1. **Problem Statement:** We want to find the surface area of a solid formed by revolving a curve $y=f(x)$ around the x-axis from $x=a$ to $x=b$. 2. **Concept:** The surface area of a frustum (a truncated cone) formed by revolving a small segment of the curve can be approximated and summed to find the total surface area. 3. **Formula for Surface Area of Frustum:** $$S = 2\pi \times \text{average radius} \times \text{slant height}$$ where the average radius $R_{av} = \frac{r_1 + r_2}{2}$. 4. **Applying to the curve:** For a small segment between $x$ and $x + \Delta x$, the radii are $r_1 = f(x)$ and $r_2 = f(x + \Delta x)$. The average radius is approximately $f(x)$ for very small $\Delta x$. 5. **Slant height $L$:** Using Pythagoras theorem, $$L = \sqrt{(\Delta x)^2 + (\Delta y)^2} = \sqrt{1 + \left(\frac{\Delta y}{\Delta x}\right)^2} \Delta x$$ In the limit as $\Delta x \to 0$, this becomes $$L = \sqrt{1 + \left(\frac{dy}{dx}\right)^2} dx$$ 6. **Surface area of small frustum segment:** $$dS = 2\pi f(x) \times \sqrt{1 + \left(\frac{dy}{dx}\right)^2} dx$$ 7. **Total surface area:** Summing all small segments from $a$ to $b$ gives the integral $$S = \int_a^b 2\pi f(x) \sqrt{1 + \left(\frac{dy}{dx}\right)^2} dx$$ 8. **Conclusion:** We have proved that the surface area of the solid of revolution formed by revolving $y=f(x)$ about the x-axis from $x=a$ to $x=b$ is $$\boxed{S = 2\pi \int_a^b f(x) \sqrt{1 + \left(\frac{dy}{dx}\right)^2} dx}$$ This matches the classical formula for surface area of revolution about the x-axis.