1. **Problem statement:** Find the volume generated by revolving the curve $f(x) = 4 - x^2$ bounded by $x_1 = -2$ and $x_2 = 2$ around the x-axis. 2. **Formula used:** The volume $V$ of a solid of revolution around the x-axis is given by the disk method: $$V = \pi \int_{a}^{b} [f(x)]^2 \, dx$$ where $a = -2$ and $b = 2$. 3. **Set up the integral:** $$V = \pi \int_{-2}^{2} (4 - x^2)^2 \, dx$$ 4. **Expand the integrand:** $$(4 - x^2)^2 = 16 - 8x^2 + x^4$$ So, $$V = \pi \int_{-2}^{2} (16 - 8x^2 + x^4) \, dx$$ 5. **Integrate term-by-term:** $$\int_{-2}^{2} 16 \, dx = 16x \Big|_{-2}^{2} = 16(2) - 16(-2) = 32 + 32 = 64$$ $$\int_{-2}^{2} -8x^2 \, dx = -8 \int_{-2}^{2} x^2 \, dx = -8 \left[ \frac{x^3}{3} \right]_{-2}^{2} = -8 \left( \frac{8}{3} - \frac{-8}{3} \right) = -8 \times \frac{16}{3} = -\frac{128}{3}$$ $$\int_{-2}^{2} x^4 \, dx = \left[ \frac{x^5}{5} \right]_{-2}^{2} = \frac{32}{5} - \frac{-32}{5} = \frac{64}{5}$$ 6. **Sum the integrals:** $$64 - \frac{128}{3} + \frac{64}{5}$$ Find common denominator 15: $$64 = \frac{960}{15}, \quad -\frac{128}{3} = -\frac{640}{15}, \quad \frac{64}{5} = \frac{192}{15}$$ Sum: $$\frac{960}{15} - \frac{640}{15} + \frac{192}{15} = \frac{512}{15}$$ 7. **Calculate volume:** $$V = \pi \times \frac{512}{15} = \frac{512\pi}{15}$$ **Final answer:** $$\boxed{\frac{512\pi}{15}}$$ This is the volume of the solid formed by revolving the curve $f(x) = 4 - x^2$ from $x = -2$ to $x = 2$ around the x-axis.