies x = \pm 1$$ - Find where $y = 6 - x^2$ intersects $y = 3$: $$6 - x^2 = 3 \implies x^2 = 3 \implies x = \pm \sqrt{3}$$ 3. **Analyze the regions:** - For $x$ between $-1$ and $1$, the line $y=3$ lies above $y = x^2 + 2$ and below $y = 6 - x^2$. - For $x$ between $-\sqrt{3}$ and $-1$, and between $1$ and $\sqrt{3}$, the region is bounded between $y=3$ and $y=6 - x^2$. 4. **Set up integrals for the area:** The total area is the sum of two parts: - Area between $x=-1$ and $x=1$ bounded by $y=3$ and $y=x^2+2$: $$A_1 = \int_{-1}^1 (3 - (x^2 + 2)) \, dx = \int_{-1}^1 (1 - x^2) \, dx$$ - Area between $x=-\sqrt{3}$ and $x=-1$ and between $x=1$ and $x=\sqrt{3}$ bounded by $y=6 - x^2$ and $y=3$: $$A_2 = 2 \times \int_1^{\sqrt{3}} ((6 - x^2) - 3) \, dx = 2 \times \int_1^{\sqrt{3}} (3 - x^2) \, dx$$ 5. **Calculate $A_1$:** $$A_1 = \int_{-1}^1 (1 - x^2) \, dx = \left[ x - \frac{x^3}{3} \right]_{-1}^1 = \left(1 - \frac{1}{3}\right) - \left(-1 + \frac{1}{3}\right) = \left(\frac{2}{3}\right) - \left(-\frac{2}{3}\right) = \frac{4}{3}$$ 6. **Calculate $A_2$:** $$A_2 = 2 \times \int_1^{\sqrt{3}} (3 - x^2) \, dx = 2 \times \left[ 3x - \frac{x^3}{3} \right]_1^{\sqrt{3}}$$ Evaluate inside the bracket: $$3\sqrt{3} - \frac{(\sqrt{3})^3}{3} - \left(3(1) - \frac{1^3}{3}\right) = 3\sqrt{3} - \frac{3\sqrt{3}}{3} - (3 - \frac{1}{3}) = 3\sqrt{3} - \sqrt{3} - \frac{8}{3} = 2\sqrt{3} - \frac{8}{3}$$ Multiply by 2: $$A_2 = 2 \times \left(2\sqrt{3} - \frac{8}{3}\right) = 4\sqrt{3} - \frac{16}{3}$$ 7. **Total area:** $$A = A_1 + A_2 = \frac{4}{3} + 4\sqrt{3} - \frac{16}{3} = 4\sqrt{3} - \frac{12}{3} = 4\sqrt{3} - 4$$ **Final answer:** $$\boxed{4\sqrt{3} - 4}$$