1. **State the problem:** Calculate the surface area formed 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. **Find the derivative $f'(x)$:** Given $f(x) = (1-x)^{1/2}$, $$f'(x) = \frac{1}{2}(1-x)^{-1/2} \times (-1) = -\frac{1}{2\sqrt{1-x}}$$ 4. **Square the derivative:** $$\left(f'(x)\right)^2 = \left(-\frac{1}{2\sqrt{1-x}}\right)^2 = \frac{1}{4(1-x)}$$ 5. **Set up the integral:** $$S = 2\pi \int_0^{1/2} \sqrt{1-x} \sqrt{1 + \frac{1}{4(1-x)}} \, dx$$ 6. **Simplify inside the square root:** $$1 + \frac{1}{4(1-x)} = \frac{4(1-x) + 1}{4(1-x)} = \frac{4 - 4x + 1}{4(1-x)} = \frac{5 - 4x}{4(1-x)}$$ 7. **Rewrite the integral:** $$S = 2\pi \int_0^{1/2} \sqrt{1-x} \times \sqrt{\frac{5 - 4x}{4(1-x)}} \, dx = 2\pi \int_0^{1/2} \sqrt{\frac{(1-x)(5 - 4x)}{4(1-x)}} \, dx$$ 8. **Cancel $\sqrt{1-x}$ terms:** $$\sqrt{\frac{(1-x)(5 - 4x)}{4(1-x)}} = \sqrt{\frac{5 - 4x}{4}} = \frac{\sqrt{5 - 4x}}{2}$$ 9. **Integral becomes:** $$S = 2\pi \int_0^{1/2} \frac{\sqrt{5 - 4x}}{2} \, dx = \pi \int_0^{1/2} \sqrt{5 - 4x} \, dx$$ 10. **Evaluate the integral:** Let $u = 5 - 4x$, then $du = -4 dx$, so $dx = -\frac{du}{4}$. When $x=0$, $u=5$; when $x=\frac{1}{2}$, $u=5 - 4 \times \frac{1}{2} = 3$. $$S = \pi \int_{u=5}^{3} \sqrt{u} \left(-\frac{1}{4}\right) du = -\frac{\pi}{4} \int_5^3 u^{1/2} du = \frac{\pi}{4} \int_3^5 u^{1/2} du$$ 11. **Integrate:** $$\int u^{1/2} du = \frac{2}{3} u^{3/2}$$ 12. **Apply limits:** $$S = \frac{\pi}{4} \times \frac{2}{3} \left(5^{3/2} - 3^{3/2}\right) = \frac{\pi}{6} \left(5^{3/2} - 3^{3/2}\right)$$ 13. **Final answer:** $$\boxed{S = \frac{\pi}{6} \left(5^{3/2} - 3^{3/2}\right)}$$ This is the surface area of the solid formed by revolving the curve $f(x) = \sqrt{1-x}$ from $x=0$ to $x=\frac{1}{2}$ about the x-axis.