1. **Problem Statement:** Calculate the surface area generated by revolving the curve $f(x) = \sqrt{1-x}$ from $x=0$ to $x=\frac{1}{2}$ around the x-axis. 2. **Formula Used:** The surface area $S$ of a solid of revolution about the x-axis is given by: $$S = 2\pi \int_a^b f(x) \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \, dx$$ 3. **Given Function and Derivative:** $$f(x) = \sqrt{1-x} = (1-x)^{1/2}$$ $$f'(x) = \frac{1}{2}(1-x)^{-1/2}(-1) = -\frac{1}{2\sqrt{1-x}}$$ 4. **Square of the Derivative:** $$(f'(x))^2 = \frac{1}{4(1-x)}$$ 5. **Substitute into the Surface Area Integral:** $$S = 2\pi \int_0^{1/2} \sqrt{1-x} \sqrt{1 + \frac{1}{4(1-x)}} \, dx$$ 6. **Simplify the expression inside the square root:** $$\sqrt{1 + \frac{1}{4(1-x)}} = \sqrt{\frac{4(1-x) + 1}{4(1-x)}} = \frac{\sqrt{5 - 4x}}{2\sqrt{1-x}}$$ 7. **Simplify the integral:** $$S = 2\pi \int_0^{1/2} \sqrt{1-x} \cdot \frac{\sqrt{5 - 4x}}{2\sqrt{1-x}} \, dx = \pi \int_0^{1/2} \sqrt{5 - 4x} \, dx$$ 8. **Substitution:** Let $u = 5 - 4x$, then $$du = -4 dx \implies dx = -\frac{1}{4} du$$ Change limits: When $x=0$, $u=5$; When $x=\frac{1}{2}$, $u=5 - 4 \cdot \frac{1}{2} = 3$ 9. **Rewrite the integral:** $$S = \pi \int_5^3 \sqrt{u} \left(-\frac{1}{4}\right) du = -\frac{\pi}{4} \int_5^3 u^{1/2} du = \frac{\pi}{4} \int_3^5 u^{1/2} du$$ 10. **Integrate:** $$\int u^{1/2} du = \frac{2}{3} u^{3/2}$$ 11. **Evaluate definite integral:** $$S = \frac{\pi}{4} \cdot \frac{2}{3} \left[u^{3/2}\right]_3^5 = \frac{\pi}{6} \left(5^{3/2} - 3^{3/2}\right)$$ 12. **Calculate values:** $$5^{3/2} = 5 \cdot \sqrt{5} \approx 11.1803$$ $$3^{3/2} = 3 \cdot \sqrt{3} \approx 5.1962$$ 13. **Final surface area:** $$S = \frac{\pi}{6} (11.1803 - 5.1962) = \frac{\pi}{6} \times 5.9841 \approx 3.1323$$ **Answer:** The surface area is approximately $3.1323$.