1. **Problem Statement:** Find the total surface area formed by revolving a curve $f(x)$ around the x-axis from point $a$ to $b$. 2. **Formula and Explanation:** The surface area of a frustum formed by revolving a small segment of the curve is given by: $$S = 2\pi R_{av} L$$ where $R_{av} = \frac{r_1 + r_2}{2}$ is the average radius of the frustum, and $L$ is the slant height. 3. **Slant Height Calculation:** Using the Pythagorean theorem for a small change: $$\Delta L = \sqrt{\Delta x^2 + \Delta y^2} = \sqrt{1 + \left(\frac{\Delta y}{\Delta x}\right)^2} \Delta x$$ In the limit as $\Delta x \to dx$ and $\Delta y \to dy$: $$L = \sqrt{1 + \left(\frac{dy}{dx}\right)^2} dx$$ 4. **Surface Area of Small Segment:** For a small segment, the surface area is: $$dS = 2\pi \left(\frac{r_1 + r_2}{2}\right) L$$ Approximating $\frac{r_1 - r_2}{2} \approx f(x)$, and substituting $L$: $$dS = 2\pi f(x) \sqrt{1 + \left(\frac{dy}{dx}\right)^2} dx$$ 5. **Total Surface Area:** Summing all small frustums from $a$ to $b$ gives the integral: $$S = 2\pi \int_a^b f(x) \sqrt{1 + \left(\frac{dy}{dx}\right)^2} dx$$ This completes the proof of the formula for the surface area of revolution around the x-axis. **Final formula:** $$S = 2\pi \int_a^b f(x) \sqrt{1 + \left(\frac{dy}{dx}\right)^2} dx$$